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Jim Randell 8:10 am on 29 July 2025 Permalink | Reply
Tags: by: C Higgins ( 38 ), by: H Bradley ( 38 )
Teaser 2502: [Swimming lengths]
From The Sunday Times, 5th September 2010 [link] [link]

John and Pat were swimming at the pool. John started at the deep end at the same time as Pat started at the shallow end, each swimming lengths at their own steady speed. They passed each other on their first lengths and then passed each other again on their second lengths. It turned out that the length of the pool was a multiple of the distance between those first two passing points. John got out of the water after completing 48 lengths. At that moment Pat had also completed a whole number of lengths.

How many?

This puzzle was originally published with no title.

[teaser2502]
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Frits and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:11 am on 29 July 2025 Permalink | Reply
  (See also: Enigma 799).
  
  Suppose J takes 1 unit of time to swim a length, and P takes p units of time to swim a length.
  
  And after J has completed 48 lengths, P has completed n = 48/p lengths, which is a whole number.
  
  For them to pass on their first and then second lengths we have:
  
  p > 1/2
  p < 2
  
  Hence:
  
  24 < n < 96
  
  For a given n value, we can calculate the crossing positions using a time/distance graph:
  
  And then look for values where the distance between the positions divides exactly into the pool length.
  
  This Python program runs in 74ms. (Internal runtime is 4.6ms).
```
from enigma import (irange, rdiv, line_intersect, catch, printf)

# positions for J after lengths 0, 1, 2; P after length 0
(J0, J1, J2, P0) = ((0, 1), (1, 0), (2, 1), (0, 0))

# consider number of lengths completed by P at t=48
for n in irange(25, 95):
  p = rdiv(48, n)
  # positions for P after lengths 1, 2
  (P1, P2) = ((p, 1), (2*p, 0))

  # calculate crossing position during first two lengths
  i1 = catch(line_intersect, J0, J1, P0, P1, internal=1, div=rdiv)
  i2 = catch(line_intersect, J1, J2, P1, P2, internal=1, div=rdiv)
  if i1 is None or i2 is None: continue
  ((t1, d1), (t2, d2)) = (i1.pt, i2.pt)

  # pool length is an exact multiple of the distance between the positions
  if d1 == d2: continue
  k = rdiv(1, abs(d1 - d2))
  if k % 1 != 0: continue

  # output solution
  printf("n={n} [p={p} k={k}]")
```
  Solution: Pat had completed 80 lengths.
  
  So Pat completes a length in 48/80 = 3/5 the time it takes John. And the distance between the crossing positions is exactly 1/2 the length of the pool.
  
  We have:
  
  (t₁, d₁) = (3/8, 5/8)
  (t₂, d₂) = (9/8, 1/8)
  
  LikeLike
- Frits 2:27 pm on 30 July 2025 Permalink | Reply
```
# L = length of the pool

# 1st passing
# 0     j1*L                  p1*L            L   j1 + p1 = 1
# |---------------+---------------------------|

# speed ratio r1 = p1 / j1

# 2nd passing
# 0                                           L  
# |-------------------------------------------|
#                  j2*L                  p2*L   j2 + p2 = 1
# |------------------------------------+------|

# speed ratio Pat / John
# r = n / 48
# j1 = 1 / (r + 1)
# 2nd passing: distance that P swam = r * (distance that J swam)
# L + j2.L = r.(L + p2.L) or 1 + j2 = r.(2 - j2) or j2 = (2.r - 1) / (r + 1)
# j2 = (2 * r - 1) / (r + 1)

#  24 < n < 96
# n must be a multiple of 4 as denominator in formula below is a multiple of 4
for n in range(28, 96, 4):
  # distance between those first two passing points > 0
  if n == 48: continue
  # L = multiple of the distance passing points = m * (abs(j1 - j2) * L)
  # if (r + 1) % (2 * r - 2): continue  
  if (n + 48) % (2 * n - 96): continue  
  print("answer:", n)
```
  LikeLike
  - Frits 3:05 pm on 30 July 2025 Permalink | Reply
    
    “n” can even be proven to be a multiple of 16.
    
    LikeLike
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Jim Randell 1:56 am on 27 July 2025 Permalink | Reply
Tags: by: Stephen Hogg ( 64 )

Teaser 3279: It’s a doddle!

From The Sunday Times, 27th July 2025 [link] [link]

The Holmes’s new maths tutor, Dr Moriarty, wrote several two-digit by three-digit whole number multiplications on the board (with no answer repeated). Young Sherlock whispered to his brother, “That first one’s a doddle and the three-digit number is prime.”

Mycroft replied, “Yes, it is a doddle, but the answer is also a DODDLE”. Sherlock looked puzzled. Mycroft explained that a a DODDLE is a number with “Descending-Order Digits left to right — all different — DivisibLe Exactly by each digit”.

Mycroft continued, “All of the above applies to the second multiplication as well”. Sherlock noticed that just one other answer, to the final problem, was a DODDLE, with the three-digit value being odd, but not prime.

What is the answer to the final problem?

[teaser3279]

Jim Randell 1:57 am on 27 July 2025 Permalink | Reply

This Python program collects numbers that are DODDLEs resulting from a multiplication with a 3-digit prime (in n1s) and those that result from a multiplication with a 3-digit odd non-prime (in n2s).

We then look for a pair of results from n1s, and a different result from n2s to give a viable solution.

It runs in 138ms. (Internal runtime is 59ms).

from enigma import (irange, primes, cproduct, nsplit, is_decreasing, subsets, printf)

# check if n is DODDLE
def is_doddle(n):
  ds = nsplit(n)
  if 0 in ds: return False
  return all(n % d == 0 for d in ds) and is_decreasing(ds, strict=1)

# collect results where the 3-digit number is prime or non-prime
(n1s, n2s) = (set(), set())
for (a, b) in cproduct([irange(10, 99), irange(101, 999, step=2)]):
  n = a * b
  if not is_doddle(n): continue
  if b in primes:
    printf("[1] {a} * {b} = {n}")
    n1s.add(n)
  else:
    printf("[2] {a} * {b} = {n}")
    n2s.add(n)

# consider possible results for the first 2 problems
for r12 in subsets(n1s, size=2):
  # find a final problem with a different result
  for rf in n2s.difference(r12):
    printf("r12={r12} -> rf={rf}", r12=sorted(r12))

Solution: The answer to the final problem is: 6432.

There are only 2 possible multiplications where the 3-digit number is prime that give a DODDLE.

72 × 131 = 9432
72 × 137 = 9864

And these give the first two problems (in some order).

There are only 3 possible multiplications where the 3-digit number is odd but not prime that give a DODDLE:

24 × 393 = 9432
24 × 411 = 9864
32 × 201 = 6432

But as no results are repeated the final multiplication on this list gives the required answer to the puzzle.

In fact the only multi-digit DODDLEs are:

3-digit: 432, 864
4-digit: 6432, 9432, 9864

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Jim Randell 3:44 pm on 27 July 2025 Permalink | Reply

A faster approach is to generate possible DODDLEs (as there are not very many), and then factorise them into 2- and 3-digit numbers.

The following Python program runs in 61ms. (Internal runtime is 535µs).

from enigma import (irange, subsets, nconcat, chain, divisors_pairs, is_prime, printf)

# generate k-digit doddles
def doddle(k):
  for ds in subsets(irange(9, 1, step=-1), size=k):
    n = nconcat(ds)
    if all(n % d == 0 for d in ds):
      yield n

# record results using 3-digit numbers that are prime and non-prime
(n1s, n2s) = (set(), set())
# consider 4- and 5-digit doddles
for n in chain(doddle(4), doddle(5)):
  # look for 2-digit, 3-digit divisor pairs
  for (a, b) in divisors_pairs(n):
    # a is a 2-digit number
    # b is a 3-digit odd number
    if a > 99 or b < 100: break
    if a < 10 or b > 999 or b % 2 == 0: continue
    if is_prime(b):
      printf("[1] {a} * {b} = {n}")
      n1s.add(n)
    else:
      printf("[2] {a} * {b} = {n}")
      n2s.add(n)

# consider possible results for the first 2 problems
for r12 in subsets(n1s, size=2):
  # find a final problem with a different result
  for rf in n2s.difference(r12):
    printf("r12={r12} -> rf={rf}", r12=sorted(r12))

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GeoffR 12:31 pm on 28 July 2025 Permalink | Reply

The teaser description states that there are three Doddle type numbers, which my posting is based upon.

from enigma import is_prime, nsplit

cnt = 0
DL = []

def is_doddle(num):
  # 4-digit Doddle numbers
  if 1000 < num < 10000:
    a, b, c, d = nsplit(num)
    # cannot divide by zero
    if 0 not in (b, c, d):
      if len(str(num)) == len(set(str(num))):
        if a > b > c > d and all(num % n == 0 \
        for n in (a, b, c, d)):
          return True
        
    # 5-digit Doddle numbers
    if 10000 < num < 99999:
      a, b, c, d, e = nsplit(num)
      if 0 not in (b, c, d, e):
        if len(str(num)) == len(set(str(num))) :
          if a > b > c > d > e and all(num % n == 0 \
          for n in (a, b, c, d, e)):
            return True
                                 
    return False

# find the first two DODDLE numbers
# ... where the three digit number(b) is prime
for a in range(10, 99):
  for b in range(101, 999):
    if not is_prime(b): continue
    if not is_doddle(a * b): continue
    cnt += 1
    if a * b not in DL:
      DL.append(a * b)
    if cnt == 2: 
      # we are specifically told there are
      # ... only 2 Doddle numbers with b as prime
      print(f"First two Doddle numbers are {DL[0]} and {DL[1]}")
           
# find the unique third DODDLE number
# this three digit number (b) is not prime, but is of odd parity
for a in range(10, 99):
  for b in range(101, 999):
    if not is_prime(b) and b % 2 == 1:
      if is_doddle(a * b) and a * b not in DL:
        print("The answer to the final problem is", a * b)

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Jim Randell 9:43 am on 24 July 2025 Permalink | Reply
Tags: by: C Higgins ( 38 ), by: H Bradley ( 38 )
Teaser 2477: [Shamrock in a box]
From The Sunday Times, 14th March 2010 [link]

At the St Patrick’s Day Feis, Pat exhibits his painting “Shamrock in a Box”. This shows a square containing two touching circles of the same diameter (less than a metre). One circle also touches two adjacent sides of the square; the other touches the remaining two sides. For the middle leaf of the shamrock, Pat drew a third circle in the square: it touches each of the other circles, is the largest that will fit in the available space, and its radius is almost exactly a whole number of centimetres, that number being the same as the radius of the larger circles, but with the two digits in reverse order.

What is the radius of the larger circles?

This puzzle was originally published with no title.

[teaser2477]
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Frits and Jim Randell are discussing. Toggle Comments
- Jim Randell 9:43 am on 24 July 2025 Permalink | Reply
  The layout looks like this:
  
  Suppose the large circles have a radius 1.
  
  The semi-diagonal of the square is: 1 + √2.
  
  And if the smaller circle has a radius f (where 0 < f < 1), then we can also write the length of the semi-diagonal as: f√2 + h.
  
  Where: f + 1 = hypot(h, 1).
  
  And so we have two expressions for h:
  
  h = 1 + √2 − f√2
  h^2 = (f + 1)^2 − 1
  
  We can solve these equations to find the value of f, and then consider the actual radius of the large circle, R (a 2-digit integer, less than 50). And then calculate the radius of the smaller circle r = f . R.
  
  And we are look for cases where r is very close to the reverse of R.
  
