Consider 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?
Since first two and last two digits are equal, let the four-digit number be XXYY
This number can be expressed as:
1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k2 (perfect square)
In order for this to be true, 100X + Y must be the product of 11 and a perfect square, and looks like X0Y.
11 x 16 = 176; 11 x 25 = 275; 11 x 36 = 396; 11 x 49 = 593; 11 x 64 = 704; 11 x 81 = 891
The only one that fits is 704. This means there is only one four-digit number that works, and it is 7744.
The correct option is A.