
#1
Jan1510, 05:55 PM

P: 5

Can anyone give me hints to how to prove this??
Prove that for any positive integer n, n^5 and n have the same units digit in their base 10 representations; that is, prove that n^5 = n (mod 10). Thanks! 



#2
Jan1510, 06:30 PM

P: 403

What does the EulerFermat theorem tells you, when applied to a congruence mod 10?




#3
Jan1510, 08:47 PM

P: 5

I'm sorry, I'm still a bit lost. Can you please explain what the EulerFermat theorem is and how I can apply that to this problem?
Thanks 



#4
Jan1510, 09:31 PM

P: 403

Congruence Modulo n Proving
See the following link:
http://planetmath.org/encyclopedia/E...atTheorem.html And notice that you only have to prove that: [tex]n^{4}\equiv 1 \left(mod 10\right)[/tex] For n coprime with 10. 



#5
Jan1610, 12:30 AM

Mentor
P: 20,939

You can show this by proving that n^{5}  n is even, and is divisible by 5. The first part is easy, since two of the factors of n^{5}  n are n and n + 1, one of which has to be even for any value of n. The second part, showing that n^{5}  n is divisible by 5 can be done by math induction, and isn't too tricky. 


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