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Jim Randell 9:43 am on 10 February 2026 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 2333: [Continuous ledger]
From The Sunday Times, 10th June 2007 [link]

The firm I work for has kept a ledger since its foundation. One fresh page is used each day, and these pages have been numbered consecutively since the firm was founded. During a weekend last year I noticed that the page number was a perfect square. Then, during a weekend earlier this year, I noticed that the page number was a perfect cube. On another day earlier that month, I had noticed that the page number was a perfect fourth power.

When will the number next be palindromic?

This puzzle was originally published with no title.

There are now 1300 Teaser puzzles available on the site — which is 25 years worth of weekly puzzles. (And between S2T2 and Enigmatic Code there are 3866 puzzles available in total).

[teaser2333]
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- Jim Randell 9:44 am on 10 February 2026 Permalink | Reply
  You need to take into account the date the puzzle was originally published when solving this puzzle.
  
  This Python program runs in 72ms. (Internal runtime is 1.12ms).
```
from datetime import (date, timedelta)
from enigma import (powers, inf, first, le, repeat, inc, peek, irange, is_npalindrome, printf)

# date of puzzle (all dates must be earlier than this)
pub = date(2007, 6, 10)

# consider numbers up to 200_000
(sqs, cbs, p4s) = (first(powers(1, inf, k), count=le(200000)) for k in [2, 3, 4])

# is a date a weekend (= Saturday (6) or Sunday (7))
is_weekend = lambda d: d.isoweekday() in {6, 7}

# consider powers of 4
for p4 in p4s:

  # look for a higher cube, potentially within the same month
  for cb in reversed(cbs):
    if not (cb > p4): break
    if cb - p4 > 30: continue

    # consider possible dates for the cube in the current year (before publication)
    for dcb in repeat(inc(timedelta(days=-1)), pub):
      if dcb.year < pub.year: break
      # must be on a weekend
      if not is_weekend(dcb): continue
      # calculate date for the power of 4
      dp4 = dcb - timedelta(days=cb - p4)
      # must be in the same month
      if not (dp4.month == dcb.month): continue

      # look for earlier perfect squares
      for sq in sqs:
        if not (sq < p4): break
        # calculate date for the square
        dsq = dp4 - timedelta(days=p4 - sq)
        y = dp4.year - dsq.year
        if not (0 < y < 2): continue
        # must be on a weekend
        if not is_weekend(dsq): continue

        # when is the next palindromic date (after cb)
        pl = peek(n for n in irange(cb + 1, inf) if is_npalindrome(n))
        dpl = dp4 + timedelta(days=pl - p4)

        # calculate founding date
        df = dp4 - timedelta(days=p4 - 1)

        # output solution
        printf("p4 = D{p4} = {dp4}; cb = D{cb} = {dcb}; sq = D{sq} = {dsq}; pl = D{pl} = {dpl}; D1 = {df}")
```
  Solution: The next palindromic number will occur on 20th June 2007.
  
  The square number is day 50176 (= 224^2) on 2006-01-07 (Saturday).
  
  The perfect cube is day 50653 (= 37^3) on 2007-04-29 (Sunday).
  
  The power of 4 is day 50625 (= 15^4) on 2007-04-01 (Sunday).
  
  And the next palindromic number after 50653 is 50705 on 2007-06-20 (Wednesday), i.e. 10 days after the puzzle was published.
  
  If the ledger has been in continual operation from the day of founding, then day 1 was 1868-08-23 (Sunday).
  
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Jim Randell 7:12 am on 8 February 2026 Permalink | Reply
Tags: by: Stephen Hogg ( 64 )
Teaser 3307: A dish best served cold
From The Sunday Times, 8th February 2026 [link] [link]

King Otto’s birthday cake was a giant mousse. Its tiers were two concentric regular octagons. The upper tier’s span equalled the lower tier’s side-length (the diagrams show top and side views). Each tier’s height equalled the base tier octagon’s side-length (a two-figure whole number of inches under 20) divided by the square root of two (chef used a 45 degree right triangle gauge for this). When all tables at the party (arranged in three equal lines) were fully occupied (each set for the same number of couples) each person got an identical portion of eight cubic inches of cake (not necessarily in the shape of a cube). The small amount left over was less than this.

How many tables were there?

[teaser3307]
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- Jim Randell 7:57 am on 8 February 2026 Permalink | Reply
  (See also: Teaser 2976, Teaser 2921).
  
  This Python program considers possible side lengths of the octagonal lower tier, calculates the total volume of both tiers, and then divides it into 8 cu in portions.
  
  The number of portions is then allocated to three equal rows of tables, with the same number of guests at each table. (I allowed between 2 and 20 couples at each table, which gives a single answer).
  
  It runs in 70ms. (Internal runtime is 87µs).
```
from enigma import (irange, sqrt, fdiv, intr, div, divisors_pairs, printf)

r2 = sqrt(2)

# area of an octagon with side x
area = lambda x: 2 * x * x * (1 + r2)

# consider possible side lengths of the larger octagon = x (2-digit numbers less than 20)
for x in irange(10, 19):
  # calculate the height of the tiers = h
  h = fdiv(x, r2)
  # volume of the lower tier
  V1 = area(x) * h
  # volume of the upper tier
  z = fdiv(x, 1 + r2)
  V2 = area(z) * h
  # total volume of the cake (to the nearest cu in)
  V = intr(V1 + V2)

  # divide into portions = k; remainder = r
  (k, r) = divmod(V, 8)
  # there is a small amount remaining
  if r == 0: continue
  # there are 3 rows of tables, and some couples on each table
  p = div(k, 6)
  if p is None: continue
  # each seating the same number of couples = a, with tables per row = b
  for (a, b) in divisors_pairs(p, every=1):
    if not (a > 1): continue  # more than 1 couple per table
    if a > 20: break  # but no more than 20
    # output scenario
    printf("x={x} h={h:.2f} -> V={V} k={k} r={r}")
    printf("-> 3 rows of: {p} portions = {a} couples on {b} tables", p=2*p)
    printf("--> total tables = {T}", T=3*b)
    printf()
```
  Solution: There were 183 tables.
  
  The tables are arranged in 3 rows, each with 61 tables, and each table seats 3 couples (= 6 guests), for a total of 1098 guests, which requires 8784 cu in of cake.
  
  The lower tier of the cake has side length of 13 in, which makes the total volume of the cake 8788 cu in (leaving 4 cu in left over).
  
  Without restrictions on the number of guests at each table we could seat 1098 guests using:
  
  3 rows, each with 1 table, and each table seats 183 couples (366 guests)
  3 rows, each with 3 tables, and each table seats 61 couples (122 guests)
  3 rows, each with 61 tables, and each table seats 3 couples (6 guests)
  3 rows, each with 183 tables, and each table seats 1 couple (2 guests)
  
  I decided to keep the tables a manageable size with 3 couples (= 6 guests) at each table.
  
  But perhaps it would have been better if the total number of guests had been asked for (or the allowable number of guests per table specified).
  
  If you determine the total volume of cake in terms of x (or let Wolfram Alpha determine it for you [ link ]), you will find:
  
  V = 4 x³
  
  Which can be used instead of lines 10-18 to directly to calculate the volume of the cake (without using floating point approximations).
  
  And can also be used in a manual solution, as follows:
  
  We see that for even x the volume will be an exact multiple of 8, and there will be no unused portion, so we can consider odd values for x, and calculate V (which will have leave a remainder of 4 cu in). The number of portions p must then be divisible by 6 (for each couple in each of the 3 rows).
  
  x = 11: V = 5324 = 665×8 + 4; p/6 = 110.83…
  x = 13: V = 8788 = 1098×8 + 4; p/6 = 183
  x = 15: V = 13500 = 1687×8 + 4; p/6 = 281.16…
  x = 17: V = 19652 = 2456×8 + 4; p/6 = 409.33…
  x = 19: V = 27436 = 3429×8 + 4; p/6 = 571.5
  
  Only x = 13, gives a viable number of portions. And we can further split 183 to give the four candidate scenarios given above:
  
  183 = 183×1 → 3 rows, each with 183 tables, and each table seats 1 couple (2 guests)
  183 = 61×3 → 3 rows, each with 61 tables, and each table seats 3 couples (6 guests)
  183 = 3×61 → 3 rows, each with 3 tables, and each table seats 61 couples (122 guests)
  183 = 1×183 → 3 rows, each with 1 table, and each table seats 183 couples (366 guests)
  
  Of these the first seems unlikely (1 couple per table), and we can reduce the remaining three to a single candidate by placing an upper limit (of 120 or fewer) on the number of guests per table.
  
  Giving a solution of 3 rows, each with 61 tables, so 183 tables tables in total.
  
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  - Jim Randell 10:55 am on 17 February 2026 Permalink | Reply
    If you don’t want to do the analysis yourself (or let Wolfram Alpha do it), and you don’t want to wheel out the big guns (like SymPy), then the following class allows you to do simple arithmetic operations involving an irrational number.
```
from enigma import enigma

Q = enigma.Rational()

# manipulate numbers of the form: a + b*sqrt(n) where n is an integer, and a and b are rational
class Surd(object):

  def __init__(self, n, a=0, b=1):
    self.n = n
    self.a = a
    self.b = b

  def __repr__(self):
    return str.format("Surd[{a} + {b}*sqrt({n})]", a=self.a, b=self.b, n=self.n)

  # evaluate to a standard numeric type (you might get a float, or a complex, or a Q object)
  def val(self):
    (n, a, b) = (self.n, self.a, self.b)
    return (a if b == 0 else a + b * n**0.5)

  # extract (n, a, b, c, d) parameters
  def _params(self, other):
    if not isinstance(other, Surd): other = Surd(self.n, other, 0)
    if not (self.n == other.n): raise ValueError("Surd: incompatible values")
    return (self.n, self.a, self.b, other.a, other.b)

  # calculate self + other
  def __add__(self, other):
    (n, a, b, c, d) = self._params(other)
    return Surd(n, a + c, b + d)

  __radd__ = __add__

  def __neg__(self):
    return Surd(self.n, -self.a, -self.b)

  def __sub__(self, other):
    (n, a, b, c, d) = self._params(other)
    return Surd(n, a - c, b - d)

  __rsub__ = lambda self, other: -self + other

  # calculate self * other
  def __mul__(self, other):
    (n, a, b, c, d) = self._params(other)
    return Surd(n, a * c + b * d * n, a * d + b * c)

  __rmul__ = __mul__

  # calculate self / other
  def __truediv__(self, other):
    (n, a, b, c, d) = self._params(other)
    D = c * c - d * d * n
    return Surd(n, Q(a * c - b * d * n, D), Q(b * c - a * d, D))

  def __rtruediv__(self, other):
    if not isinstance(other, Surd): other = Surd(self.n, other, 0)
    return other.__truediv__(self)

  # Python 2 support
  __div__ = __truediv__
  __rdiv__ = __rtruediv__
```
    My program can then be modified to use this class, and we find that the irrational falls out of the expression and always leaves us with an exact integer volume for the cake.
```
from enigma import (irange, as_int, div, divisors_pairs, printf)

# representation of sqrt(2)
r2 = Surd(2)

# area of an octagon with side x
area = lambda x: 2 * x * x * (1 + r2)

# consider possible side lengths of the larger octagon = x (2-digit numbers less than 20)
for x in irange(10, 19):
  # calculate the height of the tiers = h
  h = x / r2
  # volume of the lower tier
  V1 = area(x) * h
  # volume of the upper tier
  z = x / (1 + r2)
  V2 = area(z) * h
  # total volume of the cake
  V = V1 + V2
  # retrieve the actual numeric value
  V = as_int(V.val())

  # divide into portions = k; remainder = r
  (k, r) = divmod(V, 8)
  # there is a small amount remaining
  if r == 0: continue
  # there are 3 rows of tables, and some couples on each table
  p = div(k, 6)
  if p is None: continue
  # each seating the same number of couples = a, with tables per row = b
  for (a, b) in divisors_pairs(p, every=1):
    if not (a > 1): continue  # more than 1 couple per table
    if a > 20: break  # but no more than 20
    # output scenario
    printf("x={x} h={h:.2f} -> V={V} k={k} r={r}", h=float(h.val()))
    printf("-> 3 rows of: {p} portions = {a} couples on {b} tables", p=2*p)
    printf("--> total tables = {T}", T=3*b)
    printf()
```
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Jim Randell 8:39 am on 6 February 2026 Permalink | Reply
Tags: by: Angela Newing ( 39 )

Teaser 2459: [Queue for tea]

From The Sunday Times, 8th November 2009 [link]

At today’s tea break only one of the six workers in the canteen queue actually had tea, the other five having different drinks. Terry was directly behind the one who had hot chocolate, while Tony (who was not at the front of the queue) was immediately in front of the sparkling-water drinker (who was not Terry). The milk drinker was three places in front of the person who had white coffee. Tilly was directly behind Tommy. Timmy was two places ahead of the hot-chocolate drinker. The person at the front of the queue had black coffee.

Who had tea? And where was Trudy in the queue?

This puzzle was originally published with no title.

[teaser2459]

Jim Randell 8:41 am on 6 February 2026 Permalink | Reply

Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

The following run file executes in 79ms. (Internal runtime of the generated code is 87µs).

