Sunday Times Teaser 3200 – Puzzling Sum

by BRG on 19 January 2024

by Andrew Skidmore

Published Friday January 19 2024 (link)

Liam was given a set of addition sums as part of his homework. The teacher had cleverly chosen sums where the answer, together with the two numbers to be added, used all the digits from 1 to 9 just once (and there were no zeros). For instance, one of the sums was 193 + 275 = 468. I noticed that one of them was particularly interesting in that the two numbers to be added were perfect powers (squares, cubes etc.).

In the interesting sum, what are the two numbers that were to be added?

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BRG permalink


# powers of integers with four or less digits, none duplicated nor zero
pwrs = sorted(set(p for n in range(2, 100) for x in range(2, 14)
          if (p := n ** x) < 10000 and '0' not in (q := str(p))
                      and len(q) == len(set(q))))

for i, a in enumerate(pwrs):
  for b in pwrs[i:]:
    dgts = str(a) + str(b) + str(a + b)
    if '0' not in dgts and len(dgts) == len(set(dgts)) == 9:
      print(f"{a} + {b} == {a + b}")

# powers of integers with four or less digits, none duplicated nor zero

pwrs = sorted(set(p for n in range(2, 100) for x in range(2, 14)

if (p := n ** x) < 10000 and '0' not in (q := str(p))

and len(q) == len(set(q))))

for i, a in enumerate(pwrs):

for b in pwrs[i:]:

dgts = str(a) + str(b) + str(a + b)

if '0' not in dgts and len(dgts) == len(set(dgts)) == 9:

print(f"{a} + {b} == {a + b}")

Frits permalink


from math import log
# powers below 100 cause duplicate digits or zero digits 
# so we must have abc + def = ghi

m = 123   # minimum possible power
M = 876   # maximum possible power

pwrs = sorted(set(p for n in range(2, int(log(M, 2)) + 1) 
              for b in range(2, int(M**(1 / n)) + 1)
              if (p := b ** n) >= m and len(s := str(p)) == len(set(s)) and
              '0' not in s))

for i, p1 in enumerate(pwrs[:-1]):
  for p2 in pwrs[i + 1:]:
    dgts = str(p1) + str(p2) + (rhs := str(p1 + p2))
    if len(dgts) == len(set(dgts)) == 9 and '0' not in dgts:
      print(f"{p1} + {p2} == {rhs}")

from math import log

# powers below 100 cause duplicate digits or zero digits

# so we must have abc + def = ghi

m = 123 # minimum possible power

M = 876 # maximum possible power

pwrs = sorted(set(p for n in range(2, int(log(M, 2)) + 1)

for b in range(2, int(M**(1 / n)) + 1)

if (p := b ** n) >= m and len(s := str(p)) == len(set(s)) and

'0' not in s))

for i, p1 in enumerate(pwrs[:-1]):

for p2 in pwrs[i + 1:]:

dgts = str(p1) + str(p2) + (rhs := str(p1 + p2))

if len(dgts) == len(set(dgts)) == 9 and '0' not in dgts:

print(f"{p1} + {p2} == {rhs}")

John Z permalink


from itertools import combinations

# get powers with up to 4 digits (not worrying about repeats here)
# i.e. a one digit number plus a four digit number can give a four digit sum
# for a total of nine digits. But a one digit number plus a five digit number
# will give at least a five digit sum for a total of eleven digits
a = set(x for n in range(2, 99) for e in range(2, 14) if (x := n **e) < 9999)
     
for a1, a2 in combinations(a, 2):
  
  ss = str(a1) + str(a2) + str(a1 + a2) + '0'
  
  # are there 10 different digits in the sum including the added zero?
  if len(ss) == 10 == len(set(ss)):
    print(f'{a1} + {a2} = {a1 + a2}')

from itertools import combinations

# get powers with up to 4 digits (not worrying about repeats here)

# i.e. a one digit number plus a four digit number can give a four digit sum

# for a total of nine digits. But a one digit number plus a five digit number

# will give at least a five digit sum for a total of eleven digits

a = set(x for n in range(2, 99) for e in range(2, 14) if (x := n **e) < 9999)

for a1, a2 in combinations(a, 2):

ss = str(a1) + str(a2) + str(a1 + a2) + '0'

# are there 10 different digits in the sum including the added zero?

if len(ss) == 10 == len(set(ss)):

print(f'{a1} + {a2} = {a1 + a2}')

John Z permalink

With printout of the base b and power n resulting in the two numbers of the solution


from itertools import combinations

# get powers with up to 4 digits; base b and power n stored for later printout
xne = [(x, b, n) for b in range(2, 99)
       for n in range(2, 14) if (x := b ** n) < 9999]

for (a1, b1, n1), (a2, b2, n2) in combinations(xne, 2):
  
  ss = str(a1) + str(a2) + str(a1 + a2) + '0'
  
  # are there 10 different digits in the sum including the added zero?
  if len(ss) == 10 == len(set(ss)):
    print(f'{a1} + {a2} = {a1 + a2}')
    print(f'{b1} ** {n1}, {b2} ** {n2}')

from itertools import combinations

# get powers with up to 4 digits; base b and power n stored for later printout

xne = [(x, b, n) for b in range(2, 99)

for n in range(2, 14) if (x := b ** n) < 9999]

for (a1, b1, n1), (a2, b2, n2) in combinations(xne, 2):

ss = str(a1) + str(a2) + str(a1 + a2) + '0'

# are there 10 different digits in the sum including the added zero?

if len(ss) == 10 == len(set(ss)):

print(f'{a1} + {a2} = {a1 + a2}')

print(f'{b1} ** {n1}, {b2} ** {n2}')

Sunday Times Teaser 3200 – Puzzling Sum

by Andrew Skidmore

Published Friday January 19 2024 (link)

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