Sunday Times Teaser 3302 – Challenging Grandson

by BRG on January 4, 2026

by Howard Williams

Published Sunday January 04 2026 (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 (eg, 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?

From → Uncategorized

4 Comments Leave one →

BRG permalink


from itertools import permutations
from collections import defaultdict

# number abcd -> add all digit pairs:
# ab + ba + ca + da )
# ac + bc + cb + db ) = 33.(a + b + c + d) 
# ad + bd + cd + dc )
#
# similarly for abcde -> 44.(a + b + c + d + e)

# map the five digit multiple length to possible solutions
d = defaultdict(list)

# consider the possible four digits of the number
for c4 in permutations(range(10), 4):
  if c4[0] == 0:
    continue
  n4 = 1000 * c4[0] + 100 * c4[1] + 10 * c4[2] + c4[3]                
  # which are mulliples of the sum of all two digit numbers
  # formed from all permutations of two from the four digits
  if n4 % (33 * sum(c4)) == 0:
    # now add a different digit to the end of the number    
    for d5 in set(range(10)).difference(c4):
      n5 = 10 * n4 + d5
      # and find its multiple of the same sum over five digits
      m, r = divmod(n5, 44 * sum(c4 + (d5,)))
      if not r:
        d[len(str(m))].append(n5)

for k, v in d.items():
  if len(v) == 1:
    print(f"His five digit number was {v[0]}.")

from itertools import permutations

from collections import defaultdict

# number abcd -> add all digit pairs:

# ab + ba + ca + da )

# ac + bc + cb + db ) = 33.(a + b + c + d)

# ad + bd + cd + dc )

# similarly for abcde -> 44.(a + b + c + d + e)

# map the five digit multiple length to possible solutions

d = defaultdict(list)

# consider the possible four digits of the number

for c4 in permutations(range(10), 4):

if c4[0] == 0:

continue

n4 = 1000 * c4[0] + 100 * c4[1] + 10 * c4[2] + c4[3]

# which are mulliples of the sum of all two digit numbers

# formed from all permutations of two from the four digits

if n4 % (33 * sum(c4)) == 0:

# now add a different digit to the end of the number

for d5 in set(range(10)).difference(c4):

n5 = 10 * n4 + d5

# and find its multiple of the same sum over five digits

m, r = divmod(n5, 44 * sum(c4 + (d5,)))

if not r:

d[len(str(m))].append(n5)

for k, v in d.items():

if len(v) == 1:

print(f"His five digit number was {v[0]}.")

John Z. permalink


sols = [[] for i in range(5)]

# the five digit number must be divisible by 44, cannot start with 0
for n5 in range(10000 // 44 * 44 + 44, 100000, 44):
  # four digit number must be divisible by 33, 5 different digits
  if (n4 := n5 // 10) % 33  == 0 and len(set(d5 := tuple(str(n5)))) == 5:
    #separate out the five digits
    d5 = tuple(map(int, d5))
    # apply the division by combinations of digits criterion
    if n4 % (33 * sum(d5[:-1])) == 0 and n5 % (44 * sum(d5)) == 0:
      # get the multiplier
      mp = n5 // (44 * sum(d5))
      # save possible solutions indexed by number of digits in multiplier 
      sols[len(str(mp))] += [n5]

# the solution has only one possible multiplier 
print(*min(sols, key = lambda x: abs(len(x) - 1)))

sols = [[] for i in range(5)]

# the five digit number must be divisible by 44, cannot start with 0

for n5 in range(10000 // 44 * 44 + 44, 100000, 44):

# four digit number must be divisible by 33, 5 different digits

if (n4 := n5 // 10) % 33 == 0 and len(set(d5 := tuple(str(n5)))) == 5:

#separate out the five digits

d5 = tuple(map(int, d5))

# apply the division by combinations of digits criterion

if n4 % (33 * sum(d5[:-1])) == 0 and n5 % (44 * sum(d5)) == 0:

# get the multiplier

mp = n5 // (44 * sum(d5))

# save possible solutions indexed by number of digits in multiplier

sols[len(str(mp))] += [n5]

# the solution has only one possible multiplier

print(*min(sols, key = lambda x: abs(len(x) - 1)))

John Z. permalink

line 19 can be written more succinctly as:

Python

print(*(s[0] for s in sols if len(s) == 1))

1
2
3

print(*(s[0] for s in sols if len(s) == 1))

Reply

Frits permalink

Only considering 7 values for the four-digit number, reusing some code of John.


