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Question:

What are the possible value of the last digit of $6^n?$

I know it has to do with Euler's theorem and we have to find the number modulus 10, but I can't get any further.

YuiTo Cheng
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TheHedge
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2 Answers2

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In $\mathbb{Z}/{10\mathbb{Z}}$ we have that $6^2 = 6$, so it's an idempotent.

It follows that all powers of $6$ (except the $0$-th power) are $6$, so all integers that have remainder $6$ when divided by $10$ (so last digit is $6$) stay that way in all their powers.

Henno Brandsma
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You have

  • $6^n \equiv_2 0$
  • $6^n \equiv_5 1^n \equiv_5 1$

$$x \equiv_2 0, x \equiv_5 1 \Rightarrow \boxed{x\equiv_{10} 6}$$

trancelocation
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  • Correct answer but which concrete rules did you apply here? – nilo de roock Apr 01 '19 at 10:57
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    $10 = 2\cdot 5$. So, calculate with respect to the modules $2$ and $5$. The Chinese remainder theorem ( https://en.wikipedia.org/wiki/Chinese_remainder_theorem ) gives then the answer wrt. the module $10$. – trancelocation Apr 01 '19 at 11:03
  • So you solved x=0 mod 2 and x=1 mod 5. With solution x=6 mod 10. Got it. Thanks – nilo de roock Apr 01 '19 at 11:08
  • Exactly. You are welcome. – trancelocation Apr 01 '19 at 11:08
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    @niloderoock $\large \bmod 10!:\ ,6^{\large 1+N} \equiv 6(\underbrace{6^{\large N} \bmod 5}_{\Large 1^{\large N}\ \equiv\ 1 })\equiv 6(1)\ $ by the mod Distributive Law, is a simple way to do it without CRT – Bill Dubuque Apr 01 '19 at 17:07
  • One can also simply reason that even numbers which give a remainder of $1$ when divided by $5$ are those of the form $6+10\cdot k$. :-) – trancelocation Apr 01 '19 at 17:56
  • @trancelocation But that's essentially the same as applying CRT (unless one uses some ad hoc method such as brute-force search - which won't scale well to large numbers). The mod Distributive Law is actually equivalent to CRT, but being operational it is often more convenient to apply in computations. – Bill Dubuque Apr 01 '19 at 20:08