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Jim Randell 10:00 am on 4 November 2025 Permalink | Reply
Tags: by: Graham Smithers ( 84 )
Teaser 2512: [A square, a cube, a prime]
From The Sunday Times, 14th November 2010 [link] [link]

Using each of the digits 0 to 9 exactly once, I have written down two three-digit numbers and a four-digit number.

In no particular order, one of them is a perfect square, one is a perfect cube, and the other is a prime.

If I told you which of the square, the cube and the prime was the smallest then it would be possible for you to work out what those three numbers are.

What are they?

This puzzle was originally published with no title.

[teaser2512]
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ruudvanderham and Jim Randell are discussing. Toggle Comments
- Jim Randell 10:01 am on 4 November 2025 Permalink | Reply
  I wondered if we are meant to exclude numbers that fall into more than one category (i.e. 729 and 4096 are both squares and cubes). It turns out you get the same answer if you exclude them or not.
  
  I recently extended the [[ choose() ]] function in enigma.py, so that you can specify different collections for each choice, and we can use that functionality to solve this puzzle.
  
  The following Python program runs in 88ms. (Internal runtime is 11ms).
```
from enigma import (powers, primes, is_duplicate, choose, filter_unique, printf)

# find 3 and 4 digit, squares, cubes, primes with no repeated digits
def numbers(seq):
  (n3s, n4s) = (list(), list())
  for n in seq:
    if n < 100: continue
    if n > 9999: break
    if is_duplicate(n): continue
    (n3s if n < 1000 else n4s).append(n)
  return (n3s, n4s)

(sq3, sq4) = numbers(powers(10, 99, 2))  # squares
(cb3, cb4) = numbers(powers(5, 21, 3))  # cubes
(pr3, pr4) = numbers(primes.between(100, 9999))  # primes

# find combinations of (square, cube, prime) that use 10 different digits
check = lambda *ns: not is_duplicate(*ns)
ss = list()
for nss in [(sq3, cb3, pr4), (sq3, cb4, pr3), (sq4, cb3, pr3)]:
  ss.extend(choose(nss, check, multi_vs=1, multi_fns=0))

# find candidates unique by the category of the smallest number
rs = filter_unique(ss, (lambda ss: ss.index(min(ss)))).unique

# output solution
for (sq, cb, pr) in rs:
  printf("square = {sq}, cube = {cb}, prime = {pr}")
```
  Solution: The numbers are: 324 (square), 6859 (cube), 701 (prime).
  
  There are 19 candidate sets of numbers (or 16 if you remove those containing numbers that are both squares and cubes), but only one of them is uniquely identified if we are told that the square is the smallest number.
  
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- ruudvanderham 12:09 pm on 4 November 2025 Permalink | Reply
```
import istr
import collections

squares = [i for i in istr.range(100, 10000) if i.all_distinct() and i.is_square()]
cubes = [i for i in istr.range(100, 10000) if i.all_distinct() and i.is_cube()]
primes = [i for i in istr.range(100, 10000) if i.all_distinct() and i.is_prime()]

solutions = collections.defaultdict(list)
for square in squares:
    for cube in cubes:
        for prime in primes:
            if len(square | cube | prime) == 10 and (square | cube | prime).all_distinct():
                solutions[(square == min(square, cube, prime), cube == min(square, cube, prime), square == min(square, cube, prime))].append(
                    (square, cube, prime)
                )

for solution in solutions.values():
    if len(solution) == 1:
        print(solution[0])
```
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Jim Randell 7:02 am on 2 November 2025 Permalink | Reply
Tags: by: Victor Bryant ( 147 )
Teaser 3293: Three sisters
From The Sunday Times, 2nd November 2025 [link] [link]

I have three granddaughters Jay, Kay and Elle. I set Jay and Kay a test and asked each of them to produce a list of positive numbers that used just nine digits [in total] with no digit occurring more than once. I wanted the majority of the numbers in each list to be odd and the majority of the numbers in each list to be perfect squares.

Jay’s list (which contained more numbers than Kay’s) added up to give the year of Elle’s birth, whereas Kay’s list added up to give the year in which Elle will be 25.

