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

Answer: (A)

$3630=2^1{3}^1{5}^{1}11^2=2^a{3}^b{5}^{c}11^d$
$\therefore$ Number of odd divisors of $3630$ are $(b+1)(c+1)(d+1)=\mathrm{2.2.3}$
$\therefore$ Total number of odd divisors of $3630=12$
These divisors are the factors of $(3\times 5\times 11^2)=1815$
which are $1,3,5,11,15,33,55,121,165,363,605,1815$ From these divisors w e need those divisor w hich leaves the remainder 1, when divided by 4
$\therefore$ Those divisors are $1,5,33,121,165,605$
$\therefore$ Number of such divisors $=6$
$\therefore$ Mean of such divisors =
$\frac{\text{ Sum of all odd divisors which leave remainder 1 when divided by 4}}{\text{ Number of divisors of the form}(4k+1)}$
$=\frac{1+5+33+121+165+605}{6}$
$=\frac{930}{6}=155$