#! python3 -m enigma -rr

SubstitutedExpression

# we assign queue positions 1 (= front) to 6 (= back)
--digits="1-6"
--macro="@Terry = A"
--macro="@Tony = B"
--macro="@Tilly = C"
--macro="@Tommy = D"
--macro="@Timmy = E"
--macro="@Trudy = F"
--macro="@Te = G"  # Tea
--macro="@HC = H"  # Hot Chocolate
--macro="@SW = I"  # Sparkling Water
--macro="@Mk = J"  # Milk
--macro="@WC = K"  # White Coffee
--macro="@BC = L"  # Black Coffee
--distinct="ABCDEF,GHIJKL"

# "Terry was directly behind Hot Chocolate"
"@HC + 1 = @Terry"

# "Tony (who is not at the front of the queue) ..."
"@Tony > 1"
# "... was immediately in front of Sparkling Water ..."
"@SW - 1 = @Tony"
# "... (who was not Terry)"
"@SW != @Terry"

# "Milk was 3 places in front of White Coffee"
"@WC - 3 = @Mk"

# "Tilly was directly behind Tommy"
"@Tommy + 1 = @Tilly"

# "Timmy was 2 places ahead of Hot Chocolate"
"@HC - 2 = @Timmy"

# "The person at the front of the queue had Black Coffee"
"@BC = 1"

--template="(A B C D E F) (G H I J K L)"
--solution=""
--verbose="1-W"

And here is a Python program that runs the solver, and then formats the output nicely:

from enigma import (SubstitutedExpression, irange, printf)

# load the puzzle
p = SubstitutedExpression.from_file(["{dir}/teaser2459.run"])

# link symbols to values
s2n = dict(A='Terry', B='Tony', C='Tilly', D='Tommy', E='Timmy', F='Trudy')
s2d = dict(G='tea', H='hot chocolate', I='sparkling water', J='milk', K='white coffee', L='black coffee')

# solve the puzzle and format solutions
for s in p.solve(verbose="0"):
  # for each position 1-6 fill out a corresponding name and drink
  (name, drink) = (dict((s[k], v) for (k, v) in m.items()) for m in (s2n, s2d))
  # output the queue
  for i in irange(1, 6):
    (n, d) = (m.get(i, '???') for m in (name, drink))
    printf("{i}: {n}, {d}")
  printf()

Solution: Terry had tea. Trudy was 6th in the queue.

The positions in the queue and drinks required are:

1: Timmy, black coffee
2: Tommy, milk
3: Tilly, hot chocolate
4: Terry, tea
5: Tony, white coffee
6: Trudy, sparkling water

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Jim Randell 9:34 am on 3 February 2026 Permalink | Reply
Tags: by: Nick MacKinnon ( 57 ), flawed ( 55 )
Teaser 2461: [Football league]
From The Sunday Times, 22nd November 2009 [link]

The saying “Most teams lose most of the time” was true of our league last season. Five teams, Albion, Broncos, City, Devils and Eagles, play each other once, earning three points for a win, one for a draw. Albion won and the teams ended in alphabetical order, the positions of teams tying on points determined by goal difference. Albion’s goal difference was four times the Broncos’. No match scores were the same; 32 goals were scored; City scored none against Albion; and Devils scored four different consecutive numbers of goals.

What were the scores in the four Devils’ games (DvA, DvB, DvC, DvE)?

Although this puzzle can be solved there are two possible answers (although only one of them was given when the solution was published in The Sunday Times).

This puzzle was originally published with no title.

[teaser2461]
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- Jim Randell 9:35 am on 3 February 2026 Permalink | Reply
  There are 5 teams in the league, so each team plays 4 matches. For a team to “lose most of the time”, they would have to lose at least 3 of their 4 matches. And for “most teams to lose most of the time”, there have to be at least 3 teams that lost at least 3 of their matches.
  
  So, of the 10 matches in the tournament, at least 9 of them must be lost/won, i.e. there can be at most 1 drawn match.
  
  All the scores in the matches must be different, so in increasing numbers of total goals scored in lost/won match we have the possible following scorelines:
  
  1: 1-0
  2: 2-0
  3: 3-0, 2-1
  4: 4-0, 3-1
  5: 5-0, 4-1, 3-2
  
  At this point, we have identified 9 matches, and the total number of goals scored in them is 32. So all these must take place, and the remaining match must be a 0-0 draw. And we have identified the scores in all of the 10 matches.
  
  This Python program allocates the scores to the matches. It runs in 20s (using PyPy).
```
from enigma import (Football, subsets, diff, cproduct, ordered, is_consecutive, printf)

# scoring system
football = Football(games="wdl", points=dict(w=3, d=1))

# scores in the 10 matches; all different, and 32 goals scored in total
scores = [(1, 0), (2, 0), (3, 0), (2, 1), (4, 0), (3, 1), (5, 0), (4, 1), (3, 2), (0, 0)]
assert sum(x + y for (x, y) in scores) == 32

# allocate <k> scores from <scs>
def allocate(k, scs):
  for ss in subsets(scs, size=k, select='P'):
    rs = diff(scs, ss)
    for xys in cproduct({(x, y), (y, x)} for (x, y) in ss):
      yield (xys, rs)

# calculate the table and goal difference from a sequence of scores
def table(ss, ts):
  T = football.table(football.outcomes(ss), ts)
  (f, a) = football.goals(ss, ts)
  return (T, f - a)

# X did better than Y
def better(ptsX, ptsY, gdX, gdY):
  return ptsX > ptsY or (ptsX == ptsY and gdX > gdY)

# choose the scores in D's matches
for ((AD, BD, CD, DE), scs1) in allocate(4, scores):
  # D's goals in each match must be consecutive (in some order)
  if not is_consecutive(ordered(AD[1], BD[1], CD[1], DE[0])): continue
  # calculate the table for D
  (D, gdD) = table([AD, BD, CD, DE], [1, 1, 1, 0])

  # choose the scores in C's matches
  for ((AC, BC, CE), scs2) in allocate(3, scs1):
    # C scored 0 goals in the A v C match
    if not (AC[1] == 0): continue
    # calculate the table for C
    (C, gdC) = table([AC, BC, CD, CE], [1, 1, 0, 0])
    # C did better than D
    if not better(C.points, D.points, gdC, gdD): continue

    # choose the scores in B's matches
    for ((AB, BE), scs3) in allocate(2, scs2):
      # calculate the table for B
      (B, gdB) = table([AB, BC, BD, BE], [1, 0, 0, 0])
      # B did better than C
      if not better(B.points, C.points, gdB, gdC): continue

      # allocate the remaining match
      for ([AE], _) in allocate(1, scs3):
        # calculate the table for A
        (A, gdA) = table([AB, AC, AD, AE], [0, 0, 0, 0])
        # A did better than B
        if not better(A.points, B.points, gdA, gdB): continue
        # A's goal difference is 4 times B's
        if not (gdA == 4 * gdB): continue

        # calculate the table for E
        (E, gdE) = table([AE, BE, CE, DE], [1, 1, 1, 1])
        # D is higher than E
        if not better(D.points, E.points, gdD, gdE): continue

        # at least 3 teams lost at least 3 matches
        if not sum(X.l >= 3 for X in [A, B, C, D, E]) >= 3: continue

        # output scenario
        printf("AB={AB} AC={AC} AD={AD} AE={AE} BC={BC} BD={BD} BE={BE} CD={CD} CE={CE} DE={DE}")
        printf("A={A} gd={gdA}")
        printf("B={B} gd={gdB}")
        printf("C={C} gd={gdC}")
        printf("D={D} gd={gdD}")
        printf("E={E} gd={gdE}")
        printf()
```
  Solution: There are two possible answers:
  
  DvA = 0-3; DvB = 2-3; DvC = 1-4; DvE = 3-1
  DvA = 1-4; DvB = 2-3; DvC = 0-3; DvE = 3-1
  
  Only the first of these was given in the published solution.
  
  These correspond to the following 4 scenarios:
  
  AvB = 0-0; AvC = 4-0; AvD = 3-0; AvE = 5-0; BvC = 2-1; BvD = 3-2; BvE = 1-0; CvD = 4-1; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 3-0; AvE = 5-0; BvC = 1-0; BvD = 3-2; BvE = 2-1; CvD = 4-1; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 4-1; AvE = 5-0; BvC = 2-1; BvD = 3-2; BvE = 1-0; CvD = 3-0; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 4-1; AvE = 5-0; BvC = 1-0; BvD = 3-2; BvE = 2-1; CvD = 3-0; CvE = 0-2; DvE = 3-1
  
  i.e. the order of {AvD, CvD} = {3-0, 4-1} and {BvC, BvE} = {2-1, 1-0} is not determined.
  
  But in each case the final table is:
  
  1st: A → 3w1d = 10 points; goal diff = 12
  2nd: B → 3w1d = 10 points; goal diff = 3
  3rd: C → 1w3l = 3 points; goal diff = −4
  4th: D → 1w3l = 3 points; goal diff = −5
  5th: E → 1w3l = 3 points; goal diff = −6
  
  With teams C, D, E (i.e. “most teams”) having lost 3 of their matches (i.e. “lost most of the time”).
  
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Jim Randell 6:53 am on 1 February 2026 Permalink | Reply
Tags: by: Victor Bryant ( 147 )

Teaser 3306: Pin-up

From The Sunday Times, 1st February 2026 [link] [link]

Over the years I have used seven different PINs. The first consisted of a four-digit number (the first digit not being zero), and thereafter each new PIN was obtained by swapping over two of the digits of the previous PIN; each time this increased the PIN.

The first PIN (not surprisingly!) was divisible by 1, the second was divisible by 2, the third by 3, and so on, with the seventh divisible by 7 (in fact the seventh PIN was the only one divisible by 7).

What was the third PIN?

[teaser3306]

Jim Randell 7:06 am on 1 February 2026 Permalink | Reply
It is quicker (and neater) to build the list in reverse order. We can start with a final PIN that is a multiple of 7 (and 3), and then work backwards, undoing the swaps, to get back to the initial PIN.

This Python program runs in 76ms. (Internal runtime is 6.8ms).
```
from enigma import (irange, subsets, nsplit, nconcat, swap, printf)

# generate possible swaps by index (on a 4-digit sequence)
swap_inds = list(subsets(irange(4), size=2))

# solve the puzzle for a sequence of (<number>, <digit>) pairs
# at each stage finding the _previous_ PIN
def _solve(ss):
  k = len(ss)
  # are we done?
  if k == 7:
    yield ss
  else:
    (n, ds) = ss[0]
    # swap 2 digits of the following PIN
    for (i, j) in swap_inds:
      ds1 = swap(ds, i, j)
      # no leading zeros
      if ds1[0] == 0: continue
      n1 = nconcat(ds1)
      # previous PINs must be: smaller; not be divisible by 7; divisible by their position (= 7 - k)
      if n1 < n and n1 % 7 > 0 and n1 % (7 - k) == 0:
        yield from _solve([(n1, ds1)] + ss)

# solve the puzzle for final PIN <n>
def solve(n):
  # final PIN is divisible by 7
  if n % 7 > 0: return
  # find previous PINs
  for ss in _solve([(n, nsplit(n))]):
    # return the list of numbers
    yield tuple(n for (n, ds) in ss)

# consider the final PIN (divisible by 7 (and 3))
for n in irange.round(1000, 9999, step=21, rnd='I'):
  for ns in solve(n):
    printf("{ns}")
```
Solution: The third PIN is 5286.

There are 2 possible sequences of PINs that differ at the second PIN:

2568 → {2586, 5268} → 5286 → 8256 → 8265 → 8562 → 8652

LikeLike
- Ruud 11:19 am on 1 February 2026 Permalink | Reply
  
  How do you know that the final PIN does not start with 0 ?
  
  LikeLike
  - Jim Randell 11:47 am on 1 February 2026 Permalink | Reply
    
    @Ruud: The first PIN has no leading zero, so is greater than 999. And each subsequent PIN is larger than the one preceding it, so all the PINs are greater than 999, hence none of them have a leading zero.
    
    LikeLike
- ruudvanderham 11:27 am on 1 February 2026 Permalink | Reply
  
  Ah, I see: because it has to be greater than the first PIN, which is >=1000 .
  But, how do you know that the final PIN is divisible by 3?
  
  LikeLike
  - Jim Randell 11:48 am on 1 February 2026 Permalink | Reply
    
    @Ruud: If a number is divisible by 3 than any number formed from a rearrangement of the digits is also divisible by 3. Hence all the PINs must be divisible by 3.
    
    So if the final PIN is divisible by both 7 and 3, then it must be divisible by 21 (= LCM(7, 3)).
    
    LikeLike
    - Jim Randell 12:25 pm on 2 February 2026 Permalink | Reply
      
      @Ruud: Did you try this? It produces no output for me.
      
      If you want to go in steps of 21 you need to start with the final PIN and work backwards, because the final PIN is the only one that is divisible by 7.
      
      LikeLike
      - Ruud 1:10 pm on 2 February 2026 Permalink | Reply
        
        I thought I did, but apparently I did not. Can you remove this?
        
