Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 such that each of EEN, VIER and NEGEN is a perfect square. None of the numbers can contain any leading zero.

Disregarding the non leading zero condition, if we additionally impose the restriction that GIVEN is divisible by 23, then what will be the corresponding substitution?

A brief program gives the two solutions in well under a second of execution time.

Original: (EEN, VIER, NEGEN) 441, 3249, and 14641 are perfect squares.

Expanded: (GIVEN added, leading zeroes allowed) 004, 2601, 40804, with GIVEN = 86204 divisible by 23.

Posted by ed bottemiller on 2010-08-17 15:12:56