  This Python program runs in 61ms. (Internal runtime is 193µs).
```
from enigma import (Polynomial, sqrt, sq, irange, nrev, printf)

r2 = sqrt(2)  # "root 2"

# we have 2 expressions for h^2, which we can express as polynomials
p1 = sq(Polynomial([1 + r2, -r2]))  # h^2 = (1 + sqrt(2) - f.sqrt(2))^2
p2 = sq(Polynomial([1, 1])) - 1  # h^2 = (f + 1)^2 - 1

# find real roots of p(f) = p1 - p2
for f in (p1 - p2).roots(domain='F'):
  if not (0 < f < 1): continue
  printf("[f = {f}]")

  # consider 2-digit numbers for R
  for R in irange(10, 49):
    # the reverse should be a smaller 2-digit number
    rR = nrev(R)
    if rR < 10 or rR >= R: continue

    # check size of smaller circle is close to rev(R)
    r = f * R
    if abs(r - rR) < 0.05:
      printf("R={R} -> r={r}")
```
  Solution: The larger circles have a radius of 32 cm.
  
  And the smaller circles have a radius of 22.998 cm, which is very close to 23 cm.
  
  f is a root of the following quadratic equation:
  
  f^2 − (6 + 2√2)f + (3 + 2√2) = 0
  
  And this allows us to give a formula for f:
  
  f = 3 + √2 − 2√(2 + √2)
  f = 0.718695432327948…
  
  Hence:
  
  R = 32
  ⇒
  r = f . R = 22.99825383449434…
  
  LikeLike
- Frits 12:49 pm on 24 July 2025 Permalink | Reply
  Using Jim’s formulas.
```
# h = 1 + sqrt(2) - f.sqrt(2)
# h^2 = (f + 1)^2 - 1 
h1 = lambda f: 1 + (1 - f) * 2**.5 
h2 = lambda f: ((f + 1)**2 - 1)**.5

sols = set()
for a in range(1, 5):
  for b in range(1, a):
    R, r = 10 * a + b, 10 * b + a
    f = r / R
    # store difference between 2 calculations for "h"
    sols.add((abs(h1(f) - h2(f)), R))

if (mn := min(sols))[0] < 0.01:
  print("answer:", mn[-1])
```
  LikeLike
Jim Randell 8:50 am on 22 July 2025 Permalink | Reply
Tags: by: Keith Austin ( 6 )
Brainteaser 1263: I do apologise
From The Sunday Times, 16th November 1986 [link]

The artist Pussicato has a new painting which is a 6×6 array of squares and each square is red or blue or green.

(A) Label the horizontal rows of the painting U, V, W, X, Y, Z from top to bottom and the vertical columns 1, 2, 3, 4, 5, 6 from left to right. Then U1, U3, V5 are red; W2, Z2 are blue; and U2, V4, X2, X3, Y3, Y4 are green. Also V1 and W1 are the same colour, as are Y1 and Z1, and also X6 and Y6. Each row and each column contains two red squares, two blue and two green.

(B) I do apologise. The information in (A) gives you a painting which is wrong in every square.

(C) Use the information in (B), together with the following facts. Any two squares which are next to each other, horizontally or vertically, have different colours. There are more green squares than blue squares.

(D) Look, I am so sorry. The information in (C) gives you a painting which is wrong in every square.

(E) To be serious, the information in (B) and (D) is correct.

Give (in order) the colours of the squares in the top row of Pussicato’s painting.

[teaser1263]
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- Jim Randell 8:52 am on 22 July 2025 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression() ]] solver from the enigma.py library.
  
  The program first uses the information in (A) to produce a grid (= “layout 1”). It then uses layout 1, and the information in (B) and (C) to produce another layout (= “layout 2”). It then uses layout 1, layout 2 and the information in (B) and (D) to produce a final layout, that provides the answer to the puzzle.
  
  Unfortunately, the program below fails with the standard CPython implementation (due to limitations in the CPython interpreter), but works fine with PyPy.
  
  (The code on codepad includes a workaround for CPython [*]).
  
  The following Python program runs in 132ms (under PyPy). (Internal runtime is 37.9ms).
  
  Run: [ @codepad ]
```
from enigma import (enigma, SubstitutedExpression, first, sprintf as f, join, printf)

# symbols for the 36 cells
syms = first(enigma.str_upper + enigma.str_lower, 36)

# map cells to symbols
(rs, cs) = ("UVWXYZ", "123456")
(v, vs) = (dict(), iter(syms))
for r in rs:
  for c in cs:
    v[r + c] = next(vs)

# rows and cols
rows = list(list(r + c for c in cs) for r in rs)
cols = list(list(r + c for r in rs) for c in cs)

# colours
(red, green, blue) = (0, 1, 2)


# find layout 1
def solve1():

  # construct expressions
  exprs = list()

  # U1, U3, V5 are red
  exprs.extend(f("{{{x}}} = {red}", x=v[x]) for x in ["U1", "U3", "V5"])
  # W2, Z2 are blue
  exprs.extend(f("{{{x}}} = {blue}", x=v[x]) for x in ["W2", "Z2"])
  # U2, V4, X2, X3, Y3, Y4 are green
  exprs.extend(f("{{{x}}} = {green}", x=v[x]) for x in ["U2", "V4", "X2", "X3", "Y3", "Y4"])
  # V1 = W1; Y1 = Z1; X6 = Y6
  exprs.extend(f("{{{x}}} = {{{y}}}", x=v[x], y=v[y]) for (x, y) in [("V1", "W1"), ("Y1", "Z1"), ("X6", "Y6")])

  # each row/column contains 2 red, 2 green, 2 blue
  check = lambda *vs: all(vs.count(x) == 2 for x in (red, green, blue))
  for r in rows:
    exprs.append(join((f("{{{x}}}", x=v[x]) for x in r), sep=", ", enc=["check(", ")"]))
  for c in cols:
    exprs.append(join((f("{{{x}}}", x=v[x]) for x in c), sep=", ", enc=["check(", ")"]))

  # make an alphametic puzzle to solve the grid
  p = SubstitutedExpression(exprs, base=3, distinct=0, env=dict(check=check))
  for s in p.solve(verbose=0):
    yield list(s[k] for k in syms)


# find layout 2
def solve2(g1):

  # no colour from layout 1 is correct
  d2i={ red: [], green: [], blue: [] }
  for (k, v) in zip(syms, g1):
    d2i[v].append(k)

  # no adjacent squares are the same colour
  distinct=list()
  for (i, k) in enumerate(syms):
    (r, c) = divmod(i, 6)
    if c < 5: distinct.append(k + syms[i + 1])
    if r < 5: distinct.append(k + syms[i + 6])

  # there are more green than blue squares
  check = lambda *vs: vs.count(green) > vs.count(blue)
  expr = join(syms, sep="}, {", enc=["check({", "})"])

  # make an alphametic puzzle to solve the grid
  p = SubstitutedExpression([expr], base=3, distinct=distinct, d2i=d2i, env=dict(check=check))
  for s in p.solve(verbose=0):
    yield list(s[k] for k in syms)


# solve for layout 1
for g1 in solve1():
  printf("g1={g1}")
  # solve for layout 2
  for g2 in solve2(g1):
    printf("g2={g2}")
    # generate the grid where each cell is different from g1 and g2
    g3 = list(3 - (c1 + c2) for (c1, c2) in zip(g1, g2))
    printf("g3={g3}")

    # output the grid
    d = { red: 'R', green: 'G', blue: 'B' }
    printf()
    for (i, v) in enumerate(g3):
      if i % 6 == 0: printf("[ \\")
      printf("{v} \\", v=d[v])
      if i % 6 == 5: printf("]")
```
  Solution: The squares in the top row are: blue, blue, green, red, blue, red.
  
  The requirements in (A) give a single layout (shown below on the left).
  
  The requirements (B) and (C) also give a single layout (shown below in the centre).
  
  And finally, the requirements (D) give a single layout (where each square is different colour to the corresponding positions the first two grids). (This is shown below on the right).
  
  [*] When producing layout 2 there is only a single expression, involving all 36 variables, so the “denesting” code cannot split the expressions into blocks that use fewer expressions. To fix this we add in the following expression, which mentions half the symbols (and is always true), and this allows the denesting code to do its job:
```
true(A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R)
```
  LikeLike
Jim Randell 7:20 am on 20 July 2025 Permalink | Reply
Tags: by: Mark Valentine ( 17 )
Teaser 3278: Page weight
From The Sunday Times, 20th July 2025 [link] [link]

An A5 booklet is produced by printing the pages onto sheets of A4 (both sides), then binding and folding along the middle. The booklet contains two chapters. The longer first chapter runs to 34 pages.

The pages in each chapter are numbered sequentially from 1 upwards. The front cover is page 1 of chapter one. Page 1 of chapter two follows directly after page 34 of chapter one. Any blank pages at the end of the booklet are not numbered.

Without changing their order, the sheets can be split into two piles, where the sums of the page numbers in each pile are equal.

How many pages does the second chapter have?

[teaser3278]
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- Jim Randell 7:52 am on 20 July 2025 Permalink | Reply
  Although not explicitly stated, I am assuming the booklet consists of the minimum number of A4 sheets required to fit all pages of both chapters.
  
  Here is a constructive solution, that constructs the pairs of pages and organises them into A4 sheets, and then looks for a scenario where the sheets can be split as required. (This allows easy output of the page numbers on each A4 sheet, if desired).
  
  The following Python program runs in 60ms. (Internal runtime is 539µs).
  
  Run: [ @codepad ]
```
from enigma import (irange, divc, div, chunk, chain, printf)

# pages for chapter 1
n1 = 34

# consider the number of pages in chapter 2 (< n1)
for n2 in irange(1, n1 - 1):

  # pair up the pages
  pairs = list(chunk(chain(irange(1, n1), irange(1, n2)), 2))
  # total sum of all page numbers divided by 2
  s = div(sum(map(sum, pairs)), 2)
  if s is None: continue

  # total number sheets of A4 required
  t = divc(n1 + n2, 4)

  # collect the page numbers for each sheet
  sheets = [()] * t
  for (i, p) in zip(chain(irange(t), irange(t, step=-1)), pairs):
    sheets[i] += p

  # can we find some sheets that sum to s ?
  x = 0
  for k in irange(t):
    x += sum(sheets[k])
    if x == s:
      # output solution
      printf("chapter 2 has {n2} pages [{t} sheets in total; first {k} sheets sum to {x}]", k=k + 1)
    if x >= s:
      break
```
  Solution: Chapter 2 has 25 pages.
  
  There are 15 A4 sheets, and the page numbers are as follows:
  
  1: (1, 2) (25, -)
  2: (3, 4) (23, 24)
  3: (5, 6) (21, 22)
  4: (7, 8) (19, 20)
  5: (9, 10) (17, 18)
  6: (11, 12) (15, 16)
  7: (13, 14) (13, 14)
  8: (15, 16) (11, 12)
  9: (17, 18) (9, 10)
  
  10: (19, 20) (7, 8)
  11: (21, 22) (5, 6)
  12: (23, 24) (3, 4)
  13: (25, 26) (1, 2)
  14: (27, 28) (33, 34)
  15: (29, 30) (31, 32)
  
  The sum of the page numbers of sheets 1-9 and sheets 10-15 both come to 460.
  
  If we can construct the booklet with more than the minimum required number of A4 sheets, then it is possible to construct a further solution:
  
  If chapter 2 has 14 pages and the booklet was printed from 17 sheets, then the sum of the page numbers on sheets 1-12 and sheets 13-17 would both come to 350 (and there would be 20 blank pages at the end of the booklet).
  
  This justifies the initial assumption.
  
  The program can be used to investigate similar puzzles where the number of pages in the first chapter is other than 34.
  
  An interesting case is when chapter 1 has 19 pages. There are then 2 solutions:
  
  Chapter 2 has 4 pages.
  The booklet consists of 6 sheets.
  The first 4 (and last 2) sheets have page numbers which sum to 100.
  
  Chapter 2 has 16 pages.
  The booklet consists of 9 sheets.
  The first 5 (and last 4) sheets have page numbers which sum to 163.
  