        LikeLike

Ruud 9:34 am on 1 February 2026 Permalink | Reply

import istr


def solve(pins):
    for i0, i1 in istr.combinations(range(4), r=2):
        n = len(pins) + 1
        pin = istr.join(pins[-1][int(i0)] if i == i1 else pins[-1][int(i1)] if i == i0 else pins[-1][i] for i in range(4))
        if pin > pins[-1] and pin.is_divisible_by(n):
            if n == 7:
                yield pins + [pin]
            elif not pin.is_divisible_by(7):
                yield from solve(pins + [pin])


for pin in istr(range(1000,10000)):
    for pins in solve([pin]):
        print(pins)

LikeLike

Frits 1:32 pm on 1 February 2026 Permalink | Reply

from itertools import combinations

# generate and check the k-th PIN (decreasing k)
def check(s, k):
  if not k:
    yield int(''.join(s[4]))      # third PIN
  else:
    # swap 2 digits
    for i, j in combinations(range(4), 2):
      # swap should result in a lower number > 999
      if s[-1][i] <= s[-1][j] or (i == 0 and s[-1][j] == "0"): continue
      new = [s[-1][j] if y == i else s[-1][i] if y == j else s[-1][y]
           for y in range(4)]  
      # the seventh PIN was the only one divisible by 7
      if (n := int(''.join(new))) % k == 0 and n % 7: 
        yield from check(s + [new], k - 1)       

sols = set()
# first possible seventh PIN divisible by 21 (div 7, div 3)
f = next(i for i in range(2001, 2024, 3) if i % 7 == 0)

for p7 in range(f, 10000, 21):
  # div 5 case
  if "5" in (s := str(p7)):
    # we need an even digit for the div 4 case
    if set("02468").isdisjoint(s): continue
  elif "0" in s:
    # we need at least 2 even digits for the div 4 case 
    if sum(int(x) % 2 for x in s) > 2: continue
  else: 
    continue  
    
  # we need at least 7 different permutations  
  if len(set(s)) < 3: continue
  
  # generate all PINs and store the third PIN
  for p3 in check([s], 6):
    sols.add(p3)
    
print(f"answer: {' or '.join(str(s) for s in sols)}")

LikeLike

Ruud 5:43 am on 2 February 2026 Permalink | Reply
ChatGPT solved it in 47 seconds:
```
from itertools import permutations, combinations

def swaps(a, b):
    """Return True if b can be obtained from a by swapping exactly two digits."""
    diffs = [i for i in range(4) if a[i] != b[i]]
    return len(diffs) == 2 and a[diffs[0]] == b[diffs[1]] and a[diffs[1]] == b[diffs[0]]

def valid_chain(chain):
    """Check divisibility, increasing order, and swap condition."""
    for i in range(7):
        if chain[i] % (i+1) != 0:
            return False
        if i > 0:
            if chain[i] <= chain[i-1]:
                return False
            if not swaps(str(chain[i-1]), str(chain[i])):
                return False
    # only the 7th is divisible by 7
    return all(chain[i] % 7 != 0 for i in range(6))

solutions = []

# choose 4 distinct digits, first not zero
digits = [str(d) for d in range(10)]

for digs in combinations(digits, 4):
    for perm in permutations(digs):
        if perm[0] == "0":
            continue
        nums = sorted(
            int("".join(p)) for p in permutations(perm) if p[0] != "0"
        )
        for chain in permutations(nums, 7):
            if valid_chain(chain):
                solutions.append(chain)

# print solutions and the third PIN
for sol in solutions:
    print(sol, "→ third PIN:", sol[2])
```
Scary …

LikeLike
- Ruud 6:10 am on 2 February 2026 Permalink | Reply
  
  Maybe less scary than I thought.
  Although, ChatGPT gave the proper answer, it can’t have been done with this pure brute force program. It takes for ever, even for one PIN.
  
  LikeLike
- Jim Randell 3:48 pm on 3 February 2026 Permalink | Reply
  
  This code looks like it might solve the puzzle, although it has decided to generate all possible sequences of 7 numbers and the check to see if it can find any that work (which itself quite a slow approach). But the code it has written for doing this is terrible.
  
  First it selects 4 digits from 10 (= C(10, 4) = 210 arrangements, and also means it is assuming that all the digits are different), then for each of these it considers all possible permutations of the digits (= P(4, 4) = 24 arrangements), discarding those that begin with a zero. It then makes a sorted list of all arrangements of these digits that do not have a leading zero (up to P(4, 4) = 24 arrangements), and then considers all possible 7 digit permutations of this list (not combinations so they are already correctly sorted), which accounts for up to 1,744,364,160 checks for each of the possible 210 × 24 = 5,040 sequences of digits. So in total it is trying about 8.7 × 10^12 possible sequences, including many repeated checks.
  
  It is no wonder it does not run in a reasonable time. For quite a straightforward problem ChatGPT has made a real mess of it.
  
  As Samuel Johnson might say, “[it] is like a dog walking on his hind legs. It is not done well; but you are surprised to find it done at all”.
  
  In my experience ChatGPT is good at producing something that looks plausible. So you need to know enough about what you are asking it, so that you can determine if the answer is actually reasonable. And this case, it is a massive failure.
  
  The really scary thing is that some people treat its responses as though it knows what it is talking about.
  
  LikeLiked by 1 person

Ruud 7:06 am on 2 February 2026 Permalink | Reply

I asked ChatGPT again (mentioning that it didn’t finish in reosonable time). And now it comes with a solution. Amazing …

from collections import defaultdict

SWAP_PAIRS = [(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)]

def swap_digits(s, i, j):
    if i == j:
        return s
    lst = list(s)
    lst[i], lst[j] = lst[j], lst[i]
    return "".join(lst)

def build_neighbors():
    """
    neighbors[n] = sorted list of all valid 4-digit numbers obtainable by swapping two digits of n
                   (leading digit cannot become 0).
    """
    neighbors = [[] for _ in range(10000)]
    for n in range(1000, 10000):
        s = f"{n:04d}"
        seen = set()
        for i, j in SWAP_PAIRS:
            t = swap_digits(s, i, j)
            if t[0] == "0":
                continue
            m = int(t)
            if m != n and m not in seen:
                seen.add(m)
        neighbors[n] = sorted(seen)
    return neighbors

def find_chains():
    neighbors = build_neighbors()

    # Precompute which numbers satisfy divisibility by k (k=1..7)
    div_ok = {k: [False]*10000 for k in range(1, 8)}
    for n in range(1000, 10000):
        for k in range(1, 8):
            div_ok[k][n] = (n % k == 0)

    solutions = []

    def dfs(chain):
        step = len(chain) + 1  # next step index (1..7)
        if step == 8:
            solutions.append(chain[:])
            return

        prev = chain[-1]

        for nxt in neighbors[prev]:
            if nxt <= prev:
                continue
            if not div_ok[step][nxt]:
                continue
            # only the 7th is divisible by 7
            if step < 7 and nxt % 7 == 0:
                continue
            dfs(chain + [nxt])

    # Step 1: any 4-digit number works for divisibility by 1,
    # but it must allow an increasing chain of swaps.
    # Still, searching all 9000 starts is fine with pruning.
    for start in range(1000, 10000):
        # (Optional) tiny prune: if start has no larger neighbors, skip
        if not neighbors[start] or neighbors[start][-1] <= start:
            continue
        dfs([start])

    return solutions

if __name__ == "__main__":
    sols = find_chains()
    print(f"Found {len(sols)} solution chain(s).")
    for ch in sols:
        print(" -> ".join(map(str, ch)), " | third PIN:", ch[2])

LikeLike

Ruud 4:26 pm on 3 February 2026 Permalink | Reply

I fully agree with you. But when told ChatGPT that this code does not run in reasonable time, it immediately came up with a way better solution (see my reply). And I find that amazing.

LikeLike

GeoffR 10:09 am on 4 February 2026 Permalink | Reply

Interesting reply from ChatGPT, after I put to it that it was not very efficient at coding teasers.

It looks as though the problem is recognised and things may improve in the future. Quite a lot of detailed comments – things can only get better!

https://chatgpt.com/share/698313ad-02fc-8000-ab4c-2a3873afb266

LikeLike

Jim Randell 8:33 am on 30 January 2026 Permalink | Reply
Tags: by: Danny Roth ( 88 )

Teaser 2331: Everyone’s invited

From The Sunday Times, 27th May 2007 [link]

The letters A, B, C, D, E, F, G, H and I stand for the digits 1 to 9 in some order. Thus AB, CD and EF are two-digit numbers; [and] GHI is a three-digit number. George and Martha arranged a big wedding for their granddaughter. The top table consisted of AB members (fewer than 60) of their extended family. There were CD other tables each seating EF guests. The total attendance was GHI (fewer than 500).

What was the total attendance?

[teaser2331]

Jim Randell 8:34 am on 30 January 2026 Permalink | Reply
Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

It runs in 92ms. (Internal runtime of the generated code is 14.8ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

--digits="1-9"

"A < 6"  # or: "AB < 60"
"G < 5"  # or: "GHI < 500"
"AB + CD * EF = GHI"

--template="(AB CD EF GHI)"
--answer="GHI"
```
Solution: The total attendance was 489.

The top table seated 57.

And then there were either 12 additional tables, each seating 36; or 36 additional tables, each seating 12.

The total attendance is thus: 57 + 12 × 36 = 489.

LikeLike

ruudvanderham 9:46 am on 30 January 2026 Permalink | Reply

import peek
import istr

for A, B, C, D, E, F, G, H, I in istr.permutations(range(1, 10)):
    if A <= 5 and G <= 4 and (A | B) + (C | D) * (E | F) == (G | H | I):
        peek((A | B), (C | D), (E | F), (G | H | I))

LikeLike

GeoffR 5:52 pm on 30 January 2026 Permalink | Reply

from itertools import permutations
dgts = set('123456789')

for p1 in permutations(dgts, 2):
    A, B = p1
    AB = int(A + B)
    if AB < 60:
        q1 = dgts.difference({A, B})
        for p2 in permutations(q1):
            C, D, E, F, G, H, I = p2
            CD, EF = int(C + D), int(E + F)
            GHI = int(G + H + I)
            if GHI == AB + CD * EF:
                if GHI < 500:
                    print(f"Total attendance = {GHI}")

# Total attendance = 489

LikeLike

BRG 9:30 am on 3 February 2026 Permalink | Reply

# 10.A + B + (10.C + D).(10.E + F) == 100.G + 10.H + I < 500
# note: the order of C and E is undetermined (use C < E)

from itertools import permutations

# start with C and E: 100.C.E < 500  E < 4 // C + 1
for C in range(1, 10):
  for E in range(C + 1, 4 // C + 1):
    s8 = set(range(1, 10)) - {C, E}
    # complete the number of tables (CD) and the number on tables (EF)
    for D, F in permutations(s8, 2):
      CD, EF = 10 * C + D, 10 * E + F
      s6 = s8 - {D, F}
      # now find the number on the top table (AB < 60)
      for A, B in permutations(s6, 2):
        if (AB := 10 * A + B) < 60:
          # find the attendance (< 500) and ensure correct digit allocations
          GHI = (AB := 10 * A + B) + CD * EF
          s_ghi = {int(x) for x in str(GHI)}
          if GHI < 500 and len(s_ghi) == 3 and s_ghi.issubset(s6 - {A, B}):
            print(f"({AB = }) + ({CD = }) x ({EF = }) == ({GHI = })")

LikeLike

Jim Randell 10:05 am on 27 January 2026 Permalink | Reply
Tags: by: Derek Emley
Teaser 2415: [Double shift]
From The Sunday Times, 4th January 2009 [link]

I have written down a large number in which no digit occurs more than twice. If I were to move the last digit of the number from the end to the front of the number, then the effect would be to double the number. If I repeated the process with the new number, then the effect would be to double it again. And if I repeated the process again, the effect would be to double the number yet again.

What is the number that I have written down?

This puzzle was originally published with no title.

[teaser2415]
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Frits and Jim Randell are discussing. Toggle Comments
- Jim Randell 10:06 am on 27 January 2026 Permalink | Reply
  If we suppose the original number n is of the form:
  
  n = 10a + z
  where z is the final digit, and a is a k-digit number
  
  Then moving the final digit to the front gives us:
  
  2n = (10^k)z + a
  
  From which we derive that a is a k-digit number, such that:
  
  19a = (10^k − 2)z
  
  If the number contains no digit more than twice, then the longest the number can be is 20 digits long, and so we can consider k up to 19.
  
  This Python program runs in 70ms. (Internal runtime is 719µs).
```
from enigma import (irange, div, nsplit, rotate, zip_eq, seq2str, printf)

# check a candidate number <n> (with digits <ds>)
def check(n, ds):
  # check no digit occurs more than twice
  if any(ds.count(x) > 2 for x in set(ds)): return
  # record shifts that result in doublings
  ns = [n]
  # collect doubling sequences
  while True:
    n *= 2
    ds = rotate(ds, -1)
    if not zip_eq(ds, nsplit(n)): break
    ns.append(n)
  return ns

