Determine all possible 8-digit perfect square(s) (in base 10) having the form ababbbcc, each containing no leading zeroes, that are divisible by 7.

Note: While a solution may be trivial with the aid of a computer program, show how to derive it without one.

The only perfect square of the form ababbbcc (leading zeroes not permitted*) is 57577744. It is also divisible by 7.

In the following analysis, I shall assume leading zeroes are not permitted and, thus, a cannot be 0.

All perfect squares that end with the same two digits end with either 00 or 44. Thus, in order for ababbbcc to be a perfect square, c must be 0 or 4. If c were 0, the pattern for the tentative perfect square would be ababbb00. As ababbb00 is divisible by the perfect square 100, the pattern ababbb would also need be a perfect square for ababbbcc to be a perfect square. By the same token that all perfect squares that end with the same two digits end with either 00 or 44, b would need be a 0 or 4. With b as 0, the pattern a0a000 would then need be that of a perfect square. And, likewise, as a0a000 is divisible by the perfect square 100, the pattern a0a0 would need be a perfect square.
It can can readily be seen, that a0a0 has the factors: a, 2, 5 and 101. It can also readily be seen that a, being a single digit, cannot, itself, have 2, 5 and 101 as factors, thus, a0a0 cannot be a perfect square. Therefore, ababbb, such that b is 0, cannot be a perfect square. If b were 4, then at least one of the following numbers would need be a perfect square: 141444, 242444, 343444, 444444, 545444, 646444, 747444, 848444 and 949444. In addition, as we are only concerned with a perfect square divisible by 7, the number modulo 7 must be 0. As none other than 444444 are either and 444444 is not a perfect square,  b cannot be 4. As b cannot be 0 or 4, c cannot be 0, and must, in order for ababbbcc to be a perfect square, be a 4. Therefore, it follows, the pattern for the tentative perfect square must be ababbb44.

All numbers of the pattern ababbb44 are divisible by the perfect square 4. Of the resulting quotient (which must be a perfect square in order for ababbb44 to be a perfect square), each will end in one of the following digit pair: 11 (when b is 0, 4, or 8), 36 (when b is 3 or 7), 61 (when b is 2 or 6), or 86 (when b is 1, 5 or 9). As no perfect square can end in 11 (as stated previously, only 00 or 44 exist as the ending digit twins for perfect squares) or 86, b must either be 2, 3, 6 or 7 for the tentative perfect square. This leaves the following patterns to be the tentative perfect square(s): a2a22244, a3a33344, a6a66644, and a7a77744,

Substituting each digit, 1 through 9, as a in the four patterns and taking the square root of each of the resulting 36 numbers, it is found that only 57577744 is a perfect square.

*Leading zeroes permitted, 00000000, where a, b and c are each zero, is both of the form ababbbcc and a perfect square, as well as divisible by 7.

Edited on September 21, 2008, 2:39 am

Posted by Dej Mar on 2008-09-20 12:15:50