Question

Solve this:
Q.92. Find the sum of divisors of 544 which are perfect squares.
(A) 42
(B) 21
(C) 34
(D) 68

Answer 1

Dear student
$$\begin{gathered} 544=2\times 2\times 2\times 2\times 2\times 17 \\ \mathrm{Sum}\mathrm{of}\mathrm{divisors}\mathrm{of}544\mathrm{which}\mathrm{are}\mathrm{perfect}\mathrm{square}\mathrm{is} \\ \left(2^0+2^2+2^4\right)\left({17}^0\right) \\ =(1+4+16)\left(1\right) \\ =21 \end{gathered}$$
Regards