Factorials

What is the smallest positive integer n n such that n ! ( n + 1 ) \large \frac{{n!}}{{(n+1)}} is a positive integer?


The answer is 5.

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2 solutions

Arul Kolla
Mar 28, 2017

5 ! 6 \Huge \frac{\color{#3D99F6}{5!}}{\color{#D61F06}{6}} is 20. Checking the smaller ones, they doesn't work. Thus, the answer is 5 \boxed{5} .

In general, any n > 3 n \gt 3 that is not of the form p 1 p - 1 for some prime p p will be such that n ! n + 1 \dfrac{n!}{n + 1} is an integer.

Brian Charlesworth - 4 years, 2 months ago

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Nice generalization!

Syed Hamza Khalid - 2 years, 7 months ago
Blan Morrison
Oct 9, 2018

Although n + 1 n+1 can't be a prime factor of n ! n! , n + 1 n+1 can share factors with n ! n! . A good way to start is to take the first two whole numbers, 2 and 3, and set their product equal to n + 1 n+1 ; this will definitely give us a positive integer. Indeed, 5 ! 6 = 20 ; n = 5 \frac{5!}{6}=20;~n=\boxed{5}

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