Question

The number of divisors of the form $2n-1(n\ge 2)$ of the number $2^p{3}^q{4}^r{5}^s$, where $p,q,r,s$ belong to N, is

• A. $qs+q+s+1$
• B. $(p+1)(q+1)(r+1)(s+1)-1$
• C. $qs+q+s$
• D. $qs$

Answer 1

Answer: (C)

$qs+q+s$

Total number of divisors for $2^p{.3}^q{.4}^r{.5}^s=(p+1)(q+1)(r+1)(s+1)-1$
But since it is of the form, 2n-1 it should be an odd divisor.
so,
total$=(q+1)(s+1)-1$
$=qs+q+s+1-1$
$=qs+q+s$