November 15th, 2018, 07:06 AM  #1 
Member Joined: Oct 2012 Posts: 70 Thanks: 0  Large number divisible
Find the largest integer k such that k divides n^55n for all integer n?
Last edited by skipjack; November 16th, 2018 at 05:07 AM. 
November 15th, 2018, 08:32 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,251 Thanks: 519 
Are you sure that this is the entire question? $\text {Find the largest integer k that divides evenly into } x \text { given that } x = n^{55}  n \text { and } n \in \mathbb Z.$ If $n \le 1$, there is no largest integer that divides into x because x will equal 0. Otherwise, the largest integer that can divide evenly into x is x. Last edited by JeffM1; November 15th, 2018 at 08:40 AM. 
November 15th, 2018, 10:35 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 
$\displaystyle k=n^{55} n$ 
November 15th, 2018, 10:36 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  
November 15th, 2018, 10:50 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,251 Thanks: 519  Actually, that does not work. Consider n = 1. Then $1^{55}  1 = 1  1 = 0.$ And 0 does not divide evenly into 0. And $n = \ 2 \implies 2 > (\ 2)^{55}  (\ 2) \text { and } 2 \  \ (\ 2)^{55}  (\ 2).$ Really stupid problem as posted. Last edited by JeffM1; November 15th, 2018 at 10:57 AM. 
November 16th, 2018, 05:11 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
n^55  n is divisible by n  1, n and n + 1, so it's divisble by 2 × 3 = 6. Hence the largest k is at least 6. 

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