# look for up to 19 digits prefixes
for k in irange(1, 19):
  x = 10**k - 2
  # consider possible final digits
  for z in irange(1, 9):
    a = div(x * z, 19)
    if a is None: continue
    n = 10*a + z
    ds = nsplit(n)
    if len(ds) != k + 1: continue
    ns = check(n, ds)
    if ns and len(ns) >= 4:
      printf("{ns}", ns=seq2str(ns, sep=" -> ", enc=""))
```
  Solution: The number is: 105263157894736842.
  
  The number is 18 digits long, and uses each digit twice except 0 and 9 (which are used once each).
  
  Shifting the final digit to the front multiple times we get:
  
  105263157894736842 = n
  210526315789473684 = n × 2
  421052631578947368 = n × 4
  842105263157894736 = n × 8
  
  A run of 4 is the longest run we can get in base 10 (as n × 16 will not have the same number of digits as n), but there is another run of 3 we can make:
  
  157894736842105263 = n
  315789473684210526 = n × 2
  631578947368421052 = n × 4
  
  These numbers are the repetends of fractions of the form k / 19.
  
  LikeLike
- Frits 8:51 am on 31 January 2026 Permalink | Reply
```
'''
  n = xbcd where x is a k-digit number and b,c and d are digits 
2.n = dxbc
4.n = cdxb
8.n = bcdx with x, b and c all even

8.n = bcdx also has k+3 digits so b = 8 and first digit of x = 1
   --> c = 4 (not 9) and last digit of x must be 6
       d = 2 (as 2.n must be 2.....)
       
n = 1....6842      

  n = xbcd = 1000.x + bcd  (x has k digits)
8.n = bcdx = 10^k . bcd + x

8000.x + 8.bcd = 10^k . bcd + x
7999.x = 19.421.x = (10^k - 8).842
19.x = 2.(10^k - 8)
'''
      
for k in range(2, 18):  
  x, r = divmod(2 * (10**k - 8), 19)
  if r: continue
  n = str(x) + "842"
  # no digit occurs more than twice
  if all(n.count(str(i)) <= 2 for i in range(10)): 
    print("answer method 1:", n)
  
# in order for 2 * (10**k - 8) % 19 = 0 we must have
# 2 * 10**k % 19 = 16 
# 2 * 10**k = 19.10**(k - 1) + 10**(k - 1)
# so 10**(k - 1) % 19 = 16

# 10**(k - 1) % 19 follows a logical pattern
k = 2
a = 10**(k - 1) % 19
while a != 16: 
  a = (a + 19 * (a % 2)) // 2
  k += 1

n = str(2 * (10**k - 8) // 19) + "842"
# no digit occurs more than twice
if all(n.count(str(i)) <= 2 for i in range(10)): 
  print("answer method 2:", n)
```
  LikeLike
Jim Randell 8:11 am on 25 January 2026 Permalink | Reply
Tags: by: Colin Vout ( 38 )
Teaser 3305: Things with scales and things to scale with
From The Sunday Times, 25th January 2026 [link] [link]

Rummaging through a cupboard I came across an old Snakes and Ladders set. The ladders (label them A to E) went from square 4 to 26, 18 to 28, 19 to 73, 22 to 60, and 41 to 98. The snakes (label them W to Z) went from square 53 to 24, 58 to 5, 72 to 33, and 90 to 66. For old times’ sake I couldn’t resist playing a game on my own. Starting from square 1, rolling the die and moving appropriately, travelling travelling up ladders and down snakes when I chanced to land on their starting squares, I happened to finish exactly on square 100. The total of all the numbers I had thrown on the die was 51.

What was the order of all my up and down jumps? (e.g. A, Y, D).

[teaser3305]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 8:56 am on 25 January 2026 Permalink | Reply
  Here is a program that attempts to constructively play games that finish exactly on square 100 and hve a throw total of 51.
  
  We assume that once a possible game is found, all other possible games have the same sequence of jumps. (I’m thinking about better ways to show the answer to the puzzle is unique).
  
  It runs in 77ms. (Internal runtime is 1.1ms).
```
from enigma import (tuples, seq2str, printf)

# snakes and ladders
jumps = {
  # ladders (lower -> higher)
  (4, 26): 'A', (18, 28): 'B', (19, 73): 'C', (22, 60): 'D', (41, 98): 'E',
  # snakes (higher -> lower)
  (53, 24): 'W', (58, 5): 'X', (72, 33): 'Y', (90, 66): 'Z',
}

# linked squares
link = dict(jumps.keys())

# possible throws
throws = [6, 5, 4, 3, 2, 1]

# play the game
# p = current position
# t = target position
# b = throw budget (must not be exceeded)
# ds = dice throws
# ps = positions
def play(p, t, b, ds=[], ps=[]):
  ps = ps + [p]
  # are we done
  if p == t or p >= 100 or b <= 0:
    yield (ds, ps, b)
  elif b > 0:
    # make the next roll
    for d in throws:
      if not (d > b):
        # move the player
        p1 = p + d
        # perform any jump
        p2 = link.get(p1)
        # continue play
        if p2 is None:
          yield from play(p1, t, b - d, ds + [d], ps)
        else:
          yield from play(p2, t, b - d, ds + [d], ps + [p1])

# find a possible game
for (ds, ps, r) in play(1, 100, 51):
  if r == 0 and ps[-1] == 100:
    printf("throws = {ds} (sum = {t}) -> positions = {ps}", t=sum(ds))
    # find the jumps involved
    js = tuple(filter(None, (jumps.get(k) for k in tuples(ps, 2))))
    printf("jumps = {js}", js=seq2str(js, enc="[]"))
    break
```
  Solution: The jumps made are: C, Z, Y, E.
  
  So the game proceeds as follows:
  
  dice throws of 18: move 1 → 19 (avoiding 4, 18)
  take ladder C from 19 → 73
  dice throws of 17: move 73 → 90
  take snake Z from 90 → 66
  dice throws of 6: move 66 → 72
  take snake Y from 72 → 33
  dice throws of 8: move 33 → 41
  take ladder E from 41 → 98
  dice throws of 2: move 98 → 100
  total dice throws = 18 + 17 + 6 + 8 + 2 = 51
  
  LikeLike
  - Jim Randell 1:28 pm on 25 January 2026 Permalink | Reply
    Here is a better solution. It is shorter, faster, can be used to experiment with different throw budgets, and it finds all solutions (thereby showing the answer to the original puzzle is unique).
    
    We don’t care about the actual dice throws, a game is made up of a sequence of contiguous blocks (that are moved with throws of the dice) with jumps (i.e. a snake or a ladder) between the blocks. The blocks themselves can usually be made by lots of dice throws (and the starts of the jumps are sufficiently fragmented that we can always avoid any unwanted jumps).
    
    This Python program makes a collection of possible contiguous segments, and then joins a sequence of them together with jumps inbetween to make a game that starts at 1, ends at 100, and requires a total of 51 to be scored on the dice throws during the segments.
    
    It runs in 71ms. (Internal runtime is 88µs).
```
from enigma import (defaultdict, tuples, seq2str, printf)

# snakes and ladders
jumps = {
  # ladders (lower -> higher)
  (4, 26): 'A', (18, 28): 'B', (19, 73): 'C', (22, 60): 'D', (41, 98): 'E',
  # snakes (higher -> lower)
  (53, 24): 'W', (58, 5): 'X', (72, 33): 'Y', (90, 66): 'Z',
}

# linked squares
link = dict(jumps.keys())

# possible contiguous segments; segs: start -> ends
segs = defaultdict(list)
seg = lambda x, y: segs[x].append(y)
seg(1, 100)  # no jumps
for (a, b) in jumps.keys():
  # add in 1 to the start of a jump
  seg(1, a)
  # add in end of a jump to 100
  seg(b, 100)
  # add in end of a jump to start of a jump
  for (x, y) in jumps.keys():
    if b < x: seg(b, x)

# find contiguous segments from <s> to <t> with throw budget <b>
def solve(s, t, b, ss=[]):
  # are we done?
  if s == t and b == 0:
    yield ss
  elif s < 100 and b > 0:
    # add in the next segment
    for x in segs[s]:
      d = x - s
      if not (d > b):
        yield from solve(link.get(x, x), t, b - d, ss + [(s, x)])

# find games from 1 to 100 with a throw budget of B
B = 51
for ss in solve(1, 100, B):
  # find jumps used
  js = tuple(jumps[(b, x)] for ((a, b), (x, y)) in tuples(ss, 2))
  printf("{B}: {ss} -> {js}", js=seq2str(js, enc="[]"))
```
    LikeLiked by 1 person
Jim Randell 9:48 am on 23 January 2026 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 2462: [What NOW?]
From The Sunday Times, 29th November 2009 [link]

Puzzlers are often confused about whether the number 1 is prime or square, so perhaps this Teaser will clarify the issue!

• ONE is not prime, but it is an odd perfect square;
• TWO is divisible by 2 but not 4;
• The number of odd primes less than SIX is TWO.

In those statements digits have consistently been replaced by capital letters, with different letters being used for different digits.

What is the number NOW?

This puzzle was originally published with no title.

[teaser2462]
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Jim Randell, Frits, and ruudvanderham are discussing. Toggle Comments
- Jim Randell 9:49 am on 23 January 2026 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  The following run-file executes in 98ms. (Internal runtime of the generated code is 14.4ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

# no leading zeros (ONE, TWO, SIX, NOW)
--invalid="0,OTSN"

# collect primes < 1000
--code="primes.expand(999)"

# ONE is not prime, but it is an odd perfect square
"ONE not in primes"
"E % 2 = 1"
"is_square(ONE)"

# TWO is divisible by 2, but not 4
"O % 2 = 0"
"WO % 4 != 0"

# the number of odd primes less than SIX is TWO
"icount(primes.between(3, SIX - 1)) = TWO"

--template="(ONE) (TWO) (SIX)"
--answer="NOW"
```
  Solution: NOW = 820.
  
  There are two possible assignments of digits to letters:
  
  ONE = 289, TWO = 102, SIX = 564
  ONE = 289, TWO = 102, SIX = 567
  
  The next prime after 563 is 569, so we count 102 odd primes if X is 4 or 7.
  
  LikeLike
  - Frits 9:09 am on 24 January 2026 Permalink | Reply
    
    @Jim, your code allows N to be zero. Is this what you wanted?
    
    LikeLike
    - Jim Randell 9:29 am on 24 January 2026 Permalink | Reply
      
      @Frits: We can also disallow N from being 0 at line 6 of my code.
      
      LikeLike
- ruudvanderham 11:24 am on 23 January 2026 Permalink | Reply
```
import peek
import istr

for o, n, e, t, w, s, i, x in istr.permutations(range(10), 8):
    if o * t * s != 0:
        if not (one := istr("=one")).is_prime() and one.is_odd() and one.is_square():
            if (two := istr("=two")).is_divisible_by(2) and not two.is_divisible_by(4):
                if len(istr.primes(3, (six := istr("=six")))) == two:
                    peek(one, two, six)
```
  LikeLike
Jim Randell 9:16 am on 20 January 2026 Permalink | Reply
Tags: by: Nick MacKinnon ( 57 )
Teaser 2363: [A round number]
From The Sunday Times, 6th January 2008 [link]

We are used to thinking of 1,000 as a “round number”, but this is very much based on 10. One definition of a round number is that it should have lots of small factors, and certainly that it should be divisible by every number up to and including sixteen. There is one such round number that, when written in a certain base (of less than 10,000), looks like 1111. What is more, its base can be found very neatly.

What is that base?

This puzzle was originally published with no title.

[teaser2363]
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Frits, Jim Randell, and Ruud are discussing. Toggle Comments
- Jim Randell 9:17 am on 20 January 2026 Permalink | Reply
  Programatically, it is straightforward to just try all possible bases.
  
  This Python program runs in 75ms. (Internal runtime is 4.2ms).
```
from enigma import (irange, mlcm, call, nconcat, printf)

# M = LCM(2..16)
M = call(mlcm, irange(2, 16))

# consider possible bases
for b in irange(2, 9999):
  # n = 1111 [base b]
  n = nconcat([1, 1, 1, 1], base=b)
  # check for divisibility by 2..16
  if n % M == 0:
    # output solution
    printf("base={b} n={n}")
```
  But here is a more sophisticated approach:
  
  Another way to write 1111 [base b] is: b³ + b² + b + 1, which is an integer valued polynomial function.
  
  And this means we can use the [[ poly_roots_mod() ]] solver from the enigma.py library (described here: [ link ]) to solve the puzzle.
  
  The following Python program has an internal runtime of 572µs.
```
from enigma import (Polynomial, rev, irange, mlcm, call, poly_roots_mod, printf)

# calculate 1111 [base b]
p = Polynomial(rev([1, 1, 1, 1]))

# M = LCM(2..16)
M = call(mlcm, irange(2, 16))

# find roots of p(b) = 0 [mod M] that give a valid base
roots = poly_roots_mod(p, 0)
for b in roots(M):
  if 1 < b < 10000:
    # calculate the corresponding n value
    n = p(b)
    # output solution
    printf("base={b} n={n}")
```
  Solution: The base is 5543.
  
  And the round number is: 1111 [base 5543] = 170338568400 [decimal] = 720720 × 236345.
  
