A number AABB is the square of an integer. Find this integer, aided by pen and paper. No other calculating aids allowed.

By the problem, 1100*A + 11*B = 11(100*A + B) is a perfect square.

Since 11 is a prime, it follows iff (100*A + B)/11 is an integer.
But, 11(100*A + B) is a perfect square, and accordingly
(100*A + B)/11 must be a perfect square.

Now, (100*A + B) = 11*(9A) + (A+B) is divisible by 11 if (A+B)
is equal to either Zero Or (A+B) is equal to 11 or its multiple.

Now, A+B = 0 gives (A, B) = (0, 0), a contradiction.

Also, 0< = A, B <= 9, so that 0<= A+B< = 18, and accordingly
A + B = 11, so that B = 11 - A

Thus, (100*A + B)/11 = (100*A + 11 - A)/11 = 9*A + 1 which
is a perfect square iff A = 7, giving B = 11- 7 = 4

Consequently, the required number is 7744.

Posted by K Sengupta on 2007-10-23 05:31:11