Question

The sum of all proper divisors of $9900$ except $1$ and itself is

• A. $29351$
• B. $23951$
• C. $33851$
• D. None of these

Answer 1

The correct option is B

$23951$

Explanation for correct option

The prime factors of $9900$ are

$9900=2^2\times 3^2\times 5^2\times 11$

We know that if the prime factorization of $n$ is

$n=\prod _k{p}_k^{a_k}$

Where $p_k$ are the prime factors and $a_k$ are the positive integer exponents, the sum of all the positive integer factors is

$\prod _k\left(\sum _{i=0}^{a_k}{p}_k^i\right)$

Therefor the sum of the proper positive divisor of $9900$ is

$=\left(1+2+2^2\right)\left(1+3+3^2\right)\left(1+5+5^2\right)\left(1+11\right)-9900-1$ [Since except $1$ and itself]

$$\begin{gathered} =\frac{2^3-1}{2-1}\times \frac{3^3-1}{3-1}\times \frac{5^3-1}{5-1}\times 12-9901 \\ =33852-9901 \\ =23951 \end{gathered}$$

Hence the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$