In one of the lists the highest number was a perfect square.

What (in increasing order) were the numbers in that list?

[teaser3293]
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- Jim Randell 8:36 am on 2 November 2025 Permalink | Reply
  Assuming the puzzle is set at the current time (i.e. during 2025). Then the latest Elle can be born is 2025, and if she has not reached her 25th birthday yet, the earliest she can be born is 2000. So Jay’s numbers must sum to between 2000 and 2025. (And Kay’s numbers sum to between 2025 and 2050).
  
  I also assumed that Jay and Kay completed their test successfully, and the lists of numbers they produced satisfied the given requirements.
  
  My first program was a bit slow. This one is a bit more involved, but both find the same answer.
  
  The following Python program runs in 191ms. (Internal runtime is 121ms).
```
from enigma import (
  defaultdict, irange, powers, nsplit, seq_is_distinct, nconcat,
  divc, subsets, disjoint_union, intersect, cproduct, printf
)

# map squares to digit contents
sq = dict()
for s in powers(1, 45, 2):
  ds = nsplit(s)
  if seq_is_distinct(ds):
    sq[s] = ds

# available digits
digits = set(irange(0, 9))

# allocate remaining digits <rs> for units digits <us>
def allocate(us, rs, ns=list()):
  k = len(rs)
  if k == 1:
    # there should be 1 digit remaining unused
    yield us + ns
  elif k > 1 and us:
    # choose some digits to add to the next unit digit
    u = us[0]
    for ds in subsets(rs, max_size=3, select='P', fn=list):
      if ds and ds[0] == 0: continue
      n = nconcat(ds + [u])
      if 0 < n <= 2050:
        yield from allocate(us[1:], rs.difference(ds), ns + [n])

# record candidate sequences for J and K
(Js, Ks) = (defaultdict(set), defaultdict(set))

# consider the length of the list
for k in irange(3, 6):
  # majority size
  m = divc(k + 1, 2)

  # choose some squares
  for sqs in subsets(sq.keys(), size=m, fn=list):
    # digits used
    ds = disjoint_union(sq[s] for s in sqs)
    if not ds: continue

    # count odd numbers so far, and remaining digits
    odd = sum(s % 2 for s in sqs)
    rs = digits.difference(ds)

    # choose units digits for the remaining numbers
    for us in subsets(rs, size=k - m, fn=list):
      if odd + sum(u % 2 for u in us) < m: continue
      # allocate the remaining digits
      for ns in allocate(us, rs.difference(us)):
        ss = tuple(sorted(sqs + ns))
        t = sum(ss)
        # record candidates by birth year
        if 2000 <= t <= 2025: Js[t].add(ss)
        if 2025 <= t <= 2050: Ks[t - 25].add(ss)

# find solutions
for j in intersect([Js.keys(), Ks.keys()]):
  for (J, K) in cproduct([Js[j], Ks[j]]):
    if len(J) > len(K) and (max(J) in sq or max(K) in sq):
      printf("J={J} [{j}]; K={K} [{k}]", k=j + 25)
```
  Solution: The numbers are: 305, 784, 961.
  
  Jay’s list is:
  
  [4, 9, 625, 1387]; sum = 2025
  (4 numbers; 3 squares; 3 odd)
  
  So Elle was born this year (2025), and will be 25 in the year 2050.
  
  Kay’s list is:
  
  [305, 784, 961]; sum = 2050
  (3 numbers; 2 squares; 2 odd)
  
  In Kay’s list the highest number (961) is a square (= sq(31)), and so this list is the required answer.
  
  LikeLike
- Jim Olson 4:11 pm on 10 November 2025 Permalink | Reply
  
  I am perhaps missing something but I can’t see how the solution meets the requirements of the puzzle. Each list is supposed to be a list of positive numbers. Zero is not a positive number.
  
  LikeLike
  - Jim Randell 4:33 pm on 10 November 2025 Permalink | Reply
    
    @Jim: The numbers in the lists are all positive ((4, 9, 625, 1387) and (305, 784, 961)), but that doesn’t mean the individual digits have to be non-zero. Many positive numbers are written using a 0 digit.
    
