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Pandigital Product Crossed Square Poser (Posted on 2023-01-06) Difficulty: 3 of 5
Each of A, B, and C is a different zeroless pandigital number that satisfy this equation:

A*B=C2

Determine the minimum value of C.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 18
The minimum value for 'C' I found was 231597684

A: 164938572
B: 325196748
C: 231597684
A*B = C^2 = 53637487234163856

-----------
digs = '123456789'

pandigitals = []
squares = []
from itertools import permutations
for perm in permutations(digs):
    apan = int(''.join(perm))
    pandigitals.append(apan)
    squares.append(apan**2)
pandigitals.sort()
squares.sort()

solutions = []
found = False
for s in squares:
    c = int(s**.5)
    for a in pandigitals:
        if a > c:
            break
        b = s/a
        if a == b:
            continue
        if b%1 != 0:
            continue
        if int(b) in pandigitals:
            print(a,int(b),c,s)
            solutions.append([a,int(b),c,s])
            found = True
    if found:
        break
  Posted by Larry on 2023-01-06 10:45:51
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