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How to "show", that if x=1, then its powers will end with 1, if x=5, it's powers end with 5.

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4 Answers 4

up vote 5 down vote accepted

I am presuming you are talking about the last digit of a number in base 10 being 1 and 5 implying the powers of that number will also end with 1 and 5.

I will give you a hint:

Give two numbers whose last digits are 1, what does their product end with?

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Give two numbers whose last digits are 1, what does their product end with? - 1 –  lam3r4370 Sep 18 '10 at 17:36
    
@lam: Right. So if $x^n$ ends with 1 and $x$ ends with 1, what does $x^{n+1}$ end with? What about $x^{n+2}$, $x^{n+3}$...? –  Aryabhata Sep 18 '10 at 17:45
    
One.Again.Am I right? –  lam3r4370 Sep 18 '10 at 19:11
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@lam: Umm...what about x=4? It does not end with either 5 or 1... –  Aryabhata Sep 18 '10 at 21:33
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@lam: I suggest you try out some powers yourself and try to work it out. The hints for the main question should be enough. This looks like homework to me. –  Aryabhata Sep 19 '10 at 16:31

Use mathematical induction:

  1. Show that $x^1$ ends in 1 (or 5).
  2. Show that if $x^n$ ends in 1 (or 5), then $x^{n+1}$ also ends in 1 (or 5). (Think about multiplying a number than ends in 1 by 1 (or ends in 5 by 5).)
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If you know modular arithmetic: $\rm\; e\equiv 1,5\;\Rightarrow\; e^2\equiv e \;\Rightarrow\; e^k\equiv e\pmod{10}\;$ for all $\rm\; k>0$

If not, use induction: suppose $\rm\;\; e^2 = e + 10j\;$ and $\rm\; (e + 10 m)^k = e + 10n$

then $\rm\; (e + 10 m)^{k+1} = (e + 10n) (e + 10m) = e + 10(j+em+en+10mn)$

REMARK Note how the use of modular arithmetic has allowed us to rephrase the problem in a completely arithmetical form that makes the induction step utterly trivial - namely that $\rm 1^k \equiv 1$ or, more generally that $\rm\; e^2 \equiv e\;\Rightarrow e^k \equiv e\;$ for $\rm\: k>0$. This is but one very simple illustration of how one may often reduce problems about integers to simpler problems about integers modulo $\rm\:m\:$. Because the integers modulo $\rm\:m\:$ satisfy many of the same arithmetical laws as the integers (associative, commutative, distributive, etc) - i.e. they form a ring - we can reuse our strong intuition about integers to help discover proofs in this simpler structure integers modulo $\rm\:m\:$. And, if we're lucky, we can sometimes lift these simple solutions from the integers modulo $\rm\:m\:$ up to the integers. If you study abstract algebra you'll see many more examples of problems solved by passing to such simpler structures (technically they are known as quotient objects - structures modulo a congruence, i.e. a structure-compatible equivalence relation). Said in an informal way, such reduction to quotient objects is a method to divide and conquer mathematically.

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I don't think OP knows about modular arithmetic... –  stevenvh Sep 22 '10 at 14:14
    
That's why I gave an alternative proof by induction, which requires no knowledge of modular arithmetic. –  Bill Dubuque Sep 22 '10 at 14:55

If you know that addition and multiplication are well-defined modulo $n$, then just work (mod 10).
Let $x_0 \in$ {1,5} be the ones digit of $x$. Then for any $k$

$x^k \equiv x_0^k \equiv x_0$ (mod 10).

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