Nested Factorials

Level 1

Knowing that ( ( 3 ! ) ! ) ! 3 ! = k ⋅ n ! \dfrac{\left(\left(3!\right)!\right)!}{3!}=k\cdot n! , where k k and n n are positive integers and n n is as big as possible, find the value of k + n k+n .

This problem is not original.

719 599 739 839 120

Pablo Ruiz by Pablo Ruiz , 21, Mexico

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1 solution

Pablo Ruiz
Jul 14, 2018

( ( 3 ! ) ! ) ! 3 ! = 720 ! 3 ! = 720 ! 6 = k ⋅ n ! \frac{\left(\left(3!\right)!\right)!}{3!}=\frac{720!}{3!}=\frac{720!}{6}=k\cdot n!

As we need something in the form of n ! ∗ k n! * k on the left side with the biggest n n we can get. To assure that n n is as big as possible, we factor out 720 720 to cancel out the 6 6 in the denominator.

We get the following expression.

719 ! ⋅ 120 = k ⋅ n ! 719!\cdot120=k\cdot n!

Hence, the answer is 719 + 120 = 839 719 + 120 = 839 .

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