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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
avatar+649 
+3

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, \(719+120=\boxed{839}\)

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 Jun 12, 2020
 #2
avatar+306 
+1

thanks!

amazingxin777  Jun 12, 2020
 #3
avatar+649 
+1

no problem! :)

lokiisnotdead  Jun 13, 2020

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