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ISI2015MMA41
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Let $k$ and $n$ be integers greater than $1$. Then $(kn)!$ is not necessarily divisible by
$(n!)^k$
$(k!)^n$
$n! \cdot k! \cdot$
$2^{kn}$
isi2015mma
numericalability
numbersystem
remaindertheorem
asked
Sep 23, 2019
in
Numerical Ability
by
Arjun
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recategorized
Nov 16, 2019
by
Lakshman Patel RJIT

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1
Answer
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$D$
Let $k=n=2$
Now,
$(kn)! = (2*2)! = 4! = 24$
which is not divisible by $2^{kn} = 2^4 = 16 $
Hence, option $D$ is the right answer.
answered
Sep 24, 2019
by
`JEET
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