Consider all four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
Explanation:
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).
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