If a, b are natural numbers such that 2013+a2=b2, then the minimum possible value of a b is

Question

If a, b are natural numbers such that 2013+a2=b2, then the minimum possible value of a b is (A) 671 (B) 668 (C) 658 (D) 645

Tardigrade Admin · Accepted Answer

Given, 2013+a2=b2 ⇒ b2-a2=2013 ⇒ (b-a)(b+a)=3 × 11 × 61 a b is minimum. When b-a=33 and b+a=61 On solving, we get a=14 and b=47 ∴ Minimum value of a b=14 × 47=658

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

671

668

658

645

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

Q. If a,b are natural numbers such that 2013+a2=b2, then the minimum possible value of ab is

671

668

658

645

Given, 2013+a2=b2 ⇒b2−a2=2013 ⇒(b−a)(b+a)=3×11×61 ab is minimum. When b−a=33 and b+a=61 On solving, we get a=14 and b=47 ∴ Minimum value of ab=14×47=658

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

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