  LikeLike
- Ruud 10:52 am on 20 January 2026 Permalink | Reply
```
import istr

print(*(base for base in istr.range(2, 10000) if all(sum(base**i for i in range(4)).is_divisible_by(i) for i in range(2, 17))))
```
  LikeLike
- Frits 8:33 am on 21 January 2026 Permalink | Reply
```
# 1111 [base b] is: b^3 + b^2 + b + 1 or (b + 1)(b^2 + 1)
# I am not good at modular arithmetic but b^2 + 1 is never a multiple of
# numbers 4.k + 3. So b + 1 must be divisible by 3, 7 and 11 and thus by 231.

for b1 in range(231, 10001, 231):
  # calculate n = 1111 [base b]
  b = b1 - 1
  n = b1 * (b**2 + 1)
  # n must be divisible by numbers 2-16
  if any(n % i for i in [9, 11, 13, 14, 15, 16]): continue
  print("answer:", b)
```
  LikeLike
  - Jim Randell 9:11 am on 21 January 2026 Permalink | Reply
    @Frits: A neat bit of analysis. That is probably the kind of approach the setter had in mind.
    
    In fact as (b² + 1) is not divisible by 3, then it is also not divisible by 9, so we can proceed in steps of: 9 × 7 × 11 = 693, which only needs 14 iterations.
    
    This Python program has an internal runtime of 34µs:
```
from enigma import (irange, printf)

S = 9 * 7 * 11
R = 720720 // S
for i in irange(S, 10000, step=S):
  b = i - 1
  n = i * (b*b + 1)
  if n % R == 0:
    printf("base={b} n={n}")
```
    LikeLike
    - Jim Randell 9:40 am on 21 January 2026 Permalink | Reply
      
      In fact although (b² + 1) may be divisible by 2, it cannot be divisible by 4, so we can move in steps of:
      
      S = 8 × 9 × 7 × 11 = 5544
      
      And, as the base is limited to be less than 10000, there is only one value to check (or if we accept that there is a solution to the puzzle, then there are no values to check).
      
      LikeLike
      - Frits 1:10 pm on 21 January 2026 Permalink | Reply
        
        @Jim, good work.
        
        if b is even then b=2k and b^2 + 1 = 4k^2 + 1
        if b is odd then b = 2k+1 and b^2 + 1 = 4k^2 + 4k + 2
        
        so indeed b^2 + 1 cannot be divisible by 4.
        
        LikeLike
Jim Randell 6:38 am on 18 January 2026 Permalink | Reply
Tags: by: Mark Valentine ( 17 )
Teaser 3304: Roaming bull
From The Sunday Times, 18th January 2026 [link] [link]

Farmer Quentin keeps his prize bull in a circular field of diameter 50m. To improve the layout, he decides to switch to a square field with side length equal to the old diameter. He builds the new fencing around the old, with fence posts every metre starting at the corner. After completing the new fencing, he tethers the bull outside the fence to one of these fence posts with a rope of integer metre length, to allow the removal of the inner circular fence. As a result, the bull’s temporary roaming area is now four times what it was inside the circle.

What is the length of the rope?

[teaser3304]
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Frits and Jim Randell are discussing. Toggle Comments
- Jim Randell 7:14 am on 18 January 2026 Permalink | Reply
  If the bull were tethered to a single post with a 50 m rope in an otherwise unobstructed flat plain, then it would be able to roam a region that is 4 times the area of the original (25 m radius) circular field. So the rope must be at least this long.
  
  When tethered to a post that makes up the boundary of the square field, then bull can roam over a collection of quadrants of a circle are centred on the tether point, and the corners of the square field, as the rope wraps around the boundary of the field.
  
  I considered rope lengths allowing roaming areas of between 2 and 6 non-overlapping quadrants (of various sizes) around the outside of the field. (Which assumes the bull has no other obstructions). If the rope is longer than 100 m then the regions will overlap, as the bull can reach the opposite point on the perimeter of the square from either direction.
  
  For each possible tethering point a corresponding equation is generated and then solved to find possible integer rope lengths.
  
  The following Python code runs in 79ms. (Internal runtime is 2.1ms).
```
from enigma import (Rational, Polynomial as Poly, irange, sq, divc, sprintf as f, printf)

Q = Rational()
q = Q(1, 4)

d = 50  # size of the square
A = sq(d)  # target area (= 4 * r^2 where r = d/2)

# find roots of polynomial <p> = <v> between <lo> and <hi>
def roots(p, v, lo, hi, s=''):
  xs = p.roots(v, domain='Z', include='+', F=Q)  # integer roots
  for x in xs:
    if lo <= x <= hi:
      printf("{s}x={x}")

# suppose the bull is tethered at post a = 0 (corner) to 25 (middle)
# with a rope of length x
for a in irange(0, divc(d, 2)):

  # 2 quadrants: 0 <= x <= a
  p = 2*q * sq(Poly([0, 1]))  # x^2
  roots(p, A, 0, a, f("[2 quads] a={a} -> "))

  # 3 quadrants: a <= x <= d - a
  p += q * sq(Poly([-a, 1]))  # (x - a)^2
  roots(p, A, a, d - a, f("[3 quads] a={a} -> "))

  # 4 quadrants: d - a <= x <= d + a
  p += q * sq(Poly([a - d, 1]))  # (x + (a - d))^2
  roots(p, A, d - a, d + a, f("[4 quads] a={a} -> "))

  # 5 quadrants: d + a <= x <= 2d - a
  p += q * sq(Poly([-(a + d), 1]))  # (x - (a + d))^2
  roots(p, A, d + a, 2*d - a, f("[5 quads] a={a} -> "))

  # 6 quadrants (no overlap): 2d - a <= x <= 2d
  p += q * sq(Poly([a - 2*d, 1]))  # (x + (a - 2d))^2
  roots(p, A, 2*d - a, 2*d, f("[6 quads] a={a} -> "))
```
  Solution: The length of the rope is 58 m.
  
  And the tether is at post 2 m from a corner.
  
  The bull can roam in two quadrants with radius 58 m, one quadrant with radius 56 m, one quadrant with radius 10 m, and one quadrant with radius 6 m.
  
  Here is a diagram of the situation – the light green area (outside the square field) is 4 times the size of the dark green area (inside the square field):
  
  If Farmer Quentin only measures the rope to approximately a whole number of metres, then there are two more solutions where the rope is within 4 cm of a whole number of metres:
  
  rope = 58.99 m; tether = 6 m from a corner
  rope = 60.03 m; tether = 12 m from a corner
  
  We can see these non-integer solutions by setting [[ domain='F', F='float' ]] at line 11 in the program above.
  
  LikeLike
  - Jim Randell 11:00 am on 18 January 2026 Permalink | Reply
    Alternatively, just considering all possible rope lengths and tether points (although this doesn’t let you find non-integer rope lengths).
    
    This Python program runs in 392µs.
```
from enigma import (irange, sq, divc, printf)

d = 50  # size of the square
A = 4 * sq(d)  # target area (in quarter units)

# consider possible tether points
for a in irange(0, divc(d, 2)):

  # consider increasing rope length
  for x in irange(d, 2*d):

    # calculate total area of possible quadrants
    rs = [x, x, x - a, x + a - d, x - a - d, x + a - 2*d]
    v = sum(sq(r) for r in rs if r > 0)

    # output solution
    if v == A: printf("a={a} x={x} [quads={q}]", q=sum(r > 0 for r in rs))
    if v >= A: break
```
    LikeLike
- Frits 1:43 pm on 18 January 2026 Permalink | Reply
  3 areas (or 4 quadrants) will definitely be roamed as I don’t let the rope length start from 1.
  A 6th quadrant is not possible for diameter 50 as I have calculated a basic upper limit for the rope length.
```
d = 50 # diameter in meters

# old roaming area = 25^2.pi so new roaming area must be 50^2.pi
# southern area S = r^2 / 2 < 2500, r^2 < 5000
mx_rope = int((2 * d**2)**.5)

# the rope length must be larger than the diameter
for r in range(d + 1, mx_rope + 1):
  # tether the bull to the southern fence (on the west side)
  
  for y in range(26):   # distance from SW corner
    # quadruple areas divided by pi
    S4 = 2 * r**2        # 4 x area below the southern fence
    NW4 = (r - y)**2 
    N4 = (r - d - y)**2 if r - d - y > 0 else 0
    E4 = (r - d + y)**2 
    
    # all areas should sum to 100^2
    if S4 + NW4 + N4 + E4 == 4 * d**2:
      print("answer:", r, "meters")
```
  LikeLike
Jim Randell 8:11 am on 16 January 2026 Permalink | Reply
Tags: by: W S Graff-Baker ( 2 ), flawed ( 55 )
Teaser 2312: Backward step
From The Sunday Times, 14th January 2007 [link]

Seb had just been given his four-digit PIN, and he was afraid that he would not be able to remember it. So he broke all the rules and wrote it in his pocket diary. But to make it “more secure” he did not actually write the PIN; instead, he wrote the digits in reverse order. He was surprised to discover that the new number was a multiple of the correct PIN.

What is his PIN?

Note that without additional requirements there are many solutions to this puzzle. The intended solution seems to come from the requirement that all the digits of the PIN are non-zero.

[teaser2312]
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GeoffR, Ruud, and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:12 am on 16 January 2026 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  It runs in 82ms. (Internal runtime of the generated code is 1.7ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

# suppose the real PIN is ABCD
"ediv(DCBA, ABCD) > 1"
"D >= A" # [optional]

--distinct=""
--invalid=""
--digits="1-9"
--answer="ABCD"
```
  Solution: The PIN is: 2178.
  
  And so the reversed PIN is: 8712.
  
  And: 8712 = 4 × 2178.
  
  (The multiple solutions to the puzzle as published can be seen by removing the digit restriction on line 11).
  
  Without the restriction to non-zero digits there are 131 solutions, for example (reversed-PIN, PIN):
  
  1000 & 0001
  1010 & 0101
  …
  9801 & 1089
  …
  9990 & 0999
  
  (Although not all these would be surprising).
  
  But only one of these solutions remains if the digits of the PIN must all be non-zero.
  
  The published solution was: “1089 or 2178”.
  
  LikeLike
- Ruud 10:33 am on 16 January 2026 Permalink | Reply
```
import istr
print(*(pin for digits in istr.product(range(1, 10), repeat=4) if (pin:=istr.join(digits))[::-1].divided_by(pin) not in (1, None)))
```
  LikeLike
- GeoffR 7:43 pm on 16 January 2026 Permalink | Reply
```
from enigma import nsplit, is_pairwise_distinct