  LikeLike
  - Jim Randell 9:15 am on 20 July 2025 Permalink | Reply
    Here is a faster version of my program that works with a list of the sum of pairs of pages on a leaf.
    
    It has an internal runtime of 124µs.
```
from enigma import (irange, chunk, div, divc, seq_get, printf)

# pages for chapter 1
n1 = 34

# page pair sums from chapter 1
pairs = list(map(sum, chunk(irange(1, n1), 2)))
t = sum(pairs)
get = seq_get(pairs, default=0)

# consider the number of pages in chapter 2 (< n1)
for (i, n2) in enumerate(irange(1, n1 - 1), start=n1):
  # make the pair sum of the final page
  if i % 2:
    pairs[-1] += n2
  else:
    pairs.append(n2)
  t += n2
  s = div(t, 2)
  if s is None: continue

  # now look for sheets that sum to s, by pairing up page sums
  (x, k) = (0, len(pairs))
  if k % 2 == 0: k -= 1
  for i in irange(divc(k, 2)):
    x += get(i) + get(k - i)
    if x == s:
      # output solution
      printf("chapter 2 has {n2} pages [first {i} sheets sum to {x}]", i=i + 1)
    if x >= s:
      break
```
    LikeLike
- Frits 2:45 pm on 20 July 2025 Permalink | Reply
  One loop.
  
  The sum of 4 page numbers per sheet can only have 2, 3 or 4 different values (per number of pages in chapter 2).
  
  It took a while before I figured out how the booklet was constructed.
```
# one folded A4 sheet results in 4 pages

t = 17 * 35        # sum(1...34)
# number of pages in chapter 2
for n in range(1, 34):
  # sum of all pages
  t += n
  # both <isum> as <osum> are even so <h> must also be even
  if t % 4: continue
  h = t // 2   
  
  n_sheets = (37 + n) // 4
  n_inner_sheets = 17 - n_sheets
  
  # calculate starting page number of most inner sheet
  st = 2 * ((n + 1) // 4) + 17
  
  # calculate sum of 4 outer sheet page numbers (except most outer sheet)
  osum = (2 * (n + n % 2) - 1) + 3 if n > 1 else -1
  # calculate sum of most inner sheet page numbers
  isum = 4 * st + 6 if n_sheets < 17 else osum
  
  # can we make <h> using <k> sheets (beginning from the center)?
 
  # h = a.isum + b.osum or  b = (h - a.isum) / osum
  # use maximum <n_inner_sheets> inner sheets and possible <b> outer sheets
  d, r = divmod(h, isum)
  if not r and d <= n_inner_sheets:
    print("answer:", n)
  else:  
    # h - n_inner_sheets * inner = b * osum
    b, r = divmod(h - n_inner_sheets * isum, osum)
    if not r and 0 < b <= n_sheets - n_inner_sheets:
      if osum > 0: 
        print("answer:", n)
```
  LikeLike
Jim Randell 8:19 am on 18 July 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 2518: [Empty boxes]
From The Sunday Times, 26th December 2010 [link] [link]

For an appropriate Boxing Day exercise, write a digit in each of the empty boxes so that (counting all the digits from here to the final request) the following statements are true:

Number of occurrences of 0 = [ _ ]
Number of occurrences of 1 = [ _ ]
Total occurrences of 2s and 3s = [ _ ]
Total occurrences of 4s and 5s = [ _ ]
Total occurrences of 6s and 7s = [ _ ]
Total occurrences of 8s and 9s = [ _ ]
A digit you have written in another box = [ _ ]
The average of all your written digits = [ _ ]

What, in order, are the digits in your boxes?

This puzzle was originally published with no title.

[teaser2518]
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Frits and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:20 am on 18 July 2025 Permalink | Reply
  There are 8 boxes to be filled out with a digit (from 0 to 9), and each of the digits 0 to 9 already occurs once before the boxes are filled out.
  
  The first six boxes count the 10 given digits, plus those in the 8 boxes, so must sum to 18.
  
  The first 2 boxes must contain 1 or more, and the next 4 boxes must contain 2 or more.
  
  And the box with the mean must also be 2 or more.
  
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  The following run file executes in 95ms. (Internal runtime of the generated code is 16.8ms).
  
  Run: [ @codepad ]
```
#! python3 -m enigma -rr

SubstitutedExpression

# let the 8 empty boxes be: A B C D E F G H
--distinct=""
--digits="1-9"
--invalid="1,CDEFH"
--macro="@boxes = [A, B, C, D, E, F, G, H]"

# first 2 boxes
"@boxes.count(0) + 1 = A"
"@boxes.count(1) + 1 = B"

# next 4 boxes
"@boxes.count(2) + @boxes.count(3) + 2 = C"
"@boxes.count(4) + @boxes.count(5) + 2 = D"
"@boxes.count(6) + @boxes.count(7) + 2 = E"
"@boxes.count(8) + @boxes.count(9) + 2 = F"

# final 2 boxes
"G in [A, B, C, D, E, F, H]"
"iavg(@boxes) = H"

--template="@boxes"
--solution=""

# [optional] additional constraints to improve run time
# the sum of the first 6 boxes must be 18
"A + B + C + D + E + F = 18"
# we can't write 0 in any of the boxes, so there is only 1 zero
--assign="A,1"
```
  Solution: The digits are: 1, 2, 8, 2, 2, 3, 3, 3.
  
  The statements then become:
  
  Number of occurrences of 0 = [ 1 ]
  Number of occurrences of 1 = [ 2 ]
  Total occurrences of 2s and 3s = [ 8 ]
  Total occurrences of 4s and 5s = [ 2 ]
  Total occurrences of 6s and 7s = [ 2 ]
  Total occurrences of 8s and 9s = [ 3 ]
  A digit you have written in another box = [ 3 ]
  The average of all your written digits = [ 3 ]
  
  LikeLike
  - Frits 2:04 pm on 18 July 2025 Permalink | Reply
    
    @Jim, you can also add B in line 8 (ao because of line 32).
    
    LikeLike

Jim Randell 8:21 am on 15 July 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )

Teaser 2489: [Digit circle]

From The Sunday Times, 6th June 2010 [link] [link]

Your task today is to put more than two digits evenly spaced around a circle so that:

(a) Any three adjacent digits clockwise form a prime number.

(b) Any three adjacent digits anti-clockwise form a prime number.

(c) If a digit occurs in the circle, then it occurs a prime number of times.

(d) The total of all the digits around the circle is a prime.

What is the total of all the digits?

This puzzle was originally published with no title.

[teaser2489]

Jim Randell 8:21 am on 15 July 2025 Permalink | Reply

Each digit is the last digit of a 3-digit prime, so the only possible digits are 1, 3, 9, 7.

This Python program looks for solutions using increasing numbers of digits, and stops after the smallest possible sequences are found.

It runs in 60ms. (Internal runtime is 841µs).

Run: [ @codepad ]

from enigma import (irange, inf, primes, multiset, subsets, tuples, rev, nconcat, wrap, uniq, printf)

# up to 3-digit primes
primes.expand(999)

# possible digits
digits = [1, 3, 7, 9]

# generate numbers from <k> adjacent digits of <ns> (in both directions)
@wrap(uniq)  # skip duplicate numbers
def adj(ns, k=3):
  for ds in tuples(ns, k, circular=1):
    yield nconcat(ds)
    yield nconcat(rev(ds))

# look for solutions using <n> digits
def solve(n):
  # choose the n digits
  for ds in subsets(digits, size=n, select='R'):
    # the sum of the digits is prime
    if not (sum(ds) in primes): continue
    # each digit occurs a prime number of times
    m = multiset.from_seq(ds)
    if not all(v in primes for v in m.values()): continue

    # start with the smallest number
    n0 = m.min()
    # and choose an arrangement of the rest
    for ns in subsets(m.difference([n0]), size=len, select='mP', fn=list):
      if rev(ns) < ns: continue  # skip duplicate solutions
      ns.insert(0, n0)
      # check each 3 adjacent digits form a prime (in both directions)
      if not all(x in primes for x in adj(ns)): continue
      yield ns

# find the smallest possible sequences
for n in irange(3, inf):
  s = 0
  for ns in solve(n):
    printf("[n={n}] {ns} -> {t}", t=sum(ns))
    s += 1
  if s: break

Solution: The total of the digits is 29.

The digits in the circle are: (1, 9, 1, 9, 9). Consisting of 2 occurrences of the digit 1, and 3 occurrences of the digit 9.

The 3-digit numbers formed are: 191, 199, 919, 991. Which are all prime.

LikeLike

Ruud 8:03 am on 16 July 2025 Permalink | Reply

import istr
import collections
import itertools

primes = [i for i in istr.range(2, 1000) if i.is_prime()]
candidates = {str(prime) for prime in primes if len(prime) == 3 and all(c in "1379" for c in prime)}
for n in itertools.count(2):
    for x in itertools.product("1379", repeat=int(n)):
        s = "".join(x)
        if all(count in primes for count in collections.Counter(s).values()):
            s2 = str(s + s[0:2])
            if (
                all(s2[i : i + 3] in candidates for i in range(len(s)))
                and all(s2[i : i + 3][::-1] in candidates for i in range(len(s)))
                and (sumx:=sum(map(int, x))) in primes
            ):
                print(s,sumx)
                assert False

LikeLike

Jim Randell 6:24 am on 13 July 2025 Permalink | Reply
Tags: by: John Owen ( 33 ), flawed ( 55 )
Teaser 3277: Croquet equipment
From The Sunday Times, 13th July 2025 [link] [link]

Our non-standard croquet lawn has six hoops, at positions A to F, and a central peg at P. The equipment storage box is in the southwest corner at S, and the lawn is symmetrical in both the east-west and north-south directions. The diagram is not to scale, but D is south of the projected line from S to E.

Also AB is half the width of the lawn. When setting out the equipment, I walk along the route SEFDPCBAS. All of the distances on my route between positions are whole numbers of feet (less than 60), and both D and E are whole numbers of feet north of the south boundary.

What is the total distance of my route?

[teaser3277]
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Hugo, Jim Randell, and Frits are discussing. Toggle Comments
- Jim Randell 8:46 am on 13 July 2025 Permalink | Reply
  I get two solutions, so here is a straightforward solution to make sure I haven’t missed any thing.
  
  But if we disallow distances of 60ft or more between any two positions on the walk (not just adjacent positions), then we can eliminate the larger of these solutions.
  
  This Python program runs in 65ms. (Internal runtime is 5ms).
```
from enigma import (irange, ihypot, printf)

M = 59  # max length of a path segment

# let AB = EF = 2x
for x in irange(1, M // 2):
  # e = distance E north from southern edge
  for e in irange(1, M):
    SE = ihypot(x, e)
    if SE is None or SE > M: continue

    # d = distance D is north of EF
    for d in irange(1, e - 1):
      FD = ihypot(x, d)
      if FD is None or FD > M: continue

      # p = distance P north of D
      for p in irange(1, M):
        SA = ihypot(x, e + d + p + p + d)
        if SA is None or SA > M: continue

        # total distance walked
        T = SE + SA + 2*FD + 4*x + 2*p
        printf("T={T} [SE={SE} FD={FD} SA={SA}; x={x} e={e} d={d} p={p}]")
```
  Solution: There are two viable solutions to the puzzle: 142 ft, and 220 ft.
  
  Apparently the size of a croquet lawn is not fixed, but it should be a rectangle with sides in the ratio of 1.25.
  
  The first of the solutions involves a playing area of 48 ft × 44 ft (ratio = 1.09).
  
  And all positions on the route are within 60 ft of S (as indicated by the dotted line).
  
  The second of the solutions involves a playing area of 96 ft × 42 ft (ratio = 2.29).
  
  And B and F are further than 60 ft from S.
  
  The published solution is: “142 feet”.
  
