Question

The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is:

• A. 1
• B.
• C.
• D. 289
• E. 290

Answer 1

The correct option is C

Let the number be 'a' and 'b'.

Then, $ab=17\Rightarrow a=1andb=17orvice-versa.$

$So,\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2{b}^2}=\frac{1^2+{\left(17\right)}^2}{{(1\times 17)}^2}=\frac{290}{289}$