# 5-digit number = ABCDE
# E must be zero due to the alternating sum rule (divisibility rule for 11)
# D must be even as 10 * ABCD must be divisible by 44
# s = A + B + C + D is a multiple of 2 (as 33.s divides ABCD) and also
# a multiple of 3 (as ABCD is a multiple of 3) so s is a multiple of 6
# m1 = ABCD / (33.s)  or ABCD = 198.k.m1  with s = A + B + C + D = 6.k
# m2 = ABCDE / (44.s) or 10.ABCD = 264.k.m2
# m2 is a multiple of 15 (m2 = 7.5 * m1), m2 = 15.p
# 10.ABCD = 3960.k.p so ABCD must be a multiple of 396

# first 4-digit multiple of 396 is 1188
# if you keep adding 99 to a number 1188 (till 9900) the digit sum remains 18.
# proof from Google AI-modus: n = 1188 + 99.k (integer k, 0 <= k <= 88)
# n = 99.(k + 12), m = k + 12 (12 <= m <= 100)
#   m = 100: n = 9900 digit sum = 18
#   m < 100: m = 10.a + b
#     b = 0: n = 990.a = 1000.(a-1) + 100.9 + 10.(10-a) + 0
#            digit sum = (a-1 + 9 + 10-a + 0) = 18       valid coefficients
#     b > 0: n = 990.a + 99.b = 1000.a + 100.(b-1) + 10.(9-a) + 10-b
#            digit sum = (a + b-1 + 9-a + 10-b) = 18     valid coefficients

# so ABCD must have a digit sum of 18 --> ABCD = 594.m1 and 10.ABCD = 792.m2
# m1 must be even so ABCD is a multiple of 1188

# the solution has only one possible multiplier
print("answer:", *min([[10 * n4 for n4 in range(2 * 1188, 10000, 1188)
      if len(set('0' + str(n4))) == 5 and len(str(int(5 * n4 / 396))) == nd]
      for nd in range(1, 5)], key = lambda x: abs(len(x) - 1)))

# 5-digit number = ABCDE

# E must be zero due to the alternating sum rule (divisibility rule for 11)

# D must be even as 10 * ABCD must be divisible by 44

# s = A + B + C + D is a multiple of 2 (as 33.s divides ABCD) and also

# a multiple of 3 (as ABCD is a multiple of 3) so s is a multiple of 6

# m1 = ABCD / (33.s) or ABCD = 198.k.m1 with s = A + B + C + D = 6.k

# m2 = ABCDE / (44.s) or 10.ABCD = 264.k.m2

# m2 is a multiple of 15 (m2 = 7.5 * m1), m2 = 15.p

# 10.ABCD = 3960.k.p so ABCD must be a multiple of 396

# first 4-digit multiple of 396 is 1188

# if you keep adding 99 to a number 1188 (till 9900) the digit sum remains 18.

# proof from Google AI-modus: n = 1188 + 99.k (integer k, 0 <= k <= 88)

# n = 99.(k + 12), m = k + 12 (12 <= m <= 100)

# m = 100: n = 9900 digit sum = 18

# m < 100: m = 10.a + b

# b = 0: n = 990.a = 1000.(a-1) + 100.9 + 10.(10-a) + 0

# digit sum = (a-1 + 9 + 10-a + 0) = 18 valid coefficients

# b > 0: n = 990.a + 99.b = 1000.a + 100.(b-1) + 10.(9-a) + 10-b

# digit sum = (a + b-1 + 9-a + 10-b) = 18 valid coefficients

# so ABCD must have a digit sum of 18 --> ABCD = 594.m1 and 10.ABCD = 792.m2

# m1 must be even so ABCD is a multiple of 1188

# the solution has only one possible multiplier

print("answer:", *min([[10 * n4 for n4 in range(2 * 1188, 10000, 1188)

if len(set('0' + str(n4))) == 5 and len(str(int(5 * n4 / 396))) == nd]

for nd in range(1, 5)], key = lambda x: abs(len(x) - 1)))

Sunday Times Teaser 3302 – Challenging Grandson

by Howard Williams

Published Sunday January 04 2026 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta