Question

The natural number has prime factorization given by $N=2^3\cdot 5^x\cdot 7^y,$ where $x+y=5$ and $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}(x>y).$ Then the number of even divisors of $N($excluding $N),$ is equal to

• A. $12$
• B. $34$
• C. $36$
• D. $35$

Answer 1

The correct option is D $35$

$N=2^3\cdot 5^x\cdot 7^y$
Given, $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$
$\Rightarrow xy=6(\because x+y=5)$
$\therefore x=3,y=2(\because x>y)$
$\therefore N=2^3\cdot 5^3\cdot 7^2$
Number of even divisors of $N=3(3+1)(2+1)=36$
Hence, Number of even divisors of $N($excluding $N)$ is $35.$