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Jim Randell 12:07 am on 10 May 2026 Permalink | Reply
Tags: by: Howard Williams
Teaser 3320: Penblwydd Hapus
From The Sunday Times, 10th May 2026 [link] [link]

I’m now in my thirties and for many years I’ve treated myself to some bottles of beer on my birthday. To compensate for my increasing age, I buy the same number of bottles each year as my age. On my recent birthday, after storing my bottles, I noticed that, on the beer receipt from the supermarket, all of the nine digits 1 to 9 appeared once, and only once, in the number of bottles, the price per bottle in pence and the total cost in pence.

What was the total cost of the beer in pence?

[teaser3320]
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Ruud, Tony Smith, Hugo, and 2 others are discussing. Toggle Comments
- Jim Randell 12:20 am on 10 May 2026 Permalink | Reply
  I think the wording on this puzzle could be tightened up a little. (Specifically to clear up: “what constitutes a recent birthday?”, and “how many times can the digit 0 appear?”).
  
  I assumed we are looking for a sum consisting of 9 digits, using each of the digits 1-9 exactly once, but I find multiple solutions. (And if we allow the digit 0 to appear then there are even more solutions).
  
  Using birthdays in the last 5 years, but disallowing 0 digits, I get 3 candidate solutions. If we take “recent birthday” to mean “most recent birthday”, then we can eliminate two of these, so maybe that is what the setter intended.
  
  This Python program runs in 88ms. (Internal runtime is 19ms).
```
from enigma import (irange, inf, nsplit, flatten, printf)

# consider possible ages for a "recent" birthday
for n in irange(26, 39):
  # consider price per bottle
  for p in irange(100, inf):
    # calculate the total cost
    t = n * p
    # find the digits used in the sum
    ds = flatten(nsplit(x) for x in (n, p, t))
    # look for 9 different non-zero digits
    if len(ds) > 9: break
    if 0 not in ds and len(set(ds)) == 9:
      # output solution
      printf("{n} * {p} = {t}")
```
  Solution: The total cost of the beer was 7254p.
  
  If we suppose the “recent birthday” is the setter’s most recent birthday (that perhaps happened within the last few weeks), then the birthday is 30 .. 39, and the only solution using just the 9 digits 1-9 exactly once is:
  
  39 × 186 = 7254
  
  i.e. the beer cost 186p per bottle.
  
  However, if the recent birthday is one of the last (say) 5 (the puzzle text does say the practice has been going on “for many years”), then it could be between 26 .. 39, and this permits the following additional solutions:
  
  27 × 198 = 5346
  28 × 157 = 4396
  
  Also, the puzzle text does not place restrictions on how many occurrences of the digit 0 there may be, so we can get the following reasonable solutions:
  
  27 × 594 = 16038
  36 × 495 = 17820
  39 × 402 = 15678
  
  And if very expensive bottles of beer were purchased there are further solutions with more than one 0 digit, e.g.:
  
  39 × 1284 = 50076
  
  In fact any solution can have the price per bottle and the total price multiplied by 10 to give a further solution if no restrictions are placed on the digit 0.
  
  So, I think the intended puzzle is:
  
  I am now in my thirties. On my most recent birthday I bought a number of identical bottles of beer. That number was the same as my age in years.
  
  The number of bottles, the price per bottle in pence, and the total cost in pence, when taken together comprised 9 different non-zero digits, with each of the digits 1–9 occurring exactly once.
  
  What (in pence) was the total cost of the beer?
  
  This revised puzzle is easily solved directly from the command line using the [[ SubstitutedExpression ]] solver from the enigma.py library:
```
% python3 enigma.py SubstitutedExpression "AB * CDE = FGHI" --digits="1-9" --assign="A,3"
(AB * CDE = FGHI)
(39 * 186 = 7254) / A=3 B=9 C=1 D=8 E=6 F=7 G=2 H=5 I=4
[1 solution]
```
  (The internal runtime of the generated code is 550µs).
  
