Question

Mean value of odd divisors of $3630$ , which leaves the remainder $1$, when divided by $4$ is :

• A. $155$
• B. $6$
• C. $24$
• D. $12$

Answer 1

The correct option is A $155$

$3630=2\times 3\times 5\times {11}^2$

We need to find the odd divisors of the form $4n+1$ as it should leave a remainder $1$ upon division by $4$

As we need to find odd number, removing $2$ from factorisation, we are left with $1,3,5,{11}^2$

Possible divisors $:$

$1\times 3=3$

$1\times 5=5$

$1\times 11=11$

$1\times 121=121$

Out of these, $5$ and $121$ satisfy required condition

$3\times 5=15$

$3\times 11=33$

$3\times 121=363$

Out of these, only $33$ satisfy the required condition

$5\times 11=55$

$5\times 121=605$

Out of these, only $605$ satisfy the required condition

Also, $1$ and $3\times 5\times 11=165$ are also divisors of the form $4n+1$

Mean $=\frac{1+5+33+121+165+605}{6}=\frac{930}{6}$ $=155$