  But a correction was issued with Teaser 3280:
  
  There were in fact two possible answers: 142 and 220 feet.
  Either was accepted as correct.
  
  .
  
  LikeLike
  - Frits 11:38 am on 13 July 2025 Permalink | Reply
    
    @Jim, shouldn’t we have: e + d + p = 2.x ?
    
    LikeLike
    - Jim Randell 11:56 am on 13 July 2025 Permalink | Reply
      
      @Frits: Why?
      
      LikeLike
      - Frits 12:18 pm on 13 July 2025 Permalink | Reply
        
        I assumed that the lawn was square (but that isn’t specified).
        
        LikeLike
  - Jim Randell 4:19 pm on 13 July 2025 Permalink | Reply
    Alternatively, using the [[ pythagorean_triples() ]] function from the enigma.py library.
    
    The following Python program has an internal runtime of 122µs.
```
from enigma import (defaultdict, pythagorean_triples, subsets, div, printf)

M = 59  # maximum length of a path segment

# collect (<other>, <hypot>) sides for pythagorean triples
t = defaultdict(list)
for (x, y, z) in pythagorean_triples(M):
  t[x].append((y, z))
  t[y].append((x, z))

# consider possible x values
for x in sorted(t.keys()):
  if 2 * x > M: break
  # FD, SE, SA all have base x, and increasing vertical sides
  ts = sorted(t[x])
  for ((d, FD), (e, SE), (a, SA)) in subsets(ts, size=3):
    p = div(a - e - 2*d, 2)
    if p is None or p < 1: continue

    # total distance walked
    T = SE + SA + 2*FD + 4*x + 2*p
    printf("T={T} [SE={SE} FD={FD} SA={SA}; x={x} e={e} d={d} p={p}]")
```
    LikeLike
- Hugo 10:34 am on 21 July 2025 Permalink | Reply
  
  If hoop P has coordinates (0, 0) then it seems
  A is at (12, -13), B is at (12, 13)
  C is at (0, 8), D is at (0, -8)
  E is at (-12, -13), F is at (-12, 13)
  S is at (-24, -22).
  
  It would have been kind of them to tell us that all the distances, including x and y separations, are integers.
  
  LikeLike
Jim Randell 10:12 am on 11 July 2025 Permalink | Reply
Tags: by: C Higgins ( 38 ), by: H Bradley ( 38 )
Teaser 2475: [Stopped clock]
From The Sunday Times, 28th February 2010 [link]

“I see that our long case has stopped”, said Watson, who had been reading Holmes’s monograph on chronometers. “Is there anything significant about the time?” Holmes replied: “Apart from the obvious facts that just one hand is precisely on 12, and clockwise the hands are in the order “second” , “minute”, “hour”, I see that the time read as a 3- or 4-digit number is the product of two whole numbers, the larger of which is the number of minutes clockwise from the minute hand to the hour hand”.

At what time did the clock stop?

This puzzle was originally published with no title.

[teaser2473]
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- Jim Randell 10:12 am on 11 July 2025 Permalink | Reply
  If one of the hands is on the 12, then the clock must be showing an exact number of minutes, and so the second hand must be on 12.
  
  Then as we go clockwise from the second hand we encounter the minute hand (which is on an exact minute marking), and then the hour hand (which must also be on an exact minute marking, so the number of minutes must be a multiple of 12).
  
  This Python program considers possible hours and minutes, calculates the number of minute divisions the hour hand is ahead of the minute hand and then checks that this divides into the time (read as a 3- or 4-digit number) to give a smaller number.
  
  It runs in 60ms. (Internal runtime is 46µs).
```
from enigma import (irange, cproduct, divc, div, printf)

# possible hours and minutes (must be a multiple of 12)
for (h, m) in cproduct([irange(1, 11), irange(12, 59, step=12)]):

  # number of minutes pointed to by the hour hand
  p = (5 * h) + (m // 12)
  # minute divisions the hour hand is ahead of the minute hand
  d = p - m
  if not (d > 0): continue

  # time read as a 3- or 4-digit number
  n = 100 * h + m
  r = div(n, d)
  if r is None or not (d > r): continue

  # output solution
  printf("{h:02d}:{m:02d} -> {n} = {d} * {r}")
```
  Solution: The clock stopped at 8:12.
  
  The second hand points to 12 (= 0 minutes), the minute hand to 12 minutes, and the hour hand to 8 + 12/60 hours (= 41 minutes).
  
  The hour hand is 29 minute divisions ahead of the minute hand, and: 812 = 29 × 28.
  
  LikeLike
Jim Randell 8:25 am on 8 July 2025 Permalink | Reply
Tags: by: Bert Brooke
Brainteaser 1031: Definitive darts
From The Sunday Times, 2nd May 1982 [link]

We of the Private Bar of the Wapping Boar are an exclusive elite in darting fraternity; albeit we play the conventional darts of commoner folk and employ their customary board and rules. Our two opposing players throw three darts each, in turn. We require that the last (winning) dart of any game must score some double or the bull (50) — and must thereby raise the player’s score precisely to 501.

Last evening, two of our virtuosi played an epic exhibition; a series of games were all won by the first-to-throw and each game, required the least possible number of darts. The precision was faultless and no dart thrown by the eventual winner of any game ever exceeded the score of any preceding dart in that game. Later, our President eulogised and vouched that in the suite but recently enjoyed, every possible scoring sequence of winning games (within our definition) had been played just once.

“What about my record?” interceded that parvenu from the “public” (one we import to manage the chalks and call the 3 dart total following each throw). The low fellow claims a maximum possible of top score shouts “One hundred and eighty!!” in a demonstration match such as we’d just had; he insists his feat be recognised. The oaf must have his due if we are to use him anew, but there’s the rub — who scores for a scorer?

Readers, please say the maximum of correct “180” shouts to be credited in our circumstance.

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

[teaser1031]
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- Jim Randell 8:25 am on 8 July 2025 Permalink | Reply
  It is not possible to score 501 in fewer than 9 darts, as 60 is the maximum possible score with a single dart, and 501 / 60 > 8.
  
  But it is possible to score 501 in 9 darts, as 501 = 3 × 167, and 167 = 60 (treble 20) + 57 (treble 19) + 50 (bull). So we can finish the 9 dart run on a bull.
  
  We need to find sequences of 9 darts, with scores in non-increasing order, that score a total of 501, and finish on a double.
  
  And this corresponds to the winners throws. The winner went first, so the loser in each match only threw 6 darts, and as the scorer shouted “180” the maximum possible number of times, the loser must have thrown 6×60 for 2 shouts of “180”.
  
  This Python 3 program runs in 67ms. (Internal runtime is 1.3ms).
  
  Run: [ @codepad ]
```
from enigma import (irange, union, seq_items, chunk, printf)

# possible scores with 3 darts
single = union([irange(1, 20), [25]])
double = set(2 * x for x in single)
treble = set(3 * x for x in single if x != 25)
darts = sorted(union([single, double, treble]), reverse=1)

# score <t> in <k> darts (finishing with a double)
# using non-increasing subsets of darts[i..]
def score(t, k, i=0, ss=[]):
  x = darts[i]
  # finish on a double
  if k == 1:
    if t in double and not (x < t):
      yield ss + [t]
  elif not (x * k < t):
    # score the next dart
    for (j, x) in seq_items(darts, i):
      if x < t:
        yield from score(t - x, k - 1, j, ss + [x])

# score 501 in 9 darts, count the number of matches, and "180" shouts
n = s = 0
for ss in score(501, 9):
  n += 1
  gs = list(chunk(ss, 3))
  x = sum(sum(g) == 180 for g in gs)
  s += x
  printf("[{n}: {gs}; {x} shouts]")

# total number of shouts
t = s + 2 * n
printf("total shouts = {t} [from {n} matches]")
```
  Solution: There were 62 shouts of “180”.
  
  There were 18 matches. The winners scores as follows:
  
  1: [(60, 60, 60), (60, 60, 60), (60, 57, 24)]; 2 shouts
  2: [(60, 60, 60), (60, 60, 60), (60, 51, 30)]; 2 shouts
  3: [(60, 60, 60), (60, 60, 60), (60, 45, 36)]; 2 shouts
  4: [(60, 60, 60), (60, 60, 60), (57, 54, 30)]; 2 shouts
  5: [(60, 60, 60), (60, 60, 60), (57, 50, 34)]; 2 shouts
  6: [(60, 60, 60), (60, 60, 60), (57, 48, 36)]; 2 shouts
  7: [(60, 60, 60), (60, 60, 60), (54, 51, 36)]; 2 shouts
  8: [(60, 60, 60), (60, 60, 60), (51, 50, 40)]; 2 shouts
  9: [(60, 60, 60), (60, 60, 57), (57, 57, 30)]; 1 shout
  10: [(60, 60, 60), (60, 60, 57), (57, 51, 36)]; 1 shout
  11: [(60, 60, 60), (60, 60, 57), (54, 54, 36)]; 1 shout
  12: [(60, 60, 60), (60, 60, 57), (54, 50, 40)]; 1 shout
  13: [(60, 60, 60), (60, 60, 51), (50, 50, 50)]; 1 shout
  14: [(60, 60, 60), (60, 57, 57), (57, 54, 36)]; 1 shout
  15: [(60, 60, 60), (60, 57, 57), (57, 50, 40)]; 1 shout
  16: [(60, 60, 60), (60, 57, 54), (50, 50, 50)]; 1 shout
  17: [(60, 60, 60), (57, 57, 57), (57, 57, 36)]; 1 shout
  18: [(60, 60, 60), (57, 57, 57), (50, 50, 50)]; 1 shout
  
  This accounts for 26 of the shouts.
  
  The loser in each match accounts for another 2 shouts per match, giving 26 + 18×2 = 62 shouts in total.
  
  LikeLike
  - Frits 12:12 pm on 10 July 2025 Permalink | Reply
    
    @Jim, as “darts” is non-increasing the check in line 20 can be done outside the loop (if needed at all as x always seems to be less than t).
    
    LikeLike
    - Jim Randell 12:36 pm on 11 July 2025 Permalink | Reply
      
      @Frits: Yes, we could skip any elements that are not less than the remaining total, and then just process all the remaining elements without testing (as they are in decreasing order). But I didn’t think it was worth the additional complication.
      
      But we do need the test, for example, if we wanted to find a 2 dart finish from 37 we would call [[ score(37, 2) ]].
      