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Jim Randell 7:21 am on 31 October 2025 Permalink | Reply
Tags: by: Graham Smithers ( 84 )
Teaser 2470: [Peter’s birthday]
From The Sunday Times, 24th January 2010 [link]

Peter told me that one of his teenage birthdays had occurred on the same day of the week as the day that he was born.

Furthermore, on that birthday the 10 digits that made up his age, the date (dd/mm/yy) and the year of his birth (yy) were all different.

Knowing these facts, and that Peter was born after the summer solstice, I was able to work out his date of birth.

What is Peter’s date of birth?

This puzzle was originally published with no title.

[teaser2470]
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- Jim Randell 7:21 am on 31 October 2025 Permalink | Reply
  The birthday must be 13 – 19 (teenage), so the 1 is already used. This means the month must be chosen from 02 – 09, so the 0 is already used. And this means that the date must be 23 – 29, so the 2 is already used. So we know where the digits 0, 1, 2 are used.
  
  The following run file executes in 139ms. (Internal runtime of the generated code is 496µs).
```
#! python3 -m enigma -rr

SubstitutedExpression

--invalid=""  # allow leading zeros

# date of birth is: AB/CD/EF
# date of birthday is: AB/CD/GH
# age is: IJ (13..17)
"EF + IJ = GH"

# valid dates
"0 < AB < 32"  # day
"5 < CD < 13"  # month (in the second half of the year)
"12 < IJ < 20" # age
"(6, 20) < (CD, AB) < (12, 21)" # between solstices

# birthdate and birthday are same day of week
--code="from datetime import date"
--code="dow = lambda d, m, y: date(1900 + y, m, d).isoweekday()"
"dow(AB, CD, EF) == dow(AB, CD, GH)"

# [optional] assign known values
--assign="C,0"
--assign="I,1"
--assign="A,2"

# [optional] neaten up output
--template="AB/CD/EF -> AB/CD/GH; age = IJ"
--solution=""
--verbose="1-H"
```
  Solution: Peter’s date of birth is 23rd September 1948.
  
  Which was a Thursday.
  
  Peter’s 17th birthday was on Thursday, 23rd September 1965.
  
  An alternative that would work is Monday, 29th March 1948, but this is ruled out by the condition Peter’s birthday is after the summer solstice (in the northern hemisphere).
  
  LikeLike
- Ruud 6:56 am on 1 November 2025 Permalink | Reply
```
import datetime


day0 = datetime.datetime(1900, 1, 1)

while day0.year < 2020:
    day0 += datetime.timedelta(1)
    for age in range(10, 20):
        try:
            day1 = datetime.datetime(day0.year + age, day0.month, day0.day)
            if day0.weekday() == day1.weekday():
                s = f"{day1.year}"[2:] + f"{day1.month:02d}" + f"{day1.day:02d}" + f"{day0.year}"[2:] + f"{age:02d}"
                if len(set(s)) == 10:
                    print(f"Born: {day0:%Y-%m-%d} Age:{age}")
        except ValueError:
            ...
```
  LikeLike

Jim Randell 9:22 am on 28 October 2025 Permalink | Reply
Tags: by: Andrew Skidmore ( 88 )

Teaser 2497: [Measuring jugs]

From The Sunday Times, 1st August 2010 [link] [link]

I have two jugs that hold whole numbers of pints, and together they hold two gallons [= 16 pints]. Using only the two jugs and a tap, discarding water as necessary, I can end up with a total of any whole number of pints from 1 to 16 in the two jugs combined. If I want to end up with exactly one pint in the most economical way possible, I have to draw more water than if I want to end up with exactly four pints.

What are the capacities of the two jugs?

This puzzle was originally published with no title.

