Question 520127
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The last digit of an integer is the remainder when the number is divided by 10.  Any two natural numbers *[tex \Large x,\,y] share the same last digit exactly when *[tex \Large 10\mid(x\ -\ y)] (Note: *[tex \Large \mid] denotes "divides" and *[tex \Large \not\mid] denotes "does not divide")


First prove *[tex \Large 2\mid(n^5\ -\ n)]:  If *[tex \Large n] is even, then *[tex \Large n^5] is even and the difference of two even numbers is even.  If *[tex \Large n] is odd, then *[tex \Large n^5] is odd, and the difference of two odd numbers is even.  Thus *[tex \Large 2\mid(n^5\ -\ n)].


Next prove *[tex \Large 5\mid(n^5\ -\ n)].  Fermat's Little Theorem:  If *[tex \Large p] is prime, for any integer *[tex \Large a], *[tex \Large p\mid(a^p\ -\ a)]


Hence *[tex \Large n^5\ -\ n\ =\ 2a\ =\ 5b] for some integers *[tex \Large a,\,b].  That means *[tex \Large 2\mid\,5b].  Since *[tex \Large 2\not\mid\,5], so *[tex \Large 2\mid\,b] since 2 is prime.  Then *[tex \Large b\ =\ 2c] where *[tex \Large c] is an integer, and then *[tex \Large n^5\ -\ n\ =\ 5b\ =\ 10c].  


Therefore *[tex \Large 10\mid(n^5\ -\ 5)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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