Question

Number of 4 digit numbers where first 2 number are equal and last 2 are equal

Answer 1

If we let the four-digit number be XXYY, then this number can be expressed as:
1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2 (since it's a 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. So now our question is "which product of 11 and a perfect square looks like X0Y?" We can test them:
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 the bill is 704. This means there is only one four-digit number that works, and it's 7744

Answer 2

Answer:

Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb.