Question

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

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

Answer 1

Answer: (B)

Given number is $aabb=1000a+100a+10b+b$
$=1100a+11b$
$=11(100a+b)$
For $aabb$ to be a perfect square, $100a+b$ should be of the type
$11\times n^2$ where n is natural number
$\therefore aabb=11\times 11\times n^2$
By Hit and Trial
When $n=4$
$11\times 11\times n^2=121\times 16=1936$. This is not in the form aabb.
When $n=5$
$11\times 11\times n^2=121\times 25=3025$. This is not in form aabb.
When $n=6$
$11\times 11\times n^2=121\times 36=4356$. This is not in form aabb.
When $n=7$
$11\times 11\times n^2=121\times 49=5629$. This is not in form aabb.
When $n=8$
$11\times 11\times n^2=121\times 64=7744$. This is of the form aabb.
So, $7744$ is the four digit number.
$\therefore a+b=7+4=11$
Hence, option 'B' is correct.