[teaser2497]

Jim Randell 9:22 am on 28 October 2025 Permalink | Reply

This Python program runs in 73ms. (Internal runtime is 450µs).

from enigma import (Enumerator, irange, printf)

# solve for capacities A and B
# n = max number of steps to consider
def solve(A, B, n=50):
  k = 0
  r = dict()  # r maps (vA + vB) positions to minimal measure (volume drawn)
  p = dict()  # p maps (vA, vB) to minimal measure
  vs = {(0, 0, 0)}  # (vol in A, vol in B, vol drawn)
  while vs and k < n:
    ss = Enumerator(vs)
    vs = set()
    for (vA, vB, vD) in ss:
      # skip positions where we have already check a lower measure
      if (vA, vB) in p and p[(vA, vB)] <= vD: continue
      p[(vA, vB)] = vD

      # check current amounts
      vT = vA + vB
      # record vT -> vD in r, if vD is minimal
      if vT not in r or (vD < r[vT]): r[vT] = vD

      # move some water around:
      # fill A from the tap
      if vA < A:
        vs.add((A, vB, vD + A - vA))
      # fill B from the tap
      if vB < B:
        vs.add((vA, B, vD + B - vB))
      # discard A
      if vA:
        vs.add((0, vB, vD))
      # discard B
      if vB:
        vs.add((vA, 0, vD))
      # transfer water from A to B
      if vA and vB < B:
        if vT > B:
          vs.add((vT - B, B, vD))
        else:
          vs.add((0, vT, vD))
      # transfer water from B to A
      if vB and vA < A:
        if vT > A:
          vs.add((A, vT - A, vD))
        else:
          vs.add((vT, 0, vD))
    # move to the next step
    k += 1
  # return result
  return r

# make all values from 1 to 16 pints
ts = set(irange(1, 16))

# consider the capacity of the smaller jug
for A in irange(1, 8):
  # capacity of the larger jug
  B = 16 - A
  # find minimal configurations
  r = solve(A, B)
  # check we can make all target values
  if not ts.issubset(r.keys()): continue
  # and that making 1 pint requires drawing more water than 4 pints
  if not (r[1] > r[4]): continue

  # output configuration
  printf("capacities = [{A}, {B}]")
  for k in irange(1, 16):
    printf("  {k} pints -> {d} pints drawn", d=r[k])
  printf()

Solution: The jugs have capacities of 7 pints and 9 pints.

We can make 1 pint by drawing 28 pints (and discarding 27 pints) as follows (13 steps):

fill 7 → (7/7, 0/9; 7 total; 7 drawn)
pour 7 into 9 → (0/7, 7/9; 7 total; 7 drawn)
fill 7 → (7/7, 7/9; 14 total; 14 drawn)
pour 7 into 9 → (5/7, 9/9; 14 total; 14 drawn)
discard 9 → (5/7, 0/9; 5 total; 14 drawn)
pour 7 into 9 → (0/7, 5/9; 5 total; 14 drawn)
fill 7 → (7/7, 5/9; 12 total; 21 drawn)
pour 7 into 9 → (3/7, 9/9; 12 total; 21 drawn)
discard 9 → (3/7, 0/9; 3 total; 21 drawn)
pour 7 into 9 → (0/7, 3/9; 3 total; 21 drawn)
fill 7 → (7/7, 3/9; 10 total; 28 drawn)
pour 7 into 9 → (1/7, 9/9; 10 total; 28 drawn)
discard 9 → (1/7, 0/9; 1 total; 28 drawn)

And we can make 4 pints by drawing 18 pints (and discarding 14 pints) as follows (7 steps):

fill 9 → (0/7, 9/9; 9 total; 9 drawn)
pour 9 into 7 → (7/7, 2/9; 9 total; 9 drawn)
discard 7 → (0/7, 2/9; 2 total; 9 drawn)
pour 9 into 7 → (2/7, 0/9; 2 total; 9 drawn)
fill 9 → (2/7, 9/9; 11 total; 18 drawn)
pour 9 into 7 → (7/7, 4/9; 11 total; 18 drawn)
discard 7 → (0/7, 4/9; 4 total; 18 drawn)

There are a couple of shortcuts we could use:

If the capacities have a GCD > 1, then it will be impossible to make an amount that is not a multiple of that GCD, so we could just consider co-prime capacities.

If one of the capacities is 1 (and the other is 15), and we we can make all amounts from 1 – 16 without any wastage, so we could skip checking this arrangement.

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