  LikeLike
- Ruud 6:48 am on 10 May 2026 Permalink | Reply
  It is easy to see that the price should have 3 digits and the total cost 4 digits.
```
import peek
import istr

for age in istr.range(31, 40):
    for rest in istr.permutations({*istr.range(1, 10)} - {*age}):
        if (price := istr.join(rest[:3])) * age == (total := istr.join(rest[3:])):
            peek(age, price, total)
```
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- Frits 11:04 am on 10 May 2026 Permalink | Reply
```
# forbidden unit digits (3 is reserved for the tens digit of the age) 
fb = {0, 3}

# dictionary of price unit digits per n_bottles unit digit
d = {b: {p for p in range(10) if p not in fb and p != b and 
                                (u := (p * b) % 10) not in fb | {b, p}} 
        for b in range(10) if b not in fb}

# the number of bottles (equal to age)
for b in [n for n in range(30, 40) if n % 10 in d]:
  # unit digit
  u = b % 10
  # minimum/maximum price, price/cost should have resp. 3/4 digits
  mn, mx = 124, 9876 // b
  #pr(b, (mn, mx), (b * mn, b * mx), d[u])
  # the price per bottle in pence
  for p in [n for n in range(mn, mx + 1) if n % 10 in d[u]]:
    if '0' in (dgts := set(str(b) + str(p) + str((c := b * p)))): continue
    if len(dgts) == 9:
      print(f"answer: {c}p")
```
  LikeLike
- Hugo 12:29 pm on 12 May 2026 Permalink | Reply
  
  Keeping the pattern AB × CDE = FGHI but without the condition that A = 3, I found
  18 × 297 = 5346
  27 × 198 = 5346
  28 × 157 = 4396
  42 × 138 = 5796
  48 × 159 = 7632
  
  I didn’t try any other patterns, with or without the inclusion of 0.
  No doubt Jim can fill that gap for us.
  
  LikeLike
  - Tony Smith 11:04 am on 13 May 2026 Permalink | Reply
    
    12 x 483=5796
    
    LikeLike
  - Ruud 8:38 am on 14 May 2026 Permalink | Reply
    
    And
    4 x 1738 = 6952
    4 x 1963 =7852
    
    LikeLike
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Jim Randell 6:59 am on 1 March 2026 Permalink | Reply
Tags: by: Howard Williams
Teaser 3310: Building bridges
From The Sunday Times, 1st March 2026 [link] [link]

Chuck’s ranch is 16 km north of a narrow straight river flowing west to east. His cousin Dwayne has a ranch 72 km south and 30 km east of Chuck’s ranch. To join their ranches, they are each required to build a straight road from their ranch to a bridge that will be constructed over the river at a location yet to be decided.

Selfishly, Chuck wishes to have the bridge built at a location that would require Dwayne’s road to be as much as possible longer than the road he has to build on his side of the river. Dwayne wishes another location for the bridge, a location that would require the roads on both sides of the river to be of equal lengths.

What is the distance between these two bridge locations?

[teaser3310]
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- Jim Randell 7:12 am on 1 March 2026 Permalink | Reply
  (See also: Teaser 3286).
  
  Here is a numerical solution. (Which saves having to do any calculus).
  
  The following Python program runs in 72ms. (Internal runtime is 231µs).
```
from enigma import (hypot, find_max, find_zero, printf)

# calculate lengths of road for location of bridge x
# x = km E from C
C = lambda x: hypot(16, x)
D = lambda x: hypot(56, 30 - x)

# calculate how much more road D builds than C
f = lambda x: D(x) - C(x)

# find C's proposed bridge location (maximise f)
xC = find_max(f, -100, +130).v
printf("xC = {xC:.3f}")

# find D's proposed bridge location (f = 0)
xD = find_zero(f, -100, +130).v
printf("xD = {xD:.3f}")

# calculate the distance between the proposed locations
d = abs(xC - xD)
printf("diff = {d:.3f}")
```
  Solution: The proposed locations for the bridge are 75 km apart.
  
  C’s proposed location is 12 km west of the nearest point on the river to him. C’s road would be 20 km, and D’s road would be 70 km. (Total road length would be 90 km).
  
