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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. $3$
2. $2$
3. $4$
4. $1$

Let the four-digit number be $\text{aabb}.$

The number we can write $1000 \; \text{a} + 100 \; \text{a} + 10 \; \text{b} + \text{b}. = 1100 \; \text{a} + 11 \; \text{b} = 11(100 \; \text{a} + \text{b}) \longrightarrow(1)$

This number will be perfect square if a number in the form of $\text{a0b}$ is divisible by $11$ and when $\text{a0b}$ is divided by $11$ perfect square will be get.

$\text{a0b}$ will be divisible by $11$ when $a+b=11$

$\boxed{\text{ b=11-a }}$

put the value of $\text{a}$ in equation $(1).$

$\Rightarrow 11(100 \; \text{a} + 11 – \text{a})$

$\Rightarrow 11(99 \; \text{a} + 11)$

$\Rightarrow 11 \times 11(9 \; \text{a} + 1)$

$\Rightarrow 121 \times(9 \; \text{a} + 1)$

$\Rightarrow 121 \times 64 = 7744.$

Correct Answer $: \text{D}$
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