Question

The number of divisors of $2^2{.3}^3{.5}^4{.7}^5$ of the form $4n+1,n\in N$ is

• A. $46$
• B. $47$
• C. $96$
• D. $94$

Answer 1

The correct option is A $46$

We have

Numbers of divisors of

N = $2^2\times 3^3\times 5^3\times 7^5$

which are of the form $4n+1$

excluding.

1 = numbers of terms in product

$=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+${number of terms in product}

$=(3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1$

$=2\times 4\times 3+2\times 3\times 4-1$

$=24+24-1$

$=47$

Hence this is the answer.