Question

If $a,b,c,d\&e$ are prime integers, then the number of divisors of $a\times b^2\times c^2\times d\times e$ excluding $1$ as a factor, is?

• A. $94$
• B. $72$
• C. $36$
• D. $71$

Answer 1

The correct option is D

$71$

Explanation for correct option.

Given, $a,b,c,d\&e$ are prime integers

Now, $a\times b^2\times c^2\times d\times e$

In the above term, $a,d\&e$ comes once, $b\&c$ comes twice,

Number of of divisor is $(1+1)(2+1)(2+1)(1+1)(1+1)=72$

$1$ is excluded as factor.

Therefore, number of divisor is $72-1=71$

Hence, the correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$