Prove that, for all integers 'n', the last digit of 'n5' is the same as the last digit of 'n'

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embizze | High School Teacher | (Level 1) Educator Emeritus

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One proof: Let n=10y+x. (Any positive integer can be represented this way. e.g. 119=10(11)+9)

Then `(10y+x)^5=(10y)^5+5(10y)^4x+10(10y)^3x^2+10(10y)^2x^3+5(10y)x^4+x^5 ` I used the binomial expansion, but you will get the same result if you multiply `(10y+x)(10y+x)(10y+x)(10y+x)(10y+x)` carefully.

The key point is that every term except `x^5` must be a factor of 10; thus we can rewrite the expression as `(10y+x)^5=10z+x^5` . This means we are only concerned with the one's digit of n.

Now you can use exhaustion -- show that for every number `0<=k<=9` that the last digit of `k^5` is `k` :











Thus the result holds.

** e.g. if n=119, `119^5=23863477550+59049=10(2386347755)+59049`

`119^5=23863536599` ends in 9 as required.


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