Question

If $a, b$ are natural numbers such that $2013+a^{2}=b^{2}$, then the minimum possible value of $a b$ is

• A. 671
• B. 668
• C. 658
• D. 645

Answer 1

Answer: (C)

Given, $2013+a^{2}=b^{2}$
$\Rightarrow b^{2}-a^{2}=2013$
$\Rightarrow (b-a)(b+a)=3 \times 11 \times 61$
$a b$ is minimum.
When $b-a=33$ and $b+a=61$
On solving, we get $a=14$ and $b=47$
$\therefore$ Minimum value of $a b=14 \times 47=658$