How can I prove that the square of an even number ends in 0/4/6?

Question

I am trying to prove that the last digit of the square of an even number is either 0, 4, or 6 but I'm completely lost and have no idea how to tackle this problem.

Answer 1

Well, any number $ab....cd$ can be written as $ab....c \times 10+d$

If you square it, your square will end with the end of ... $d²$

And... that's it.

Answer 2

It is sufficient to compute the answer mod 10. Take all nonnegative even numbers less than 10 (0,2,4,6,8), square them and check the last digit. This will be the same as the last digit of the square of the number whose last digit is what you squared.

Answer 3

An even number $n$ ends either in:

• $0$, hence it is $n=10k \implies n^2=100k^2,$ which ends in $0$;
• $2$, hence it is $n=10k+2 \implies n^2=10(10k^2+4k)+4$, which ends in $4$;
• $4$, hence it is $n=10k+4 \implies n^2=10(10k^2+8k+1)+6$, which ends in $6$;
• $6$, hence it is $n=10k+6 \implies n^2=10(10k^2+12k+3)+6$, which ends in $6$;
• $8$, hence it is $n=10k+8 \implies n^2=10(10k^2+16k+6)+4$, which ends in $4$.