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Q.
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
Permutations and Combinations
Solution:
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}} $
$=\frac{ N ^{60}}{\left(\frac{ N }{25}\right)^{20}}=(5 N )^{40} \left(\Theta M =\frac{ N }{5^2}\right)$