for pin in range(1234, 9876):
    a, b, c, d = nsplit(pin)
    if 0 in (a, b, c, d): continue
    if is_pairwise_distinct(a, b, c, d):
        rev_pin = 1000*d + 100*c + 10*b + a
        q, r = divmod(rev_pin, pin)
        if r == 0 and q > 1:
           print(f"Pin = {pin}.")
```
  LikeLike

Jim Randell 3:34 pm on 14 January 2026 Permalink | Reply
Tags: by: Mary Connor ( 4 )

Brain teaser 983: Superchamp

From The Sunday Times, 24th May 1981 [link]

The five champion athletes competed against one another in their five special events to decide who was the supreme champion (the one with the highest number of points). Three points were awarded for the winner in each event. two points for second place, and one for third.

The Superchamp won more events than anyone else. All scored a different number of points, and all scored in a different number of events.

Ben and Colin together scored twice as many points as Fred. Eddie was the only one to win his own special event, and in it the Superchamp came second. In another event Eddie gained one point less than the hurdler. The hurdler only came third in the hurdles, which was won by Denis, who was last in the long jump but scored in the sprint. Fred won the long jump and was second in the walk, but came nowhere in the hurdles.

Ben scored in the long jump and Colin scored in the sprint. The thousand-metre man was third in two events. The long jumper scored more points than the sprinter.

Who was the [long] jumper, and what was his score?

This puzzle is included in the book The Sunday Times Book of Brainteasers (1994).

[teaser983]

Jim Randell 3:35 pm on 14 January 2026 Permalink | Reply

There are a lot of conditions to this puzzle, so I used the minizinc.py library to allow me to construct a declarative model for it in MiniZinc, which is then solved, and the Python program used to format the solutions.

It runs in 545ms.

from enigma import (require, printf)
from minizinc import MiniZinc

require("sys.version", "3.6") # needed to support [ use_embed=1 ]

# indices for names and events (1..5)
(B, C, D, E, F) = (1, 2, 3, 4, 5)
(hurdles, long_jump, sprint, m1000, walk) = (1, 2, 3, 4, 5)

# construct the MiniZinc model
m = MiniZinc([
  'include "globals.mzn"',

  # scores for the competitors in each event: pts[event, competitor] = score
  "array [1..5, 1..5] of var 0..3: pts",

  # scores in each event are: 3, 2, 1, 0, 0
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 3) = 1)", # 1st = 3pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 2) = 1)", # 2nd = 2pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 1) = 1)", # 3rd = 1pts
  "constraint forall (i in 1..5) (sum (j in 1..5) (pts[i, j] = 0) = 2)", # unplaced = 0pts

  # functions:
  "function var int: points(var int: j) = sum (i in 1..5) (pts[i, j])",
  "function var int: wins(var int: j) = sum (i in 1..5) (pts[i, j] = 3)",
  "function var int: scored(var int: j) = sum (i in 1..5) (pts[i, j] > 0)",

  # "the superchamp scored more points than anyone else"
  "var 1..5: S",
  "constraint forall (j in 1..5 where j != S) (points(j) < points(S))",

  # "the superchamp won more events than anyone else"
  "constraint forall (j in 1..5 where j != S) (wins(j) < wins(S))",

  # each competitor scored a different number of points
  "constraint all_different (j in 1..5) (points(j))",

  # each competitor scored in a different number of events
  "constraint all_different (j in 1..5) (scored(j))",

  # "B and C together scored twice as many points as F"
  "constraint points({B}) + points({C}) = 2 * points({F})",

  # "E was the only one to win his own event"
  # "and in that event the superchamp came second"
  "array [1..5] of var 1..5: spe",
  "constraint all_different(spe)",
  "constraint forall (j in 1..5) ((j == {E}) <-> (pts[spe[j], j] = 3))",
  "constraint pts[spe[{E}], S] = 2",

  # "in another event E gained 1 point less than the hurdler"
  "var 1..5: hurdler",
  "constraint spe[hurdler] = {hurdles}",
  "constraint exists (i in 1..5) (pts[i, {E}] + 1 = pts[i, hurdler])",

  # "the hurdler only came third in hurdles"
  "constraint pts[{hurdles}, hurdler] = 1",

  # "hurdles was won by D"
  "constraint pts[{hurdles}, {D}] = 3",

  # "D came last in long jump"
  "constraint pts[{long_jump}, {D}] = 0",

  # "D scored in the sprint"
  "constraint pts[{sprint}, {D}] > 0",

  # "F won the long jump"
  "constraint pts[{long_jump}, {F}] = 3",

  # "F was 2nd in the walk"
  "constraint pts[{walk}, {F}] = 2",

  # "F did not score in hurdles"
  "constraint pts[{hurdles}, {F}] = 0",

  # "B scored in the long jump"
  "constraint pts[{long_jump}, {B}] > 0",

  # "C scored in the sprint"
  "constraint pts[{sprint}, {C}] > 0",

  # "the 1000m man was 3rd in 2 events"
  "var 1..5: m1000er",
  "constraint spe[m1000er] = {m1000}",
  "constraint sum (i in 1..5) (pts[i, m1000er] = 1) = 2",

  # "the long jumper scored more points than the sprinter"
  "var 1..5: long_jumper",
  "constraint spe[long_jumper] = {long_jump}",
  "var 1..5: sprinter",
  "constraint spe[sprinter] = {sprint}",
  "constraint points(long_jumper) > points(sprinter)",

], use_embed=1)


# map indices back to names / events (for output)
name = {1: "Ben", 2: "Colin", 3: "Denis", 4: "Eddie", 5: "Fred"}
event = {1: "hurdles", 2: "long jump", 3: "sprint", 4: "1000m", 5: "walk"}

# solve the model and find:
# pts = results in each event
# spe = special event for each competitor
for (pts, spe) in m.solve(solver="minizinc -a", result="pts spe"):
  # collect points
  t = dict((k, 0) for k in name.keys())

  # output the results for each event
  for (i, rs) in enumerate(pts, start=1):
    (p1, p2, p3) = (rs.index(p) + 1 for p in (3, 2, 1))
    printf("{e}: 1st = {p1}, 2nd = {p2}, 3rd = {p3}", e=event[i], p1=name[p1], p2=name[p2], p3=name[p3])
    # collect points
    t[p1] += 3
    t[p2] += 2
    t[p3] += 1
  printf()

  # determine superchamp
  S = max(t.keys(), key=t.get)
  assert all(k == S or v < t[S] for (k, v) in t.items())

  # output the results for each competitor
  for j in sorted(event.keys()):
    printf("{j} [{e}]: {t} points{S}", j=name[j], e=event[spe[j - 1]], t=t[j], S=(' [SUPERCHAMP]' if j == S else ''))
  printf("--")
  printf()

Solution: Denis is the long jumper. He scored 5 points.

The program finds 2 possible scenarios:

hurdles: 1st = Denis, 2nd = Eddie, 3rd = Ben
long jump: 1st = Fred, 2nd = Ben, 3rd = Eddie
sprint: 1st = Ben, 2nd = Denis, 3rd = Colin
1000m: 1st = Eddie, 2nd = Ben, 3rd = Fred
walk: 1st = Ben, 2nd = Fred, 3rd = Eddie

Ben [hurdles]: 11 points [SUPERCHAMP]
Colin [sprint]: 1 points
Denis [long jump]: 5 points
Eddie [1000m]: 7 points
Fred [walk]: 6 points

hurdles: 1st = Denis, 2nd = Eddie, 3rd = Colin
long jump: 1st = Fred, 2nd = Ben, 3rd = Colin
sprint: 1st = Colin, 2nd = Denis, 3rd = Eddie
1000m: 1st = Eddie, 2nd = Colin, 3rd = Fred
walk: 1st = Colin, 2nd = Fred, 3rd = Eddie

Ben [sprint]: 2 points
Colin [hurdles]: 10 points [SUPERCHAMP]
Denis [long jump]: 5 points
Eddie [1000m]: 7 points
Fred [walk]: 6 points

The solution published in the book (which runs to 3 pages) only gives the first of these solutions.

I think this is due to the interpretation of “In another event Eddie gained one point less than the hurdler”. The book solution seems to infer (hurdler = 3pts, Eddie = 2pts) or (hurdler = 2pts, Eddie = 1pt), and concludes it is not possible for C to be the champ.

And in the first scenario, in the long jump, we have: the hurdler (Ben) scores 2pts (2nd place), and Eddie scores 1pt (3rd place).

However, I think the wording also allows (hurdler = 1pt, Eddie = 0pts), and in the second scenario, in the long jump, we have: the hurdler (Colin) scores 1pt (3rd place), and Eddie scores 0pts (unplaced).

LikeLike

Frits 3:06 pm on 17 January 2026 Permalink | Reply

from itertools import permutations

'''        3   2   1       
hurdles    A   F   K        
long J     B   G   L        
sprint     C   H   M              super=Superchamp
1000m      D   I   N        
Walk       E   J   O        
'''

sports = [-1, -1, -1, -1, -1]
(ben, colin, denis, eddie, fred) = range(5)

dgts, sols = set(range(5)), set()

A = denis # Denis won hurdles
B = fred  # Fred won the long jump
J = fred  # Fred was 2nd in the walk

# Fred came nowhere in the hurdles
for F, K in permutations(dgts - {A, fred}, 2):    # hurdles
  # hurdler (K) came 3rd in the hurdles, Eddie is no hurdler
  if eddie == K: continue
  for G, L in permutations(dgts - {B, denis}, 2): # Denis last in long jump
    if 0 not in {G, L} : continue                 # Ben scored in long jump
    for C, H, M in permutations(dgts, 3):         # sprint
      # Colin and Denis scored in the sprint
      if not {C, H, M}.issuperset({colin, denis}): continue 
      for D, I, N in permutations(dgts, 3):       # 1000m
        for E, O in permutations(dgts - {J}, 2):  # walk
          gold   = [A, B, C, D, E]
          silver = [F, G, H, I, J]
          bronze = [K, L, M, N, O]
          # Eddie was the only one to win his own special event
          if eddie not in gold: continue
          scores = [A, F, K, B, G, L, C, H, M, D, I, N, E, J, O]
          scored = [scores.count(i) for i in range(5)]
          # all scored in a different number of events.
          if len(set(scored)) != 5: continue
          
          pts = [sum([3 - (i % 3) for i, s in enumerate(scores) if s == j]) 
                 for j in range(5)]
          # all scored a different number of points
          if len(set(pts)) != 5: continue
          # Ben and Colin together scored twice as many points as Fred
          if pts[ben] + pts[colin] != 2 * pts[fred]: continue
          # the Superchamp won more events than anyone else ( >= 2)
          sups = sorted((f, i) for i in range(5) if (f := gold.count(i)) > 1)
          if not sups or (len(sups) == 2 and len(set(gold)) == 3): continue
          
          super = sups[-1][1]
          # once the Superchamp came 2nd after Eddie
          if super == eddie or super not in silver: continue
          # the Superchamp had the highest number of points
          if pts[super] != max(pts): continue
          
          sports[0] = K  # hurdler only came third in the hurdles
                
          # check sports where Eddie and Superchamp ended 1 and 2
          for e, (g, s) in enumerate(zip(gold, silver)):
            if (g, s) != (eddie, super): continue
            
            if e == 0: continue # Eddie is not the hurdler
            # in another event Eddie gained one point less than the hurdler
            for i, gsb in enumerate(zip(gold, silver, bronze)):
              if ((K, eddie) in zip(gsb, gsb[1:]) or 
                  (eddie not in gsb and gsb[-1] == K)): break
            else: # no break
              break    
            
            sports_ = sports.copy()
            sports_[e] = eddie  # Eddie performs sport <e>
            # assign people to remaining sports
            for p in permutations(dgts - {eddie, K}):
              sports__ = sports_.copy()
              j = 0
              for i in range(5):
                if sports_[i] != -1: continue
                sports__[i] = p[j]
                j += 1
              
              # Eddie was the only one to win his own special event
              for i, s in enumerate(sports__):
                if gold[i] == s and s != eddie: break
              else: # no break  
                # the long jumper scored more points than the sprinter
                if not (pts[sports__[1]] > pts[sports__[2]]): continue
                # the thousand-metre man was third in two events
                if bronze.count(sports__[3]) != 2: continue
                # store solution
                sols.add((sports__[1], pts[sports__[1]]))

names = "Ben Colin Denis Eddie Fred".split()                
print("answer:", ' or '.join(names[a] + ' with score ' + str(s) 
                              for a, s in sols))

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Jim Randell 6:25 am on 11 January 2026 Permalink | Reply
Tags: by: John Owen ( 33 )
Teaser 3303: Knights and knaves
From The Sunday Times, 11th January 2026 [link] [link]

On my travels, I came across a group of some knights (always truthful) and some knaves (always lying). At first, I couldn’t tell which were knights and which were knaves. However, six individuals then gave the following statements about their group:

“The number of knights is prime.”
“The number of knights is odd.”
“The number of knaves is square.”
“The number of knaves is even.”
“The total of knights and knaves is an odd number.”
“There are more knaves than knights.”

These statements enabled me to deduce the composition of the group.

How many (a) knights and (b) knaves were in the group?

[teaser3303]
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- Jim Randell 6:47 am on 11 January 2026 Permalink | Reply
  The setter has an additional piece of information – i.e. they can observe the total number of people in the group. If it is possible to break this total down in into knights and knaves in only one way that is consistent with the statements made, then the composition of the group can be determined with certainty.
  
  The following Python program runs in 71ms. (Internal runtime is 158µs).
```
from enigma import (irange, inf, decompose, filter2, is_prime, is_square_p, printf)

# consider increasing group sizes (at least 6)
for t in irange(6, inf):
  # collect candidate solutions for this total size
  sols = list()
  # decompose into (a) knights and (b) knaves
  for (a, b) in decompose(t, 2, min_v=1, increasing=0, sep=0):

    # evaluate the statements
    ss = filter2(bool, [
      # 1. "a is prime"
      is_prime(a),
      # 2. "a is odd"
      (a % 2 == 1),
      # 3. "b is square"
      is_square_p(b),
      # 4. "b is even"
      (b % 2 == 0),
      # 5. "a + b is odd"
      (t % 2 == 1),
      # 6. "b > a"
      (b > a),
    ])

    # check minimum group sizes
    if len(ss.true) > a or len(ss.false) > b: continue
    # record this candidate
    sols.append((a, b))

  printf("[{t} -> {sols}]")
  # check for unique solutions
  if len(sols) == 1:
    (a, b) = sols[0]
    printf("a = {a}; b = {b}")
    break
```
  Solution: (a) There are 5 knights; (b) There are are 2 knaves.
  
  The total size of the group is 7, and the statements made are:
  
  1. True (“the number of knights is prime” = 5)
  2. True (“the number of knights is odd” = 5)
  3. False (“the number of knaves is square” = 2)
  4. True (“the number of knaves is even” = 2)
  5. True (“the total number of knights and knaves is odd” = 7)
  6. False (“there are more knaves than knights” 5 > 2)
  
  There are 4 true statements, and 2 false statements, so both knaves made a false statement, and 4 of the 5 knights made a true statement.
  
  And this is the only way a group of 7 can be broken down consistent with the statements.
  
  If we allow the program to continue to larger group sizes, we see that the number of candidate breakdowns grows, so it would not be possible to determine the composition of a larger group.
  
  LikeLike

Jim Randell 9:58 pm on 9 January 2026 Permalink | Reply
Tags: by: Bernardo Recaman ( 6 )

Teaser 2360: [Flower arranging]

From The Sunday Times, 16th December 2007 [link]

In the following statements and instruction, digits have been consistently replaced by letters, with different letters for different digits.

F + L + O + W + E + R raised to the power P is equal to the six-digit number FLOWER.

FACTOR is divisible by A, R and T.

What is the value of FLOWER ART?

This puzzle was originally published with no title.

[teaser2360]

Jim Randell 10:03 pm on 9 January 2026 Permalink | Reply
Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

It runs in 226ms. (Internal runtime of the generated code is 126ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

"pow(F + L + O + W + E + R, P) = FLOWER"

"FACTOR % A = 0"
"FACTOR % R = 0"
"FACTOR % T = 0"

--answer="(FLOWER, ART)"
```
Solution: FLOWER ART = 390625 751.

And:

FLOWER = 390625 = 25^4 = (3 + 9 + 0 + 6 + 2 + 5) ^ 4.
FACTOR = 378105 is divisible by 7, 5, 1.

