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?

- 1
- 2
- 3
- 4

**Answer**

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) = k^{2} (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.**