Sunday Times Teaser 3255 – A Square for all Digits

by BRG on February 9, 2025

by Mark Valentine

Published Sunday February 09 2025 (link)

I have written down a set of whole numbers, each greater than 1 and less than 100. I have squared each number and written the answers down in a list. The list contains all of the digits from 0 to 9 precisely once. Interestingly, for any two of the numbers, there is no common factor other than 1.

Naturally the sum of the digits in the squared set is 45, but what is the sum of the digits in the original set of numbers that I wrote down?

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BRG permalink


from math import gcd

# map integers in 2..100 whose squares have no duplicated digits
# to the squares set of digits
d2ss = {n:t for n in range(2, 100) if len(s := str(n * n)) == len(t := set(s))}

# <ns> is the list of numbers, <nsq> is the list of numbers
# squared and <dgts> is the set of digits used so far
def solve(d2ss, ns=(), nsq=(), dgts=set()):
  # are all digits used once?
  if dgts == set('0123456789'):
    yield ns, nsq
  else:
    # consider numbers less than one hundred
    for n in range(2 if not ns else (ns[-1] + 1), 100):
      if n not in d2ss:
        continue
      # consider squares with no repeated digits, no
      # digits shared with previous squares and ...
      if not d2ss[n] & dgts:
        # ... no common factors other than one with previous squares
        if all(gcd(n * n, x) == 1 for x in nsq):
          yield from solve(d2ss, ns + (n,), nsq + (n * n,), dgts | d2ss[n])

for ns, nsq in solve(d2ss):
  ds = sum(sum(int(d) for d in str(n)) for n in ns)
  print(f"digit sum = {ds} [{ns} {nsq}]")

from math import gcd

# map integers in 2..100 whose squares have no duplicated digits

# to the squares set of digits

d2ss = {n:t for n in range(2, 100) if len(s := str(n * n)) == len(t := set(s))}

# <ns> is the list of numbers, <nsq> is the list of numbers

# squared and <dgts> is the set of digits used so far

def solve(d2ss, ns=(), nsq=(), dgts=set()):

# are all digits used once?

if dgts == set('0123456789'):

yield ns, nsq

else:

# consider numbers less than one hundred

for n in range(2 if not ns else (ns[-1] + 1), 100):

if n not in d2ss:

continue

# consider squares with no repeated digits, no

# digits shared with previous squares and ...

if not d2ss[n] & dgts:

# ... no common factors other than one with previous squares

if all(gcd(n * n, x) == 1 for x in nsq):

yield from solve(d2ss, ns + (n,), nsq + (n * n,), dgts | d2ss[n])

for ns, nsq in solve(d2ss):

ds = sum(sum(int(d) for d in str(n)) for n in ns)

print(f"digit sum = {ds} [{ns} {nsq}]")

John Z. permalink

Brilliant solution, Brian

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John Z permalink


from itertools import combinations
from math import gcd

for c in range(2, 4): # size of set of numbers

  for nc in combinations(range(2, 100), c):
    
    ncs = tuple(n * n for n in nc) # numbers squared
    if len(ad := ''.join(str(n) for n in ncs)) == 10 == len(set(ad)):
      
      if all(gcd(*r2) == 1 for r2 in combinations(nc, 2)):

        # digits of numbers in set
        dd = tuple(int(s) for s in ''.join(str(n) for n in nc))
        print(nc, ncs, 'Sum of digits:', sum(dd))

from itertools import combinations

from math import gcd

for c in range(2, 4): # size of set of numbers

for nc in combinations(range(2, 100), c):

ncs = tuple(n * n for n in nc) # numbers squared

if len(ad := ''.join(str(n) for n in ncs)) == 10 == len(set(ad)):

if all(gcd(*r2) == 1 for r2 in combinations(nc, 2)):

# digits of numbers in set

dd = tuple(int(s) for s in ''.join(str(n) for n in nc))

print(nc, ncs, 'Sum of digits:', sum(dd))

Sunday Times Teaser 3255 – A Square for all Digits

by Mark Valentine

Published Sunday February 09 2025 (link)

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