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Ruud 6:11 am on 10 January 2026 Permalink | Reply

import istr

for f, l, o, w, e, r, p in istr.permutations(range(10), 7):
    if sum(istr("=flower")) ** p == istr("=flower"):
        for a, c, t in istr.permutations(range(10), 3):
            if istr("=flowerpact").all_distinct() and all(istr("=factor").is_divisible_by(x) for x in (a, r, t)):
                print(istr("=flower"), istr("=art"))

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ruudvanderham 11:32 am on 10 January 2026 Permalink | Reply

This version is *much* faster. Note that it required the latest version of istr (1.1.16):

import istr

for p in range(2, 10):
    for flower in istr.power_ofs(p, 100000, 1000000):
        if sum(flower) ** p == flower:
            istr.decompose(flower, "flower")
            for a, c, t in istr.permutations(set(istr.range(10))-set(flower)-{p}, 3):
                if istr("=flowerpact").all_distinct() and all(istr("=factor").is_divisible_by(x) for x in (a, r, t)):
                    print(istr("=flower"), istr("=art"))

LikeLike

GeoffR 2:48 pm on 10 January 2026 Permalink | Reply

Note: P = 5 is max value for int_pow() can accomodate without an error, but OK for a solution.

% A Solution in MiniZinc
include "globals.mzn";

var 1..9:F; var 1..9:A; var 1..9:R; var 2..5:P;  
var 0..9:L; var 0..9:O; var 0..9:W; var 0..9:E;
var 1..9:T; var 0..9:C;

constraint all_different ([F, A, R, P, L, O, W, E, T, C]);

var 100000..999999:FLOWER == 100000*F + 10000*L + 1000*O
+ 100*W + 10*E + R;

var 100000..999999:FACTOR = 100000*F + 10000*A + 1000*C
+ 100*T + 10*O + R;

var 100..999:ART = 100*A + 10*R + T;

var 2..54:SUM = F + L + O + W + E + R;

constraint int_pow(SUM, P, FLOWER);

constraint FACTOR mod A == 0 /\ FACTOR mod R == 0 /\ FACTOR mod T == 0;

solve satisfy;
output["FLOWER = " ++ show(FLOWER) ++ ", ART = " ++ show(ART)];

% FLOWER = 390625, ART = 751
% ----------
% ==========

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Brian Gladman 7:40 pm on 10 January 2026 Permalink | Reply

from itertools import permutations

# convert digits to numbers and vice versa
dgts_to_nbr = lambda dgts: int(''.join(map(str, dgts)))
nbr_to_dgts = lambda nbr: tuple(int(d) for d in str(nbr))

# the digit sum of FLOWER is between the minimum 
# and maximum sums of six digits from 0..9
for sm in range(15, 40):
  # the possible powers to give sm ** P six digits
  for P in (4, 5):
    if 100_000 <= (FLOWER := sm ** P) < 1_000_000:
      t = (F, L, O, W, E, R) = nbr_to_dgts(FLOWER)
      # check that three digits remain unallocated
      if len(r3 := set(range(10)).difference(t + (P,))) == 3:
        # consider the order of the remaining digits for A, C and T
        for A, C, T in permutations(r3):
          FACTOR = dgts_to_nbr((F, A, C, T, O, R))
          # FACTOR must be divisible by A, R and T
          if FACTOR % (A * R * T) == 0:
            ART = dgts_to_nbr((A, R, T))
            print(f"FLOWER ART = {FLOWER} {ART} [{FACTOR = }, {P = }]")

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Jim Randell 11:02 am on 6 January 2026 Permalink | Reply
Tags: by: Angela Newing ( 39 )
Brain teaser 981: Bailing out
From The Sunday Times, 10th May 1981 [link]

The pirate ship “Nancy” had one leaky hold. As long as the water was not allowed to rise above a certain level there was no danger of the ship sinking; so the practice was to allow the water to rise to a mark on the side of the hold which was just below the danger level, and then to send in a team of about 15 or 20 pirates to bale out the hold until it was empty. The water came in continuously at a constant: rate, so the process had, to be repeated regularly.

There were two kinds of baling equipment: practical pans and capacious cans (which emptied the water at different rates), and each pirate who was assigned to baling duty picked up whichever of these was nearest to hand.

One baling party consisting of 11 pirates with practical pans and 6 pirates with capacious cans emptied the hold in 56 minutes. When the water had risen again, the next party of 5 pirates with practical pans and 12 with capacious cans emptied it in 35 minutes. A third party of 15 with practical pans plus 4 with capacious cans took exactly an hour to do the job.

How long would it take a party of 6 with practical pans and 10 with capacious cans?

This puzzle is included in the book The Sunday Times Book of Brainteasers (1994).

[teaser981]
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- Jim Randell 11:02 am on 6 January 2026 Permalink | Reply
  If we treat the volume of water at the point that baling starts as 1 unit, then we can write the following equation for a team with P pans and C cups as:
  
  T(P.p + C.c + f) = 1
  ⇒
  P.p + C.c + f = 1/T
  
  where:
  
  T is the total time taken to empty the hold (min);
  p is the rate at which a pan removes water (units/min);
  c is the rate at which a cup removes water (units/min);
  f is the rate at which the hold fills (units/min) (this will be negative as water is flowing into the hold).
  
  For the three teams given we get the following equations:
  
  11p + 6c + f = 1/56
  5p + 12c + f = 1/35
  15p + 4c + f = 1/60
  
  This gives us three equations in three variables, which can be solved to give the rates p, c and f.
  
  We can then calculate the time taken for the 4th team:
  
  T(6p + 10c + f) = 1
  ⇒
  T = 1/(6p + 10c + f)
  
  The following Python program runs in 74ms. (Internal runtime is 186µs).
```
from enigma import (Matrix, Rational, printf)

Q = Rational()

# construct the equations
eqs = [
  #  p   c   f   1/time
  ((11,  6,  1), Q(1, 56)), # team 1 (11p + 6c) takes 56 minutes
  (( 5, 12,  1), Q(1, 35)), # team 2 (5p + 12c) takes 35 minutes
  ((15,  4,  1), Q(1, 60)), # team 3 (15p + 4c) takes 60 minutes
]

(p, c, f) = Matrix.linear(eqs, field=Q)
printf("[p={p} c={c} f={f}]")

# calculate time for team 4 (6p + 10c)
t = Q(1, 6*p + 10*c + f)
printf("team 4 takes {t} minutes")
```
  Solution: It would take the 4th team 42 minutes to empty the hold.
  
  We get:
  
  p = 1/840
  c = 1/336
  f = −11/840
  ⇒
  1/T = 6/840 + 10/336 − 11/840
  1/T = 1/42
  T = 42
  
  So the water is coming in at 11/840 units/min, and a pirate with a pail can empty out 1/840 units/min. So 11 pirates with pails would be able to remove the water at the rate it was coming in. A pirate with a cup can empty 1/336 units/min (so a cup can bail more than a pail), and 2.5 pirates with cups could remove water at the rate it is coming in.
  
  LikeLike

Jim Randell 7:15 am on 4 January 2026 Permalink | Reply
Tags: by: Howard Williams ( 45 )

Teaser 3302: Challenging grandson

From The Sunday Times, 4th January 2026 [link] [link]

My grandson likes making puzzles; his most recent one started with a four-digit number with all the digits different. He added up all the possible two-digit combinations of these four digits (e.g., if the digits were 0, 1, 2 and 3 he would add 01 + 02 + 03 + 10 + 12 + …). This total divided exactly into his four-digit number.

Not satisfied with this, he said he had added an extra digit (different from all the others) to the end of his number to make a five-digit number. The five-digit number was an exact multiple of the total of all the two-digit combinations of these five digits. He said that if he told me how many digits were in this multiple, I would be able to work out his five-digit number.

What was his five-digit number?

[teaser3302]

Jim Randell 7:35 am on 4 January 2026 Permalink | Reply

When constructing the sum of the combinations, with 4 digits, each can be paired with 3 others, and so appears 3 times in the tens column and 3 times in the units column. The total sum is therefore:

33 × sum(digits)

And starting with 5 digits the total sum is:

44 × sum(digits)

This Python program runs in 68ms. (Internal runtime is 955µs).

from enigma import (irange, nsplit, div, ulambda, filter_unique, ndigits, printf)

# generate possible scenarios
def generate():
  digits = set(irange(0, 9))

  # the first number is a 4-digit multiple of 33 with distinct digits
  for n1 in irange.round(1000, 9999, step=33, rnd='I'):
    ds = nsplit(n1)
    if len(set(ds)) != 4: continue
    # calculate the sum of the 2-digit combinations
    t1 = sum(ds)
    s1 = 33 * t1
    # n1 is a multiple of s1
    k1 = div(n1, s1)
    if k1 is None: continue

    # add in an extra digit to form the second number
    for d in digits.difference(ds):
      n2 = 10*n1 + d
      # calculate the sum of the 2-digit combinations
      s2 = 44 * (t1 + d)
      # n2 is a multiple of s2
      k2 = div(n2, s2)
      if k2 is None: continue

      # return (number, sum, multiple) values for each number
      printf("[n1={n1} s1={s1} k1={k1} -> n2={n2} s2={s2} k2={k2}]")
      yield ((n1, s1, k1), (n2, s2, k2))

# find solutions unique by number of digits in k2
f = ulambda("((n1, s1, k1), (n2, s2, k2)): ndigits(k2)")
ss = filter_unique(generate(), f).unique

# output solutions
for ((n1, s1, k1), (n2, s2, k2)) in ss:
  printf("n2={n2} [s2={s2} k2={k2}; n1={n1} s1={s1} k1={k1}]")

Solution: The 5-digit number was 83160.

There are 6 candidate scenarios (number = sum × multiple):

2376 = 594 × 4 → 23760 = 792 × 30
3564 = 594 × 6 → 35640 = 792 × 45
4752 = 594 × 8 → 47520 = 792 × 60
7128 = 594 × 12 → 71280 = 792 × 90
8316 = 594 × 14 → 83160 = 792 × 105

In each case the sum of the digits in the 4-digit and 5-digit numbers is 18, and so the sum of the 2-digit combinations from the 4-digit number is 33 × 18 = 594, and the sum of the 2-digit combinations from the 5-digit number is 44 × 18 = 792.

Of the 5-digit numbers only the last one is a 3-digit multiple of the sum, and this provides the answer to the puzzle.

LikeLike

Ruud 8:51 am on 4 January 2026 Permalink | Reply

import peek
import istr
import collections

collect = collections.defaultdict(list)
for n4 in istr.range(1000, 10000):
    if not n4.all_distinct():
        continue
    s4 = 33 * sum(n4)
    if not n4.is_divisible_by(s4):
        continue
    for n1 in istr.range(10):
        if n1 in n4:
            continue
        n5 = n4 | n1
        s5 = 44 * sum(n5)
        if n5.is_divisible_by(s5):
            multiple = n5 / s5
            collect[len(multiple)].append(n5)

for n5s in collect.values():
    if len(n5s) == 1:
        peek(n5s[0])

LikeLike

ruudvanderham 7:15 pm on 4 January 2026 Permalink | Reply

Slightly faster:

import peek
import istr
import collections

collect = collections.defaultdict(list)
for p4 in istr.permutations(range(10), 4):
    n4 = istr.join(p4)
    s4 = 33 * sum(n4)
    if not n4.is_divisible_by(s4):
        continue
    for n1 in set(istr.range(10)) - set(p4):
        n5 = n4 | n1
        s5 = 44 * sum(n5)
        if n5.is_divisible_by(s5):
            multiple = n5 / s5
            collect[len(multiple)].append(n5)

for n5s in collect.values():
    if len(n5s) == 1:
        peek(n5s[0])

LikeLike

Frits 11:43 am on 4 January 2026 Permalink | Reply

from itertools import permutations

# the extra digit must be zero due to the alternating sum rule
d = dict()
for A, B, C in permutations(range(1, 10), 3):
  # divisibility of 11 rule (alternating digit sum is 0 or a multiple of 11)
  if (D := (A + C - B) % 11) in {0, A, B, C, 10}: continue
  # calculate sum of digits (must be a multiple of 3)
  if (s := A + B + C + D) % 3: continue
  # calculate first multiple  
  m1, r = divmod(ABCD := 1000 * A + 100 * B + 10 * C + D, 33 * s)
  if r or m1 % 2: continue  # as m2 = 7.5 * m1
  # number of digits in second multiple must be unique
  d[nd2] = 0 if (nd2 := len(str(m2 := int(7.5 * m1)))) in d else 10 * ABCD
  
for k, v in d.items():
  if v:
    print("answer:", v)

LikeLike

Frits 12:47 pm on 4 January 2026 Permalink | Reply

from itertools import permutations

# the extra digit must be zero due to the alternating sum rule
d = dict()
digits = set(range(1, 10))    
# D must be even as 10 * ABCD must be divisible by 44
for D in [2, 4, 6, 8]:
  for A, C in permutations(digits - {D}, 2):
    # divisibility of 11 rule (alternating digit sum is 0 or a multiple of 11)
    if (B := (A + C - D) % 11) in {0, A, C, D, 10}: continue
    # calculate sum of digits (must be a multiple of 3)
    if (s := A + B + C + D) % 3: continue
    # calculate first multiple  
    m1, r = divmod(ABCD := 1000 * A + 100 * B + 10 * C + D, 33 * s)
    if r or m1 % 2: continue  # as m2 = 7.5 * m1
    # number of digits in second multiple must be unique
    d[nd2] = 0 if (nd2 := len(str(int(7.5 * m1)))) in d else 10 * ABCD
      
for k, v in d.items():
  if v:
    print("answer:", v)

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Jim Randell 11:16 am on 2 January 2026 Permalink | Reply
Tags: by: Graham Smithers ( 84 )

Teaser 2463: [Single-digit multiples]

From The Sunday Times, 6th December 2009 [link]

I have used each of the digits 0 to 9 once each to make a one-digit number, a two-digit number, a three-digit number and a four-digit number. The fourth number is a single-digit multiple of the third; the third number is a single-digit multiple of the second; and the second is a single-digit multiple of the first.

