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# Factorials question

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Given that $$\frac{((3!)!)!}{3!}$$$$=\text k \times \text n!$$ where $$k$$ and $$n$$ are positive integers and $$n$$ is as large as possible, find $$k+n$$.

Jun 12, 2020

#1
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ok so

if we simplify the numerator of the fraction, we get $$((3!)!)!=(6!)!=720!$$

so our new fraction is $$\dfrac{720!}{3!}$$ which is equal to $$\dfrac{720 \cdot 719!}{6}$$

we can then cancel out the 720 and 6, and we get $$120\cdot719!$$

so, $$719+120=\boxed{839}$$

Jun 12, 2020
#2
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thanks!

amazingxin777  Jun 12, 2020
#3
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no problem! :)