Here was the solution that I stumbled upon, but I'm sure it can be improved upon:
$$
1^2=1 \\
2^2 = 4 \\
3^2 = 9 \\
\ldots \\
20^2 = 400
$$
The last digits of square numbers repeat in the following pattern: $1,4,9,6,5,6,9,4,1,0$. It seems like square numbers can end in any digit except $2,3,7, \text{and }8$. Proof that the pattern always repeats:
$$(n+10)^2=n^2+20n+100=n^2+10(2n+10)$$ A number plus a multiple of $10$ does not change its last digit, so the pattern will always repeat.
I have two questions:
- Can my solution be improved upon (e.g. by generalising it further)?
- Do you have any advice for getting to the solution more quickly when you are finding the answer to these problem solving-type questions? Even though the mathematics I used to find the solution to this problem was rather elementary, it took me half an hour of complete guess-work to find it. Is there a more structured approach one can take?
Best Answer
Consider a number $ n$ in $\mod10$. The possible representations are : $$ n \equiv 0,1,2,3,4,5,6,7,8,9\mod10$$ and $$n^2 \equiv 0,1,4,9,6,5,6,9,4,1 \mod 10$$
Hence the possible digits in the end of a square number is $\in (0,1,4,5,6,9)$
It is easier to study properties of numbers using number theory . These might be helpful: