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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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` :

`0^5=0`

`1^5=1`

`2^5=32`

`3^5=243`

`4^5=1024`

`5^5=3125`

`6^5=7776`

`7^5=16807`

`8^5=32768`

`9^5=59049`

**Thus the result holds.**

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

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