CAT 2007 — QA Question 10

Let aabb (a ≠ 0, a and b being single digits) be a perfect square.
⇒ aabb = 1000a + 100a + 10b + b = 1100a + 11b = 11(100a + b)

Also, as aabb is a perfect square, it has to be a multiple of 121.

∴ aabb = 121K, where K is also perfect square.

For K = 4, aabb is a 3 digit number, while for K > 82, K is a 5 digit number.

For 81 ≥ K ≥ 9,
121 × 9 = 1089
121 × 16 = 1936
121 × 25 = 3025
121 × 36 = 4356
121 × 49 = 5929
121 × 64 = 7744
121 × 81 = 9801

∴ There is only one number 7744 of the form aabb, which is a perfect square.

Hence, option (e).