  (This corresponds to the position where the angles that the roads make with the river are equal. So if we mirrored one of the brothers ranches in the river, so that they are both on the same side, then there would be one straight road that joins the bridge to both ranches).
  
  D’s proposed location is 33 km east of the nearest point on the river to him. Both roads would be 65 km. (Total road length would be 130 km).
  
  The minimal total road length occurs if the bridge is built 6.66 km east of the nearest point on the river to C. C’s road would be 17.33 km, and D’s road would be 60.66 km. (Total road length = 78 km).
  
  (If we were to draw a straight line between the ranches, we achieve the shortest possible combined distance, and the position of the bridge is where the line crosses the river).
  
  Note that measuring the distances between points, we have:
  
  [C→D] × [XC→XD] = 78 × 75 = 5850
  [C→XC] × [D→XD] + [C→XD] × [D→XC] = 20 × 65 + 65 × 70 = 5850
  
  Which means the quadrilateral (C, XC, D, XD) is cyclic.
  
  Analytically:
  
  The two roads are the same length when:
  
  √(56² + (30 − x)²) = √(16² + x²)
  3136 + (900 − 60x + x²) = 256 + x²
  ⇒
  60x = 3780
  x = 63
  
  And we see both roads have length 65 km.
  
  To calculate the maximum of the function f(x) we can differentiate it, at look for stationary points (i.e. points where f'(x) = 0):
  
  f(x) = √(56² + (30 − x)²) − √(16² + x²)
  f'(x) = ((x − 30) / √(x² − 60x + 4036)) − (x / √(x² + 256))
  
  And we can find values when f'(x) is 0 by looking for values where:
  
  ((x − 30) / √(x² − 60x + 4036)) = (x / √(x² + 256))
  ⇒
  (x − 30)² (x² + 256) = x² (x² − 60x + 4036)
  (x² − 60x + 900) (x² + 256) = x² (x² – 60x + 4036)
  x⁴ − 60x³ + 1156x² − 15360x + 230400 = x⁴ − 60x³ + 4036x²
  ⇒
  2880x² + 15360x − 230400 = 0
  960(x + 12)(3x − 20) = 0
  x = −12; x = 20/3
  
  x = −12 is a stationary point that gives the maximum value of f(x) (= 50 km when D’s road is 70 km and C’s road is 20 km).
  
  (And x = 20/3 is not a stationary point of f(x), but does give the position of the bridge for the minimum combined distance for the roads).
  
  LikeLike
  - Jim Randell 8:48 am on 28 March 2026 Permalink | Reply
    
    Here is a way to find the maximum value of f(x) without using calculus:
    
    We reflect ranch D in the line of the river, so it appears at D′, north of the river.
    
    For a point on the river X the road distances XD and XD′ are the same length.
    
    We then construct the triangle CXD′, and using the triangle inequality we get:
    
    XC + CD′ ≥ XD′
    ⇒
    XD′ − XC ≤ CD′
    
    with equality occurring when the points are co-linear and the triangle is degenerate (zero area).
    
    but:
    
    XC = C(x)
    XD′ = D(x)
    
    Hence:
    
    f(x) = D(x) − C(x)
    f(x) = XD′ − XC
    f(x) ≤ CD′
    
    with equality occurring when X, C, D′ are co-linear.
    
    And so when X, C, D′ are co-linear the maximum value of f(x) is achieved, and this value is the distance CD′.
    
    The distance CD′ is calculated as the hypotenuse of a right-angled triangle where the other two sides are 30 and 56 − 16 = 40, and so CD′ = 50.
    
    This is a (3, 4, 5)×10 triangle.
    
    Since the points X, C, D′ are co-linear, the triangle with hypotenuse XD′ is just a larger version of a (3, 4, 5) triangle with the 4 side measuring 56 (= 4×14).
    
    Hence the base of the triangle is 3×14 = 42, which gives a point X on the river 12 km to the W of the closest point on the river to C (i.e. x = −12).
    
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Jim Randell 7:15 am on 4 January 2026 Permalink | Reply
Tags: by: Howard Williams

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.

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