      LikeLike
- Frits 6:08 pm on 8 July 2025 Permalink | Reply
```
N = 9 # number of throws needed

# return a valid loop structure string, indent after every "for" statement
def indent(s):
  res, ind = "", 0
  for ln in s:
    res += " " * ind + ln + "\n"
    if len(ln) > 2 and ln[:3] == "for":
      ind += 2
  return res  

syms = "ABCDEFGHIJKLMNOPQR"
throws = ["X"] + list(syms[:N])
varstring = "[" + ', '.join(throws[1:]) + "]"
# possible scores with 3 darts
single = set(range(1, 21)) | {25}
double = {2 * x for x in single}
treble = {3 * x for x in single if x != 25}

# determine mininum for the first 8 throws
mn = 501 - (7 * max(treble) + max(double))
# darts to be used for the first 8 throws
darts = sorted([x for x in single | double | treble if x >= mn], reverse=True)

# generate an execution string with 8 loops
exprs = [f"X = {max(treble)}; cnt = 0"]           # X is a dummy variable
for i, t in enumerate(throws[1:], 1):
  if i < N:
    exprs.append(f"for {t} in [x for x in {darts} if x <= {throws[i - 1]}]:") 
    if i < N - 1:
      exprs.append(f"if sum([{', '.join(throws[1:i + 1])}]) + "
                   f"{N - i - 1} * {throws[i]} + {max(double)} < 501: break")
  else:
    # most inner loop
    exprs.append(f"{throws[N]} = 501 - sum([{', '.join(throws[1:-1])}])")   
    exprs.append(f"if {throws[N]} > {throws[N - 1]}: break")    
    exprs.append(f"if not ({throws[N]} in {double}): continue")   
    # the loser in each game accounts for another 2 shouts per game
    exprs.append(f"cnt += sum({varstring}[3 * i:3 * (i + 1)] == "
                 f"[max(treble)] * 3 for i in range(3)) + 2")
    #exprs.append(f"print({varstring})")
       
exprs = indent(exprs)
exprs += f"print('answer:', cnt)"
#print(exprs)
exec(exprs)
```
  The program generates (and executes) the following code:
```
X = 60; cnt = 0
for A in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= X]:
  if sum([A]) + 7 * A + 50 < 501: break
  for B in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= A]:
    if sum([A, B]) + 6 * B + 50 < 501: break
    for C in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= B]:
      if sum([A, B, C]) + 5 * C + 50 < 501: break
      for D in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= C]:
        if sum([A, B, C, D]) + 4 * D + 50 < 501: break
        for E in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= D]:
          if sum([A, B, C, D, E]) + 3 * E + 50 < 501: break
          for F in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= E]:
            if sum([A, B, C, D, E, F]) + 2 * F + 50 < 501: break
            for G in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= F]:
              if sum([A, B, C, D, E, F, G]) + 1 * G + 50 < 501: break
              for H in [x for x in [60, 57, 54, 51, 50, 48, 45, 42, 40, 39, 38, 36, 34, 33, 32] if x <= G]:
                I = 501 - sum([A, B, C, D, E, F, G, H])
                if I > H: break
                if not (I in {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 50}): continue
                cnt += sum([A, B, C, D, E, F, G, H, I][3 * i:3 * (i + 1)] == [max(treble)] * 3 for i in range(3)) + 2
print('answer:', cnt)
```
  LikeLike

Jim Randell 6:41 am on 6 July 2025 Permalink | Reply
Tags: by: Howard Williams ( 45 )

Teaser 3276: First and last multiple

From The Sunday Times, 6th July 2025 [link] [link]

Johann enjoys playing with numbers and making up numerical challenges for his elder brother Jacob. He has worked out a new challenge, which involves Jacob trying to find the whole-number square root of a six-digit number. He explained that the sum of the individual digits [of the six-digit number] is exactly five times the sum of its first and last digits. To make it easier, he tells him that no number is repeated in the six digits, there are no zeros, and the last three digits are in descending numerical order.

What is the square root of the six digit number?

[teaser3276]

Jim Randell 6:48 am on 6 July 2025 Permalink | Reply

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

It runs in 78ms. (Internal runtime of the generated program is 279µs).

Run: [ @codepad ]

#! python3 -m enigma -rr

SubstitutedExpression

# sqrt(ABCDEF) = XYZ
--distinct="ABCDEF"
--invalid="0,ABCDEFXZ"
--answer="XYZ"

# XYZ is the square root of ABCDEF
"sq(XYZ) = ABCDEF"

# the sum of the digits of the square is 5 times the
# sum of the first and last digit
"A + B + C + D + E + F == 5 * (A + F)"

# the final three digits are in descending order
"D > E > F"

# [optional] performance tweaks
--reorder="[1]"
--invalid="1|2,X"

Solution: The square root is 661.

And so the 6-digit number is: 661^2 = 436921.

And we have:

4 + 3 + 6 + 9 + 2 + 1 = 25 = 5 × (4 + 1)
9 > 2 > 1

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GeoffR 10:17 am on 6 July 2025 Permalink | Reply


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

var 1..9:a; var 1..9:b; var 1..9:c; var 1..9:d;
var 1..9:e; var 1..9:f; 
var 316..999:ans; % square root of abcdef

var 100000..999999: abcdef == 100000*a + 10000*b 
+ 1000*c + 100*d + 10*e + f;

constraint all_different([a, b, c, d, e, f]);

constraint a + b + c + d + e + f == 5 * (a + f);
constraint d > e /\ e  > f;

predicate is_sq(var int: c) =
   let {
      var 0..ceil(pow(int2float(ub(c)),(1/2))): z
   } in z*z = c;

constraint ans * ans == abcdef;
constraint is_sq(abcdef);

solve satisfy;

output["Square root of the six digit number = " ++  show(ans)];

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Frits 3:03 pm on 6 July 2025 Permalink | Reply

# last 2 digits y and z  (abcdef = xyz^2)
last2 = set(i for i in range(11, 100) 
            if i % 10 and 9 < (m := (i * i) % 100) < 90 and m // 10 > m % 10)

for yz in last2:
  # A + F <= 7 so X < 9
  for x in range(3, 9):
    xyz = 100 * x + yz
    abcdef = xyz * xyz
    if len(set(s := str(abcdef))) != 6: continue
    if s[3] <= s[4] or "0" in s: continue
    dgts6 = [int(x) for x in s]
    if sum(dgts6) != 5 * (dgts6[0] + dgts6[-1]): continue

    print("answer:", xyz)

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Frits 5:57 pm on 6 July 2025 Permalink | Reply

Processing six-digit numbers to find a valid square root.

from enigma import is_square

# last 2 digits e and f  (abcdef = xyz^2)
last2 = set(divmod(m, 10) for i in range(11, 100) 
            if i % 10 and 9 < (m := (i * i) % 100) < 90 and m // 10 > m % 10)
for (e, f) in last2:
  ef = 10 * e + f
  for d in range(e + 1, 10):
    _def = 100 * d + ef
    # 21 <= 5 * (a + f) <= 39  or  4 < a + f < 8
    for a in set(range(max(1, 5 - f), 8 - f))- {d, e, f}:
      # remaining for b + c
      if not (3 <= (bc := 5 * (a + f) - (a + d + e + f)) <= 17): continue
      adef = 10**5 * a + _def
      # determine missing numbers x and y (x + y = bc with x < y)
      for x in set(range(max(1, bc - 9), (bc + 1) // 2)) - {a, d, e, f}:
        if (y := bc - x) in {a, x, d, e, f}: continue
        for b, c in ((x, y), (y, x)):
          n = adef + 10**4 * b + 10**3 * c
          if (xyz := is_square(n)):
            print("answer:", xyz)

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Ruud 4:22 pm on 6 July 2025 Permalink | Reply

import math
import istr

for i in range(int(math.sqrt(100_000)), int(math.sqrt(1_000_000))):
    if (sum((sq := istr(i * i))) == 5 * (sq[0] + sq[-1])) and sq.all_distinct() and "0" not in sq and sq[-3] > sq[-2] > sq[-1]:
        print(sq, i)

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GeoffR 9:17 am on 9 July 2025 Permalink | Reply

sq6 = [ n**2 for n in range(317, 1000)    
        if len(set(str(n**2))) == 6 ]

for num in sq6:
  ds = str(num)
  if '0' in ds: continue
  dig = [int(x) for  x in ds]
  if (dig [3] > dig[4] > dig[5]):
    if sum(dig[:6]) == 5 * (dig[0] + dig[5]):
      print (f"Square root is {int(num ** 0.5)}.")

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Jim Randell 10:55 am on 4 July 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 2478: [Palindromic primes]
From The Sunday Times, 21st March 2010 [link]

An article in The Times of January 2 commented that 2010 was the sum of four primes. As a reader pointed out in a letter on January 5, this was hardly worthy of comment: by Goldbach’s Conjecture, any even number greater than two is the sum of just two primes.

What is worthy of note is that 2010 is the sum of four palindromic primes. I found one such set of primes and each of my two grandsons, Oliver and Sam found different sets. My set had two primes in common with Oliver’s set and two in common with Sam’s set.

What are my four palindromic primes?

This puzzle was originally published with no title.

[teaser2478]
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- Jim Randell 10:55 am on 4 July 2025 Permalink | Reply
  This Python program runs in 69ms. (Internal runtime is 902µs).
```
from enigma import (primes, is_npalindrome, subsets, intersect, printf)

# target value
T = 2010

# collect palindromic primes
ps = list(p for p in primes.between(2, T) if is_npalindrome(p))

# find 4-subsets with the required sum
ts = list(ss for ss in subsets(ps, size=4) if sum(ss) == T)

# choose a set for the setter
for V in ts:
  # find other sets with 2 values in common with V
  gs = list(ss for ss in ts if ss != V and len(intersect([ss, V])) == 2)
  if len(gs) == 2:
    # output solution
    printf("V = {V} [O,S = {gs}]")
```
  Solution: Victor’s set was: (11, 151, 919, 929).
  
  And the grandsons sets were:
  
  (11, 313, 757, 929) [shares: 11, 929]
  (11, 353, 727, 919) [shares: 11, 919]
  
  And these are the only sets of 4 palindromic primes that sum to 2010.
  
  LikeLike
- Ruud 4:32 pm on 4 July 2025 Permalink | Reply
```
import istr

solutions = [set(p) for p in istr.combinations((i for i in istr.range(2010) if i.is_prime() and i == i[::-1]), 4) if sum(p) == 2010]
print([solution for solution in solutions if sum(len(solution & solution1) == 2 for solution1 in solutions) == 2])
```
  LikeLike

Jim Randell 8:42 am on 1 July 2025 Permalink | Reply
Tags: by: J. H. Hayhurst Batty ( 2 )

Brain teaser 1029: Staff factors

From The Sunday Times, 18th April 1982 [link]

The Manager of a large company greeted his twelve new computer staff. “You have been given consecutive, five-digit, Staff Reference Numbers (SRN). I remember numbers by finding their prime factors, using mental arithmetic — pre-computer vintage. Why not try it? If you would each tell me what factors you have, without specifying them (and ignoring unity), I should be able to work out your SR numbers”.

John said: “My number is prime.”

Ted said ” I have two prime factors. Your number follows mine doesn’t it, Les?”

Ian said: “I also have two, one of which squared. Alan’s just before me on the list.”

Sam said: “One of mine is to the power four. The last two digits of my SRN give me the other prime factor.”

Pete said: “I’ve got one factor to the power four as well. The other one is my year of birth.”

Brian said: “My number has one prime factor cubed and two others, both squared.”

Chris said: “I’m the only one with four factors, one of which is squared. Fred’s number is one less than mine.”

Dave started to say: “Kevin’s SRN is the one after mine, which …” when the Manager interrupted. “I can now list all twelve!”

List the twelve people, by initials, in increasing order of SRNs. What is Sam’s SRN?

This was the final puzzle to go by the title “Brain teaser“. The next puzzle was “Brainteaser 1030“.

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

[teaser1029]

Jim Randell 8:43 am on 1 July 2025 Permalink | Reply

Instead of trying all possible 5-digit SRNs we can reduce the number of candidates considered by only considering numbers in the neighbourhood of P, as there are only a few numbers that are the 4th power of a prime multiplied by a viable prime birth year. (I considered primes between 1912 and 1982 as viable birth years).

This Python program uses the [[ SubstitutedExpression ]] solver from the enigma.py library, to solve the puzzle given a candidate start number.

