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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.


See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial