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

A. $\frac{1}{289}$

B. $\frac{289}{290}$

C. $\frac{290}{289}$

D. $289$

Solution:

Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,

\begin{aligned} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \\ = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \\ = \frac{1^{2} + {(17)}^{2}}{{(1 \times 17)}^{2}} \\ = \frac{290}{289} \end{aligned}

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