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November 15th, 2018, 07:06 AM   #1
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Large number divisible

Find the largest integer k such that k divides n^55-n for all integer n?

Last edited by skipjack; November 16th, 2018 at 05:07 AM.
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November 15th, 2018, 08:32 AM   #2
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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.
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November 15th, 2018, 10:35 AM   #3
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$\displaystyle k=n^{55} -n$
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November 15th, 2018, 10:36 AM   #4
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Quote:
Originally Posted by fahad nasir View Post
Find the largest integer k such that k divides n^55-n for all integer n?
Are you, perhaps, trying to find a function k(n)?

-Dan
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November 15th, 2018, 10:50 AM   #5
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Quote:
Originally Posted by idontknow View Post
$\displaystyle k=n^{55} -n$
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.
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November 16th, 2018, 05:11 AM   #6
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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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