The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is · ABC26GN1678
Subject: General Aptitude · Chapter: Problems on Numbers
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
Answer
Answer (as printed): C
Explanation
Let the numbers be $a$ and $b$. Then, $a b=17 \Rightarrow a=1$ and $b=17$. $$\text { So, } \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1^{2}+(17)^{2}}{(1 \times 17)^{2}}=\frac{290}{289} \text {. }$$
Explanation as extracted from the printed page; notation may be imperfect.