It runs in 308ms (although I haven’t looked at optimising the evaluation order of the expressions).

from enigma import (SubstitutedExpression, irange, primes, cache, sprintf as f, join, printf)

# prime factors of n -> ((p1, e1), (p2, e2), ...)
factors = cache(lambda n: tuple(primes.prime_factor(n)))

# I has exactly 2 prime factors; one is a power of 2
def checkI(n):
  ((p1, e1), (p2, e2)) = factors(n)
  return 2 in {e1, e2}

# S has exactly 2 prime factors; one is a power of 4, and the other is the last 2 digits of S
def checkS(n):
  ((p1, e1), (p2, e2)) = factors(n)
  return (4, n % 100) in {(e1, p2), (e2, p1)}

# P has exactly 2 prime factors; one is a power of 4, and the other is his year of birth
def checkP(n):
  ((p1, e1), (p2, e2)) = factors(n)
  return (e1 == 4 and 1911 < p2 < 1983) or (e2 == 4 and 1911 < p1 < 1983)

# B has exactly 3 prime factors; a power of 3 and two powers of 2
def checkB(n):
  ((p1, e1), (p2, e2), (p3, e3)) = factors(n)
  return sorted([e1, e2, e3]) == [2, 2, 3]

# C has exactly 4 prime factors; one is a power of 2
def checkC(n):
  ((p1, e1), (p2, e2), (p3, e3), (p3, e4)) = factors(n)
  return 2 in {e1, e2, e3, e4}

def solve(start, s2d=dict()):
  # expressions to evaluate
  exprs = [
    # J is prime
    "is_prime(J + start)",
    # T has exactly 2 prime factors
    "len(factors(T + start)) == 2",
    # all the others (except C) have < 4 prime factors
    "all(len(factors(x + start)) < 4 for x in [A, D, F, K, L])",
    # adjacent values
    "T + 1 = L",
    "I - 1 = A",
    "C - 1 = F",
    "D + 1 = K",
    # checks on the remaining values
    "checkI({I} + start)",
    "checkS({S} + start)",
    "checkP({P} + start)",
    "checkB({B} + start)",
    "checkC({C} + start)", 
  ]

  # environment
  env = dict(
    start=start,
    factors=factors, is_prime=primes.is_prime,
    checkI=checkI, checkS=checkS, checkP=checkP, checkB=checkB, checkC=checkC,
  )

  # construct the puzzle
  p = SubstitutedExpression(exprs, base=12, s2d=s2d, env=env)

  # solve the puzzle for the offsets
  for s in p.solve(verbose=''):
    # fill out the actual numbers from the offsets
    s = dict((k, v + start) for (k, v) in s.items())
    # output the numbers in order
    fmt = (lambda fs: join((f("{p}" if e == 1 else "{p}^{e}") for (p, e) in fs), sep=", "))
    for k in sorted(s.keys(), key=s.get):
      n = s[k]
      printf("{k} = {n} [{fs}]", fs=fmt(factors(n)))
    printf()

# consider possible prime birth years for P
for y in primes.between(1912, 1982):
  # consider other prime factor, raised to the 4th power
  for p in primes:
    P = p**4 * y
    if P > 99999: break
    # consider possible start numbers
    for start in irange(max(10000, P - 11), min(99988, P)):
      solve(start, dict(P=P - start))

Solution: The order is: D K A I B S T L P F C J. Sam’s SRN is 31213.

The SRNs [along with their prime factors] are given below:

D = 31208 [2^3, 47, 83]
K = 31209 [3, 101, 103]
A = 31210 [2, 5, 3121]
I = 31211 [23^2, 59]
B = 31212 [2^2, 3^3, 17^2]
S = 31213 [7^4, 13]
T = 31214 [2, 15607]
L = 31215 [3, 5, 2081]
P = 31216 [2^4, 1951]
F = 31217 [19, 31, 53]
C = 31218 [2, 3, 11^2, 43]
J = 31219 [31219]

So we see P was born in 1951, which is 31 years before the puzzle was set.

With a bit of work, we can further reduce the number of cases considered significantly.

There are only 7 candidate values for P, and there are only 11 candidate values for S. And there is only a single pair where P and S are close enough together to give a solution. This also gives a way to tackle puzzle manually.

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Frits 1:44 pm on 1 July 2025 Permalink | Reply

@Jim, is there a typo in “which is 42 years before the puzzle was set”?

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- Jim Randell 5:21 pm on 1 July 2025 Permalink | Reply
  
  @Frits: Yes. I got it mixed up with a puzzle from 1993.
  
  LikeLike
- Frits 7:39 pm on 1 July 2025 Permalink | Reply
  
  @Jim, the puzzle text says that Ian has two prime factors, one of which squared.
  
  LikeLike
  - Jim Randell 11:14 pm on 1 July 2025 Permalink | Reply
    
    @Frits: Thanks. I’ll add in the missing condition.
    
    LikeLike

Frits 7:19 pm on 1 July 2025 Permalink | Reply

from enigma import is_prime, prime_factor as pfactor, ceil, subsets

# possible SRN's for people in list <s>
SRNs = lambda s: {i for i in range(s[0] - 11, s[0] + 12) 
                  if i not in s and all(abs(x - i) < 12 for x in s)}
                  
# mandatory SRN's for people in list <s>
m_SRNs = lambda s: {i for i in range(min(s) + 1, max(s)) if i not in s}                  

mn = 1912 # minimum birth date

# set of prime numbers up to 99
P = {3, 5, 7}
P |= {2} | {x for x in range(11, 100, 2) if all(x % p for p in P)}
sP = sorted(P)

# check last item in sequence <s>
def check(s, ln, vs, fs4={}):
  if fs4:
    # mandatory missing numbers plus last item  may not have 4 factors
    if not (m_SRNs(s) | {s[-1]}).isdisjoint(fs4): return False
  
  # check length and exponents
  if len(pfs := list(pfactor(s[-1]))) != ln or {y for x, y in pfs} != vs:
    return False
  return True  

# start with P ([p1^4, 19xx])
for p1 in sP:
  if p1**4 * mn > 99999: break
  ns = [n for y in range(mn, 1965) if 9999 < (n := p1**4 * y) <= 99999]
  mn, mx = ns[0] - 11 , ns[-1] + 11
  # check S ([s1^4, s2 with s2 last 2 digits])
  # S = xxxxab and S = ab.s1^4,  100 * xxxx = ab.(s1^4 - 1) so s1^4 = ...1
  for s1 in sP[1:]:
    if (pow4 := s1**4) * 11 > 99999: break
    if pow4 % 10 != 1 or mx / pow4 > 99: continue
    # calculate possible numbers for S
    for s in [s for ab in range(ceil(mn / pow4), mx // pow4 + 1) 
              if ab in P and (s := ab * pow4) % 100 == ab]:
      # only C may have exactly 4 factors
      fs4 = {x for x in SRNs([s]) if len(list(pfactor(x))) == 4}
      for j in [j for j in SRNs([s]) if is_prime(j)]:
        for c in SRNs([s, j]):
          if not check([s, j, c], 4, {1, 2}): continue
          for b in SRNs([s, j, c]):
            if not check([s, j, c, b], 3, {2, 3}, fs4): continue
            for i in SRNs([s, j, c, b]):
              if not check([s, j, c, b, i], 2, {1, 2}, fs4): continue
              for t in SRNs([s, j, c, b, i]):
                if not check([s, j, c, b, i, t], 2, {1}, fs4): continue
                for p in SRNs([s, j, c, b, i, t]):
                  if not check([s, j, c, b, i, t, p], 2, {1, 4}, fs4): continue
                  # before I is A, after T is L, before C is F
                  a, l, f = i - 1, t + 1, c - 1
                  n10 = sorted([s, j, c, b, i, t, p, a, l, f])
                  if len(set(n10)) != 10: continue
                  # K after D, find places to insert (D, K)
                  for d, k in subsets(SRNs(n10), 2):
                    if k != d + 1: continue
                    if not {d, k}.isdisjoint(fs4): continue
                    # 12 consecutive numbers
                    if ((n12 := sorted(n10 + [d, k])) == 
                         list(range(n12[0], n12[0] + 12))):
                      # final check   
                      if not (set(n12) - {c}).isdisjoint(fs4): continue
                      map = dict(list(zip([s, j, c, b, i, t, p, a, l, f, d, k],
                                    "SJCBITPALFDK")))
                      for k, v in sorted(map.items()):
                        print(f"{v}: {k} = "
                              f"{' * '.join(['^'.join(str(y) for y in x)
                                             for x in pfactor(k)])}")

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Jim Randell 7:22 am on 29 June 2025 Permalink | Reply
Tags: by: Andrew Skidmore ( 88 )
Teaser 3275: Against the odds
From The Sunday Times, 29th June 2025 [link] [link]

I have taken a number of playing cards (more red than black) from a standard pack of 52 cards and redistributed them in a certain way in two piles, each containing both red and black cards.

A card is taken at random from the smaller pile and placed in the larger pile. The larger pile is shuffled; a card is taken at random from it and placed in the smaller pile.

There is a one in eight chance that the number of red and black cards in each pile will be the same as they were originally.

How many black cards in total were used?

[teaser3275]
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- Jim Randell 7:49 am on 29 June 2025 Permalink | Reply
  (See also: Teaser 2571).
  
  If we have two piles:
  
  pile 1, with b₁ black cards, and r₁ red cards; and:
  pile 2, with b₂ black cards, and r₂ red cards.
  
  where pile 1 (of size n₁) is smaller than pile 2 (of size n₂).
  
  Then the probability of the same colour card being chosen for both transfers is the sum of the following probabilities:
  
  P(both transfers black) = (b₁ / n₁) × ((b₂ + 1) / (n₂ + 1))
  P(both transfers red) = (r₁ / n₁) × ((r₂ + 1) / (n₂ + 1))
  ⇒
  P(both transfers same colour) = (b₁ × (b₂ + 1) + r₁ × (r₂ + 1)) / (n₁ × (n₂ + 1))
  
  The following Python program constructs all possible arrangements of cards and calculates the probability of the same coloured card moving in both directions.
  
  We can find the inverse of the combined probability, and this allows us to perform all the calculations using integers.
  
  It runs in 84ms. (Internal runtime is 22ms).
```
from enigma import (irange, subsets, Decompose, cproduct, div, printf)

# choose number of red and black cards (B < R)
for (B, R) in subsets(irange(2, 26), size=2, select='C'):

  # split into 2 piles (pile1 < pile2)
  decompose = Decompose(2, increasing=0, sep=0)
  for ((b1, b2), (r1, r2)) in cproduct([decompose(B), decompose(R)]):
    (n1, n2) = (b1 + r1, b2 + r2)
    if not (n1 < n2): continue

    # calculate probability of the same coloured card in both transfers
    # r = 1/p
    r = div(n1 * (n2 + 1), b1 * (b2 + 1) + r1 * (r2 + 1))

    # look for solution
    if r == 8:
      printf("B={B} R={R}")
      printf("  b1={b1} r1={r1} -> {n1}; b2={b2} r2={r2} -> {n2}")
      printf("    p=1/{r}")
      printf()
```
  Solution: There were 21 black cards.
  
  And 23 red cards.
  
  The cards are distributed into piles as follows:
  
  pile 1 = 20 black + 1 red (21 cards)
  pile 2 = 1 black + 22 red (23 cards)
  
  The probability of a black card moving in both directions is:
  
  pB = 20/21 × 2/24 = 40/504
  
  And the probability of a red card moving in both directions is:
  
  pR = 1/21 × 23/24 = 23/504
  
  And we have:
  
  pB + pR = 63/504 = 1/8
  
  LikeLike
  - Jim Randell 9:48 am on 30 June 2025 Permalink | Reply
    Or we can use the [[ SubstitutedExpression ]] solver from the enigma.py library, for a more declarative solution.
```
from enigma import (SubstitutedExpression, irange, printf)

# assign symbols
macro = dict(b1='W', r1='X', b2='Y', r2='Z')

# expressions to solve
exprs = [
  # more reds than blacks; no more than 26 cards per colour
  "@b1 + @b2 < @r1 + @r2 < 27",
  # pile 1 is smaller than pile 2
  "@b1 + @r1 < @b2 + @r2",
  # probability P = 1/8; P = Pn/Pd => Pd = 8 Pn
  "(@b1 + @r1) * (@b2 + @r2 + 1) == 8 * (@b1 * (@b2 + 1) + @r1 * (@r2 + 1))",
]

# construct the solver
p = SubstitutedExpression(exprs,
      base=26, digits=irange(1, 25), distinct=[], macro=macro,
      answer="(@b1, @r1, @b2, @r2)",
    )

# solve the puzzle
for (b1, r1, b2, r2) in p.answers(verbose=0):
  (B, R) = (b1 + b2, r1 + r2)
  printf("B={B} R={R} [b1={b1} r1={r1}; b2={b2} r2={r2}]")
```
    LikeLike
Jim Randell 8:28 am on 27 June 2025 Permalink | Reply
Tags: by: C Higgins ( 38 ), by: H Bradley ( 38 )
Teaser 2471: [Cutting a rod]
From The Sunday Times, 31st January 2010 [link]

In the woodwork class, students were being taught how to measure and cut accurately. Pat was given a rod that was a whole number of feet long and was told to saw it into three pieces, each a different whole number of inches long. While he was cutting his set of three pieces, Pat calculated how many different sets it would be possible to make. He discovered that the answer was equal to his father’s year of birth.

