Determine all possible triplet(s) (P, Q, N), where each of P and Q is a five digit positive base ten integer with P > Q and P*Q = N, such that:

• P and Q together contain each of the digits from 0 to 9 exactly once, and:
• N contains each of the digits from 0 to 9 exactly once, and:
• N² contains each of the digits from 0 to 9 exactly twice. Neither N nor N² can contain any leading zero.

(In reply to computer solution by Charlie)

The triplet (62037, 54981, 3410856297) is unique where the two non-leading zero 5-digit factors are — which taken together — pandigital, and the product is non-leading zero pandigital, and the square of the product contains each of the digits from 0 to 9 exactly twice (as stated in mathworld.wolfram.com reference: www.worldofnumbers.com/ninedigits.htm)


54981 * 62037 = 3410856297
3410856297² = 11633940678784552209

Posted by Dej Mar on 2009-06-03 02:02:37