# How to prove that every power of 6 ends in 6?

Yesterday I had the traditional math matriculation exam, and in it there was a question "In what digit does the number $2016^{2016}$ end in?" After the test The Matriculation Examination Board published a pdf in which they show how to basically solve all the problems in the test, and for the aforementioned question the solution was "Because every power of 6 ends in 6, so does every power of 2016, and thus the last digit is 6". I understand everything else in that problem now, except how to actually show that every power of 6 ends in 6. From there on I know how to solve the last digit if $2016^{2016}$. So,

how to prove that every power of 6 ends in 6?

• by induction on $n$ – ThePortakal Mar 24 '16 at 11:31
• I would first try to use the fact that $a_1 \equiv b_1 \pmod{10}$ and $a_2 \equiv b_2 \pmod{10}$ imply $a_1a_2 \equiv b_1b_2 \pmod{10}$ in a proof-by-induction context – tilper Mar 24 '16 at 11:33
• @user265554 Please show what you know of this problem, how you found it and how you attempted to solve it. This way people won't down vote your question. You're not a new user so you should know this by now. – Arbuja Mar 24 '16 at 11:39
• Sorry about that, I edited it. – user265554 Mar 24 '16 at 11:53
• $(10a+6)(10b + 6) = 10(10ab+6a+6b+3) +6$ – Henry Mar 24 '16 at 11:56

Use induction in order to complete the hint given by @ThePortakal.

First, show that this is true for $n=1$:

$6^1=6$

Second, assume that this is true for $n$:

$6^n=10k+6$

Third, prove that this is true for $n+1$:

$6^{n+1}=$

$6\cdot\color{red}{6^n}=$

$6\cdot(\color{red}{10k+6})=$

$60k+36=$

$60k+30+6=$

$10(6k+3)+6$

Please note that the assumption is used only in the part marked red.

A little hint: $6*6\bmod 10 = 6$.