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Q.
Product of all the even divisors of $N =1000$, is
Permutations and Combinations
Solution:
$1000=2^3 \cdot 5^3$
Number of divisors of $N$ which are $\left(2^0+2^1+2^2+2^3\right)\left(5^0+5^1+5^2+5^3\right)=16$ Product of all the 16 divisors is
$=\left(5^6\right)\left(5^6\right)\left(5^6\right)\left(5^6\right)\left(2^6\right)\left(2^6\right)\left(2^6\right)\left(2^6\right) $
$=2^{24} \cdot 5^{24}$
Now odd divisors are $1,5,25,125$
and product of odd deivisors $=5^6$
$\therefore$ Product of all the even divisors $=2^{24} \cdot 5^{18}=2^{18} \cdot 5^{18} \cdot 2^6=\left(10^{18}\right) 64$
$\text { Alternatively: Prodcut of all the even divisors } =\left(2^4 \cdot 5^6\right)\left(2^8 \cdot 5^6\right)\left(2^{12} \cdot 5^6\right) \text { (think!) } $
$=2^{24} \cdot 5^{18}=64 \cdot 10^{18}$