Question

If $aabb$ is a four digit number and also a perfect square then the value of $a+b$is:

• A. $12$
• B. $11$
• C. $10$
• D. $9$

Answer 1

The correct option is B

$11$

Solution:

Step 1: Expanding $\mathit{a}\mathit{a}\mathit{b}\mathit{b}$:

$aabb$ is a four digit number and also a perfect square. Therefore,

$$\begin{gathered} aabb \\ =1000a+100a+10b+b \\ =1100a+11b \\ =11(100a+b) \end{gathered}$$

Step 2: Finding the perfect square $\mathit{a}\mathit{a}\mathit{b}\mathit{b}$:

For $\text{'}aabb\text{'}$ to be a perfect square, $100a+b$ should be of the type $11\times k^2$, where $k$ is a natural number.

We take$100a+b=11k^2$ because, the product of two perfect squares is also perfect square.

So,$aabb=11\times 11k^2$ is a perfect square as it is product of ${11}^2$ and $k^2$.

Since, our number is a four digit number, therefore, the possible values of ${11}^2\times k^2$ at different values of $k$ are as follows:

$$\begin{gathered} 121\times 9=1089 \\ 121\times 16=1936 \\ 121\times 25=3025 \\ 121\times 36=4356 \\ 121\times 49=5629 \\ 121\times 64=7744 \\ 121\times 81=9801 \end{gathered}$$

Only $121\times 64=7744$ is in the form of $aabb$.

So, $7744$ is the required four digit number.

Step 3: Finding the sum $\mathit{a}\mathbf{+}\mathit{b}\mathbf{:}$

$$\begin{gathered} \because a=7,b=4 \\ \therefore a+b=7+4=11 \end{gathered}$$

Final answer: Hence, option(B) is correct.