What is the four-digit number?

This puzzle was originally published with no title.

[teaser2463]

Jim Randell 11:17 am on 2 January 2026 Permalink | Reply
Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library:

It runs in 85ms. (Internal runtime of the generated code is 8.8ms).
```
#! python3 -m enigma -rr

SubstitutedExpression

"div(GHIJ, DEF) < 10"
"div(DEF, BC) < 10"
"div(BC, A) < 10"

--answer="(A, BC, DEF, GHIJ)"
```
Solution: The 4-digit number is 3402.

We have:

3402 = 6 × 567
567 = 7 × 81
81 = 9 × 9
9

And between them the numbers 9, 81, 567, 3402 use each of the digits 0-9 exactly once.

LikeLike

Ruud 2:21 pm on 2 January 2026 Permalink | Reply

Here is recursive solution:

import istr

def expand(n="", used=[]):
    for d in range(1, 10):
        next_n = d * istr(n if n else 1)
        next_used = used + [next_n]
        if len(next_n) == len(n) + 1 and istr("").join(next_used).all_distinct():
            if len(next_n) == 4:
                print(*next_used)
            else:
                expand(n=next_n, used=next_used)

expand()

LikeLike

GeoffR 4:54 pm on 2 January 2026 Permalink | Reply

% A Solution in MiniZinc
include "globals.mzn";

var 1..9: a; var 0..9: b; var 0..9: c; var 0..9: d;
var 1..9: e; var 0..9: f; var 0..9: g;
var 1..9: h; var 0..9: i; var 1..9: j;

constraint all_different ([a, b, c, d, e, f, g, h, i, j]);

var 1000..9999: abcd == 1000*a + 100*b + 10*c + d;
var 100..999: efg == 100*e + 10*f + g;
var 10..99: hi == 10*h + i;
var 2..9: d1; var 2..9: d2; var 2..9: d3; 

% Four, three and two digut number constraints
constraint abcd div efg == d1 /\ d1 * efg == abcd;
constraint efg div hi == d2 /\ d2 * hi == efg;
constraint hi div j == d3 /\ d3 * j == hi;

solve satisfy;

output ["The four-digit number was " ++ show(abcd)
++ "\n" ++ "[a, b, c, d, e, f, g, h, i, j] = "
++ "\n" ++ show([a, b, c, d, e, f, g, h, i, j]) ];

% The four-digit number was 3402
% [a, b, c, d, e, f, g, h, i, j] = 
% [3, 4, 0, 2, 5, 6, 7, 8, 1, 9]
% ----------
% ==========

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Jim Randell 5:33 am on 28 December 2025 Permalink | Reply
Tags: by: Andrew Skidmore ( 88 )
Teaser 3301: Round the table
From The Sunday Times, 28th December 2025 [link] [link]

I recently attended a celebration with fewer than 100 family and friends. We sat round a large oval table with one seat for each person. I said to Liam that he could pick any seat, but it might be interesting to work out how many different arrangements were possible for people round the table. He replied: “I’ll leave that to you Grandad”, then (mischievously) added “but remember I always sit next to you and to Granny”.

The restriction that I must sit next to Liam divided the number of possible arrangements by a certain prime number. The further restriction that he must also be next to Granny caused further division by a prime number with a different number of digits.

What, in ascending order, are the number of people round the table and the two prime numbers?

This completes the archive of puzzles from 2025. There is now a complete archive of puzzles (and solutions!) from December 2009 onwards (plus a lot of earlier puzzles).

[teaser3301]
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GeoffR, Ruud, and Jim Randell are discussing. Toggle Comments
- Jim Randell 5:41 am on 28 December 2025 Permalink | Reply
  With n people there are factorial(n) free arrangements (assignments of people to seats). (Or, if we only care about who is sitting next to who, and not about the absolute positions the number of free arrangements is factorial(n − 1) / 2 (i.e. is reduced by a factor of 2n)).
  
  We can generate these by seating Liam, then Grandad, then Granny and then all the remaining (n − 3) guests.
  
  After Liam has chosen his seat, for a free arrangement, Grandad has a choice of the remaining (n − 1) seats, but if we require Grandad to sit next to Liam, there is only a choice of 2 seats.
  
  So this restriction reduces Grandad’s choice from (n − 1) to 2, so the number of arrangements is reduced by a factor of:
  
  p₁ = (n − 1) / 2
  
  And this must be a prime number.
  
  Similarly, if Granny had a free choice, she could could choose from (n − 2) seats, but if we require her to sit next to Liam, she must “choose” from only 1 seat.
  
  This further reduces the number of arrangements by a factor of:
  
  p₂ = (n − 2)
  
  And this is also a prime number (with a different number of digits to p₂).
  
  If n is less than 100, then p₂ must have 2 digits and p₁ must have 1 digit. So there are only a few cases to consider, and a manual solution is straighforward.
  
  This Python program starts from the total number of people n.
  
  It runs in 69ms. (Internal runtime is 101µs).
```
from enigma import (irange, div, is_prime, ndigits, printf)

# consider number of people around the table
for n in irange(3, 99):
  # first restriction divides possibilities by a prime number
  p1 = div(n - 1, 2)
  if not is_prime(p1): continue
  # second restriction further divides possibilities by a prime number
  p2 = n - 2
  if not is_prime(p2): continue
  # the primes have different numbers of digits
  if not (ndigits(p1) != ndigits(p2)): continue
  # output solution
  ss = sorted([n, p1, p2])
  printf("n={n} -> p1={p1} p2={p2} {ss}")
```
  Or, shorter and faster, this Python program considers possible 1-digit prime values for p₁.
  
  It has an internal runtime of 54µs.
```
from enigma import (primes, printf)

# p1 = (n - 1) / 2 is prime with fewer digits than p2 = n - 2
for p1 in primes.between(2, 9):
  n = 2 * p1 + 1
  p2 = n - 2
  if not (p2 > 9 and primes.is_prime(p2)): continue
  # output solution
  ss = [p1, p2, n]
  printf("n={n} -> p1={p1} p2={p2} {ss}")
```
  Solution: The numbers are: 7, 13, 15.
  
  There are 15 people around the table.
  
  So the total number of possible free arrangements is factorial(15) = 1307674368000.
  
  But, requiring Grandad to sit next to Liam reduces Grandad’s choices from 14 to 2, so divides the number of possibilities by 7 (to give 186810624000 arrangements).
  
  Further requiring Granny to also sit next to Liam reduces Granny’s choices from 13 to 1, so divides the number of possibilities by 13 (to give 14370048000 arrangements).
  
  (If we only care about the relative seating arrangements (who sits next to who), then the number of arrangements is reduced by a factor of 2n = 30, but this does not affect the values of the divisions).
  
  Manually:
  
  p₁ is a 1-digit prime, and p₂ is a 2-digit prime such that:
  
  p₁ = (n − 1) / 2
  p₂ = (n − 2)
  ⇒
  p₂ = 2 p₁ − 1
  n = p₂ + 2
  
  and:
  
  p₂ > 9
  ⇒
  2 p₁ − 1 > 9
  2 p₁ > 10
  p₁ > 5
  
  Considering 1-digit primes for p₁ that are greater than 5:
  
  p₁ = 7 ⇒ p₂ = 13, n = 15 [solution]
  
  LikeLike
- Ruud 8:14 am on 28 December 2025 Permalink | Reply
```
import peek
import istr
from math import factorial

for n in range(3, 100):
    f1 = istr(factorial(n) / (n * 2 * factorial(n - 2)))
    f2 = istr(factorial(n - 2) / factorial(n - 3))
    if f1.is_prime() and f2.is_prime() and len(f1) != len(f2):
        peek(sorted((n, f1, f2)))
```
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- GeoffR 9:01 am on 30 December 2025 Permalink | Reply
  An AI solution, nicely commented by AI.
```
# ST 3301 by Perplexity AI

import math

def is_prime(num):
    if num < 2:
        return False
    for i in range(2, int(math.sqrt(num)) + 1):
        if num % i == 0:
            return False
    return True

def digits(num):
    return len(str(num))

# Solve the teaser: n < 100 people around oval table (circular)
# Total arrangements: (n-1)!
# 1st restriction (Liam next to Grandad): divides by prime p1 with D1 digits
# 2nd restriction (Liam also next to Granny): further divides
# by prime p2 with D2 != D1 digits

for n in range(3, 100):  # At least 3 people (Grandad, Liam, Granny)
    # Fix Grandad's position to handle circular symmetry
    total = math.factorial(n - 1)
    
    # 1st restriction: Liam next to Grandad (2 choices for Liam's seat)
    restr1 = 2 * math.factorial(n - 2)
    div1 = total // restr1  # Should be integer prime
    if total % restr1 != 0 or not is_prime(div1):
        continue
    
    # 2nd restriction: Liam next to Grandad AND Granny
    # The three occupy 3 consecutive seats with Grandad fixed
    # 2 possible blocks (Liam-Granny or Granny-Liam to Grandad's sides)
    restr2 = 2 * math.factorial(n - 3)
    div2 = restr1 // restr2  # Further division from restr1
    if restr1 % restr2 != 0 or not is_prime(div2):
        continue
    
    # Check different number of digits
    if digits(div1) != digits(div2):
        print(f"n={n}, First prime={div1} ({digits(div1)} digits)")
        print(f"Second prime={div2} ({digits(div2)} digits)")
```
  LikeLike
Jim Randell 11:19 am on 26 December 2025 Permalink | Reply
Tags: by: C Higgins ( 38 ), by: H Bradley ( 38 )
Teaser 2465 : [Christmas bonus]
From The Sunday Times, 20th December 2009 [link]

Pat’s company employs six staff: they joined on six consecutive Christmases and have stayed ever since. This Christmas, he is giving them each a bonus in the form of vouchers, each worth a whole number of pounds, with one red voucher for each year of service for a man and one green voucher (worth £1 more) for each year of service for a woman. The value of all the year’s vouchers is £500. Ms Jones joined more than two years after Mr Smith, but their bonuses have the same total value.

What is that common value?

This puzzle was originally published with no title.

[teaser2465]
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Ruud and Jim Randell are discussing. Toggle Comments
- Jim Randell 11:20 am on 26 December 2025 Permalink | Reply
  This Python program runs in 70ms. (Internal runtime is 1.7ms).
```
from enigma import (irange, inf, subsets, div, printf)

# choose male (= 0) and female (= 1) values (longest serving -> shortest serving)
for vs in subsets([0, 1], size=6, select='M'):
  # find possible (i, j) values for Mr S and Ms J
  ijs = list((i, j) for (i, j) in subsets(irange(6), size=2) if j >= i + 3 and vs[i] == 0 and vs[j] == 1)
  if not ijs: continue

  # consider increasing values for the male voucher
  for x in irange(1, inf):
    # calculate the years of service for the longest serving employee (>= 6)
    n = 500 + sum(i * (x + v) for (i, v) in enumerate(vs))
    d = 6 * x + sum(vs)
    if n < 6 * d: break
    y = div(n, d)
    if y is None: continue

    # calculate actual gift amounts
    gs = list((y - i) * (x + v) for (i, v) in enumerate(vs))

    # find shared values for Mr S and Ms J
    for (i, j) in ijs:
      if gs[i] == gs[j]:
        # output solution
        printf("vs={vs}, y={y}, x={x}, gs={gs}, i={i}, j={j}")
```
  Solution: The common value is £80.
  
  The red vouchers (male) are worth £4 each. The green vouchers (female) are worth £5 each.
  
  The amounts given to the employees are:
  
  21 years, female → £105
  20 years, male → £80 (Mr Smith)
  19 years, female → £95
  18 years, male → £72
  17 years, male → £68
  16 years, female → £80 (Ms Jones)
  total = £500
  
  LikeLike
- Ruud 1:17 pm on 26 December 2025 Permalink | Reply
  Extreme brute force:
```
import peek
import itertools
import istr

for n in range(1, 30):
    for bonus in range(100):
        for is_womans in itertools.product((False, True), repeat=6):
            if sum(k * (bonus + is_woman) for k, is_woman in enumerate(is_womans, n)) == 500:
                for jones, smith in itertools.product(range(6), repeat=2):
                    if not is_womans[jones] and is_womans[smith] and jones > smith + 2 and (jones + n) * bonus == (smith + n) * (bonus + 1):
                        peek((jones + n) * bonus, (smith + n) * (bonus + 1))
```
  LikeLike

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