Squaring numbers that end with the digit 5

If $n$ ends in $5$, we can write $n$ in the form $10k+5$ for some integer $k$. Then

$$n^2=(10k+5)^2=100k^2+100k+25=100(k^2+k)+25\;.$$

Clearly $100(k^2+k)$ ends in $00$, so $n^2$ ends in $25$.

$$(10n + 5)^2 = 100 n^2 + 100n + 25 = 100n(n + 1) + 25.$$

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