What year was that?

This puzzle was originally published with no title.

[teaser2471]
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- Jim Randell 8:29 am on 27 June 2025 Permalink | Reply
  We assume the cuts themselves make a negligible difference to the length of the pieces.
  
  Here is a constructive program that counts the number of dissections of an n-foot rod into 3 dissimilar whole inch lengths.
  
  I decided that as the puzzle was set in 2010, viable years are in the range [1940, 1985].
  
  The program runs in 71ms. (Internal runtime is 6.1ms).
```
from enigma import (irange, inf, decompose, icount, printf)

# generate number of dissections
def generate():
  # consider number of feet
  for n in irange(1, inf):
    # divide into different sets of inches
    k = icount(decompose(12 * n, 3, increasing=1, sep=1))
    printf("[{n} ft -> {k} sets]")
    yield (n, k)

# find lengths that give up to 1985 sets
ss = list()
for (n, k) in generate():
  ss.append((n, k))
  if k > 1985: break

# output solution (between 1940 and 1985)
for (_, y) in ss:
  if 1939 < y < 1986:
    printf("year = {y}")
```
  Solution: Pat’s father was born in 1951.
  
  1951 is the number of dissections of a 13-foot rod.
  
  We can use the values accumulated by the program to determine a polynomial equation to generate the sequence:
```
% python3 -i teaser/teaser2471.py
[1 ft -> 7 sets]
[2 ft -> 37 sets]
[3 ft -> 91 sets]
[4 ft -> 169 sets]
[5 ft -> 271 sets]
[6 ft -> 397 sets]
[7 ft -> 547 sets]
[8 ft -> 721 sets]
[9 ft -> 919 sets]
[10 ft -> 1141 sets]
[11 ft -> 1387 sets]
[12 ft -> 1657 sets]
[13 ft -> 1951 sets]
[14 ft -> 2269 sets]
year = 1951

>>> from enigma import Polynomial
>>> p = Polynomial.interpolate(ss)
>>> print(p)
Polynomial[(+12)x^2 (-6)x (+1)]
>>> print(p(13))
1951
>>> print(p(5280))
334509121
```
  So, for an n-foot rod (integer n ≥ 1), the number of dissections D(n) is given by:
  
  D[n] = 12n^2 − 6n + 1
  
  This corresponds to OEIS A154105 [@oeis.org].
  
  LikeLike
  - Frits 3:32 pm on 27 June 2025 Permalink | Reply
    
    It is not too difficult to find this formula for the sum 1 + 2+4 + 5+7 + 8+10 + 11+13 + 14+16 + ….
    
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  - Frits 4:08 pm on 27 June 2025 Permalink | Reply
    
    Another formula for D(n) is:
    
    sum(6 * n – (k + 3) // 2 – k – 1 for k in range(4 * n – 1))
    
    LikeLike
- Jim Randell 8:10 am on 28 June 2025 Permalink | Reply
  
  See also: BrainTwister #25.
  
  Where we had:
  
  t[n] = intr(sq(n) / 12)
  
  Then D[n] appears as a subsequence of t[n] taking every 12th element (as there are 12 inches in 1 foot).
  
  Specifically:
  
  D[n] = t[12n − 3]
  
  And this gives rise to the quadratic equation given above.
  
  LikeLike

Jim Randell 8:26 am on 24 June 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )

Teaser 2467: [Some primes]

From The Sunday Times, 3rd January 2010 [link]

To celebrate the start of a new decade, I have written down a list of prime numbers whose sum is equal to one of the years of this decade (in other words, it is one of the numbers from 2010 to 2019, inclusive). Overall, the numbers in this list contain just eight digits, namely four different even digits and four different odd digits, each used once.

With simple logic, rather than complicated calculations, you should be able to work out the list of numbers.

What are they?

This puzzle was originally published with no title.

[teaser2467]

Jim Randell 8:26 am on 24 June 2025 Permalink | Reply

This Python 3 program looks for sets of primes with the required property.

It runs in 120ms. (Internal run time is 53ms).

from enigma import (primes, nsplit, seq_items, join, printf)

# find possible primes, and record digits
ps = list()
for p in primes.between(2, 2018):
  ds = nsplit(p)
  ss = set(ds)
  if len(ss) == len(ds):
    ps.append((p, ss))

# odd and even digits
(evens, odds) = ({0, 2, 4, 6, 8}, {1, 3, 5, 7, 9})

# solve for primes in <ps>
# i = start index in <ps>
# t = total to far
# ds = digits used so far
# ns = primes used so far
def solve(ps, i=0, t=0, ds=set(), ns=[]):
  if t > 2009:
    if len(ds) == 8:
      yield ns
  if len(ds) < 8:
    # add in another prime
    for (j, (p, ss)) in seq_items(ps, i):
      t_ = t + p
      if t_ > 2019: break
      if not ss.isdisjoint(ds): continue
      ds_ = ds.union(ss)
      if evens.issubset(ds_) or odds.issubset(ds_): continue
      yield from solve(ps, j + 1, t_, ds_, ns + [p])

# solve the puzzle
for ns in solve(ps):
  # output solution
  printf("{ns} = {t}", ns=join(ns, sep=" + "), t=sum(ns))

Solution: The numbers are: 2, 947, 1063.

And:

2 + 947 + 1063 = 2012

The unused digits are 5 (odd) and 8 (even).

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Frits 11:01 am on 24 June 2025 Permalink | Reply

from itertools import compress

def primesbelow(n):
  # rwh_primes1v2(n):
  """ Returns a list of primes < n for n > 2 """
  sieve = bytearray([True]) * (n // 2 + 1)
  for i in range(1, int(n ** 0.5) // 2 + 1):
    if sieve[i]:
      sieve[2 * i * (i + 1)::2 * i + 1] = \
      bytearray((n // 2 - 2 * i * (i + 1)) // (2 * i + 1) + 1)
  return [2, *compress(range(3, n, 2), sieve[1:])]

# decompose <t> into increasing numbers from <ns> so that <k> different
# digits have been used and the sum of the numbers is in range (t-9) ... t
def decompose(t, ns, k=8, ds=set(), s=[]):
  if k <= 0 or t < 10:
    if k == 0 and t < 10:
      # not using one odd and one even digit
      if sum(int(x) for x in set("0123456789") - ds) % 2:
        yield s 
  else:
    for i, (n, dgts) in enumerate(ns):
      if n > t: break 
      if ds.isdisjoint(dgts):
        yield from decompose(t - n, ns[i + 1:], k - len(dgts), 
                             ds | dgts, s + [n])

m = 2000
P = [(p, s) for p in primesbelow(m) if len(s := set(str(p))) == len(str(p))]

# look for prime numbers that add up to 2010 ... 2019
for ps in decompose(2019, P):
  print(f"answer: {' + '.join(str(x) for x in ps)} = {sum(ps)}")

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Jim Randell 6:33 am on 22 June 2025 Permalink | Reply
Tags: by: Colin Vout ( 38 )

Teaser 3274: Anarithm

From The Sunday Times, 22nd June 2025 [link] [link]

If you rearrange the letters of a word to form another word, it’s called an anagram. So I think that if you rearrange the digits of a number to form another number, it could be called an anarithm.

Some numbers when expressed in two different number bases have representations that are anarithms of each other. For instance, the decimal number 66 is represented as 123 in base 7 and 231 in base 5.

I’ve recently been looking at numbers in base 8 and base 5 and found that it is possible to have a number whose representations in base 8 and base 5 are anarithms of each other.

In decimal notation, what is the largest such number?

[teaser3274]

Jim Randell 6:39 am on 22 June 2025 Permalink | Reply

The following Python program produces non-trivial (i.e. >1-digit) representations that are “anarithms”, in numerical order.

It runs in 66ms. (Internal runtime is 494µs).

from enigma import (irange, inf, nsplit, join, printf)

# the two bases (A < B)
(A, B) = (5, 8)

# consider increasing numbers of digits
for k in irange(2, inf):
  (x, y) = (B**(k - 1), (A**k) - 1)
  if x > y: break
  for n in irange(x, y):
    (a, b) = (nsplit(n, base=A), nsplit(n, base=B))
    if sorted(a) == sorted(b):
      # output solution
      printf("k={k}: {n} -> {a} [base {A}], {b} [base {B}]", a=join(a), b=join(b))

Solution: The number is 540 (decimal).

In base 5 it is represented: 4130.

And in base 8 it is represented: 1034.

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Jim Randell 2:56 pm on 22 June 2025 Permalink | Reply

Or, as we want the largest possible number, we can consider the candidate values starting with the largest. Then we can stop after the first solution is found.

This Python program has an internal runtime of 227µs.

from enigma import (irange, inf, nsplit, rev, join, printf)

# find largest number for bases A, B (where A < B)
def solve(A, B):
  assert A < B
  # intervals to consider
  ivs = list()
  for k in irange(2, inf):
    (x, y) = (B**(k - 1), (A**k) - 1)
    if x > y: break
    ivs.append((x, y))

  # consider each interval from largest to smallest
  for (x, y) in rev(ivs):
    for n in irange(y, x, step=-1):
      (a, b) = (nsplit(n, base=A), nsplit(n, base=B))
      if sorted(a) == sorted(b):
        # output solution
        printf("{n} -> {a} [base {A}], {b} [base {B}]", a=join(a), b=join(b))
        return

# solve for base 5 and base 8
solve(5, 8)

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Frits 8:00 pm on 23 June 2025 Permalink | Reply

Not my fastest version but still fast and relatively simple.

# get digits of a number in base <b> (converted from <n> in base 10)
def int2base(n, b):
  ds = []
  while n:
      ds.append(int(n % b))
      n //= b
  return ds

b1, b2 = 5, 8
rng, invalid = [], set(range(b1, 10))
# get a list of valid ranges per number of digits
for n, _ in enumerate(iter(bool, 1), 2): # infinite  loop
  mn, mx = b2**(n - 1), b1**n - 1
  if mx < mn: break
  rng.append((mn, mx))

# for decreasing number of digits
for mn, mx in reversed(rng):
  # check all numbers in the valid range
  for n in range(mx, mn - 1, -1):
    # check for invalid digits
    if not invalid.isdisjoint(n_b2 := int2base(n, b2) ): continue

    # base 8 and base 5 are anarithms of each other
    if sorted(int2base(n, b1)) == sorted(n_b2):
      print(f"answer: {n}")
      exit(0)

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Jim Randell 8:12 am on 19 June 2025 Permalink | Reply
Tags: by: Graham Smithers ( 84 )
Teaser 2519: [Greek alphametic]
From The Sunday Times, 2nd January 2011 [link] [link]

This week’s letters-for-digits Teaser has a Greek theme. All words in capitals are numbers that have been coded by consistently using different letters for different digits.

Our story concerns the prime THALLO (goddess of spring buds), who was also good at sums, and produced:

ALPHA + BETA + GAMMA = OMEGA

What number is represented by OMEGA?

This puzzle was originally published with no title.

This completes the archive of Teaser puzzles originally published in 2011. There is now a complete archive of puzzles (and solutions) from the start of 2011 to present, along with many older puzzles.

[teaser2519]
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- Jim Randell 8:13 am on 19 June 2025 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  The following run file executes in 84ms. (Internal runtime of the generated code is 1.2ms).
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

"ALPHA + BETA + GAMMA = OMEGA"

--extra="is_prime(THALLO)"

--answer="OMEGA"
```
  Solution: OMEGA = 76915.
  
  The sum is:
  
  52305 + 8945 + 15665 = 76915
  
  And we have: THALLO = 405227, which is prime.
  
  LikeLike
Jim Randell 7:51 am on 17 June 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 2524: [Whispered birthday]
From The Sunday Times, 6th February 2011 [link] [link]

My birthday is on the first of a month. My three logical friends Lettice, Daisy and Bertha wanted to know which month. So I whispered to Lettice the number of letters in the spelling of the month, I whispered to Daisy the number of days in the month, and I whispered to Bertha the day of the week of my birthday this year. I explained to all three what I had done and then I asked them in turn whether they were able to work out the month yet. Their replies were:

Lettice: no
Daisy: no
Bertha: no
Lettice: no
Daisy: no
Bertha: yes

What is my birthday month?

This puzzle was originally published with no title.

[teaser2524]
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- Jim Randell 7:53 am on 17 June 2025 Permalink | Reply
  This Python program uses the [[ filter_unique() ]] function from the enigma.py library to eliminate candidate solutions at each step.
  
  It runs in 70ms. (Internal runtime is 263µs).
```
from enigma import (filter_unique, printf)

# month names
months = "January February March April May June July August September October November December".split()

# maps month names to: L = number of letters; D = number of days; B = day of week of 1st
L = dict((k, len(k)) for k in months)
D = dict(zip(months, [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]))  # for 2011
B = dict(zip(months, "Sa Tu Tu Fr Su We Fr Mo Th Sa Tu Th".split()))  # for 2011

# start with all possible months
ss = months

# answers: 1 = "no" (non_unique), 0 = "yes" (unique)
for (d, i) in [(L, 1), (D, 1), (B, 1), (L, 1), (D, 1), (B, 0)]:
  ss = filter_unique(ss, d.get)[i]

# output solution(s)
for m in ss:
  printf("month = {m}")
```
  Solution: The birthday month is March.
  
  LikeLike

Jim Randell 6:47 am on 15 June 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )

Teaser 3273: A good deal?

From The Sunday Times, 15th June 2025 [link] [link]

Delia plays a game with a stack of different pieces of card. To shuffle the stack she deals them quickly into four equal piles (face-downwards throughout) — the top card starts the first pile, the next card starts the second pile, then the third starts the third pile and likewise the fourth. She continues to place the cards on the four piles in order. Then she puts the four piles back into one big stack with the first pile on the bottom, the second pile next, then the third, with the fourth pile on the top.

She is very pleased with her shuffling method because each card moves to a different position in the stack, so she decides to repeat the shuffle a few more times. However, this is counter-productive because after fewer than six such shuffles the cards will be back in their original order.

How many cards are in her stack?

There are now 1200 Teaser puzzles available on the S2T2 site.

[teaser3273]

Jim Randell 7:00 am on 15 June 2025 Permalink | Reply

See also: Enigma 710, Teaser 2446.

This Python program runs in 69ms. (Internal runtime is 328µs).

from enigma import (irange, inf, flatten, rev, printf)

# perform a shuffle using <p> piles
def shuffle(cs, p=4):
  return rev(flatten(cs[i::p] for i in irange(p)))

# find when cards return to original order
def solve(cs, fn, M=inf):
  cs0 = list(cs)
  for k in irange(1, M):
    cs = fn(cs)
    if cs == cs0:
      return k

# consider sets of cards
for n in irange(8, inf, step=4):
  cs = list(irange(1, n))
  # check the shuffle leaves no card in its original position
  if any(x == y for (x, y) in zip(cs, shuffle(cs))): continue
  # find number of shuffles (< 6) to return to original order
  k = solve(cs, shuffle, 5)
  if k is None or k < 3: continue
  # output solution
  printf("{n} cards -> {k} shuffles")
  break  # stop at first solution

Solution: There are 52 cards in the stack.

And the stack returns to the original order after 4 shuffles. (This can be verified with a standard pack of cards).

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Jim Randell 11:26 am on 15 June 2025 Permalink | Reply

Alternatively (shorter, and doesn’t do any unnecessary shuffles):

from enigma import (irange, inf, flatten, rev, repeat, cachegen, peek, printf)

# perform a shuffle using <p> piles
def shuffle(cs, p=4):
  return rev(flatten(cs[i::p] for i in irange(p)))

# consider sets of cards
for n in irange(8, inf, step=4):
  cs = list(irange(1, n))
  # generate the result of repeated shuffles
  ss = cachegen(repeat(shuffle, cs))
  # check shuffle 1 leaves no card in its original position
  if any(x == y for (x, y) in zip(ss(1), cs)): continue
  # find number of shuffles (2 .. 5) to return to original order
  k = peek((k for k in irange(2, 5) if ss(k) == cs), default=-1)
  if k < 2: continue
  # output solution
  printf("{n} cards -> {k} shuffles")
  break  # stop at first solution

The [[ cachegen() ]] function has been in enigma.py for a while, waiting for a use to come up.

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Jim Randell 1:56 pm on 16 June 2025 Permalink | Reply

Here is a neat alternative approach.

It analyses the cycle representation of the permutation that represents the shuffle, to determine the number of shuffles required to return to the original order, without needing to actually perform the shuffles.

It has an internal runtime of 286µs.

from enigma import (irange, inf, flatten, rev, cycles, mlcm, printf)

# perform a shuffle using <p> piles
def shuffle(cs, p=4):
  return rev(flatten(cs[i::p] for i in irange(p)))

# consider sets of cards
for n in irange(8, inf, step=4):
  cs = list(irange(1, n))
  # find the length of cycles in a shuffle
  cycs = set(len(cyc) for cyc in cycles(shuffle(cs), cs))
  # there are no 1-cycles
  if 1 in cycs: continue
  # calculate the number of shuffles to return to the original order
  k = mlcm(*cycs)
  if k < 6:
    printf("{n} cards -> {k} shuffles")
    break  # stop at the first solution

LikeLiked by 1 person

Frits 10:28 am on 15 June 2025 Permalink | Reply

Also printing the first solution.

# Delia's shuffle algorithm for cards <s>
def shuffle(s, ln):
  # make 4 piles (4th pile first)
  s_ = [[s[i] for i in reversed(range(ln)) if i % 4 == (3 - j)] 
        for j in range(4)]
  # put the 4 piles back into one big stack with the 1st pile on the bottom
  return [y for x in s_ for y in x]

for m, _ in enumerate(iter(bool, 1), 1): # infinite loop, starting from 1
  n = m * 4         # number of cards in the stack
  orig, new = list(range(n)), []
  
  # Delia shuffles for the first time
  new = shuffle(orig, n)
  # each card moves to a different position in the stack
  if any(new[i] == i for i in range(n)): continue
  shuffle_count = 1
  # Delia shuffles a few more times (meaning more than once)
  while new != orig and shuffle_count < 6:
    new = shuffle(new, n)
    shuffle_count += 1
  
  if 2 < shuffle_count < 6:
    print(f"answer: {n} (first solution)")
    break

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Frits 10:50 am on 15 June 2025 Permalink | Reply

The program prints the expected answer.

The way I read the teaser there is another solution (allowing to already return back to the original order after two shuffles).

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- Jim Randell 11:07 am on 15 June 2025 Permalink | Reply
  
  I considered this, but decided the phrase “a few more times” excludes the possibility of just 1 additional shuffle reverting the original stack.
  
  LikeLike
  - Frits 11:21 am on 15 June 2025 Permalink | Reply
    
    I see. I regard this as an additional requirement. I can’t determine that for sure from the puzzle text.
    
    Delia will not check the cards after each shuffle so in my opinion nothing is to be excluded.
    
    LikeLike

Frits 9:57 am on 16 June 2025 Permalink | Reply

Using a dictionary

# Delia's shuffle algorithm for cards <s> using a dictionary
dct = lambda s: dict(enumerate(sum((s[i::4] for i in range(4)), [])[::-1]))
shuffle = lambda s: [d[x] for x in s]
  
# infinite loop, starting from 2 (avoiding trivial case of four cards)
for m, _ in enumerate(iter(bool, 1), 2): 
  n = m * 4         # number of cards in the stack
  # dictionary of position changes
  d = dct(orig := list(range(n)))
  # Delia shuffles for the first time
  new = shuffle(orig)
  # each card moves to a different position in the stack
  if any(new[i] == i for i in range(n)): continue
  shuffle_count = 1
  # Delia shuffles a few more times 
  while new != orig and shuffle_count < 6:
    new = shuffle(new)
    shuffle_count += 1
  
  if shuffle_count < 6:
    print(f"answer: {n} (first solution after trivial case n=4)")
    break

or using map()

# Delia's shuffle algorithm for cards <s> using a dictionary
shuffle = lambda s: sum((s[i::4] for i in range(4)), [])[::-1]
  
# infinite loop, starting from 2 (avoiding trivial case of four cards)
for m, _ in enumerate(iter(bool, 1), 2): 
  n = m * 4         # number of cards in the stack
  # Delia shuffles for the first time
  s1 = shuffle(orig := list(range(n)))
  # each card moves to a different position in the stack
  if any(s1[i] == i for i in range(n)): continue
  new = s1
  shuffle_count = 1
  # Delia shuffles a few more times 
  while new != orig and shuffle_count < 6:
    new = list(map(lambda x: s1[x], new))
    shuffle_count += 1
  
  if shuffle_count < 6:
    print(f"answer: {n} (first solution after trivial case n=4)")
    break

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Jim Randell 4:31 pm on 16 June 2025 Permalink | Reply

@Frits: I can’t recommend the (ab)use of [[ sum() ]] for joining lists. As a quick hack it may be acceptable to save a bit of typing, but the behaviour of [[ sum() ]] is only defined for numeric types, so it may not work at all (in CPython strings are explicitly rejected). And if it does work, it uses an inefficient algorithm for sequences (constructing a new sequence at each intermediate stage).

I would recommend using [[ enigma.flatten ]] (or [[ itertools.chain ]]) instead.

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Frits 12:03 pm on 19 June 2025 Permalink | Reply

A more efficient version. Instead of always shuffling the whole stack of cards we keep track of the position of only one specific card.

flat = lambda group: [x for g in group for x in g]

# Delia's shuffle algorithm for cards <s> using a dictionary
shuffle = lambda s: flat(s[i::4] for i in range(4))[::-1]

# determine new position of number at position <i> 
# in a sequence of 4 * m numbers
shuffle1 = lambda i, m: m * (4 - i % 4) - i // 4 - 1

# check when number 1 will return to position 1 again (numbers 0...n-1)

# infinite loop, starting from 2 (avoiding trivial case of four cards)
for m, _ in enumerate(iter(bool, 1), 2):  
  n = m * 4         # number of cards in the stack
  new, cnt = 1, 1
  while cnt < 6:
    # shuffle one card (let's say the 2nd card)
    new = shuffle1(new, m)
    cnt += 1
    # shuffle the whole stack 
    if new == 1:
      # Delia shuffles for the first time
      s1 = shuffle(orig := list(range(n)))
      # each card moves to a different position in the stack
      if any(s1[i] == i for i in range(n)): break
      new, shuffle_count = s1, 1
      # Delia shuffles a few more times 
      while new != orig and shuffle_count < 6:
        new = list(map(lambda x: s1[x], new))
        shuffle_count += 1

      if shuffle_count < 6:
        print(f"answer: {n} (first solution after trivial case n=4)")
        exit(0)
      break

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