Question

If $N=2^3\cdot 3^4\cdot 5^2\cdot 7^1$ then product of all divisors of $N$ which are divisible by 5 is

• A. $N^{20}$
• B. $N^{40}$
• C. $(5N{)}^{20}$
• D. $(5N{)}^{40}$

Answer 1

Answer: (D)

permisc $N=2^3\cdot 3^4\cdot 5^2\cdot 7^1$
Total divisors of $N=4\times 5\times 3\times 2=120$
$M=2^3\cdot 3^4\cdot 7^1$
Number of divisors of $N$ which are not divisible by $5=4\times 5\times 2=40$
$\text{ Required product }=\frac{\text{ Product of all divisors }}{\text{ Product of divisors which are not divisible by }5}$
$=\frac{N^{60}}{M^{20}}$
$=\frac{N^{60}}{{\left(\frac{N}{25}\right)}^{20}}=(5N{)}^{40}\left(\Theta M=\frac{N}{5^2}\right)$