Squaring numbers that end with the digit 5
Solution 1:
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 $25$.
Solution 2:
$$(10n + 5)^2 = 100 n^2 + 100n + 25 = 100n(n + 1) + 25.$$
$\text{ }$
Solution 3:
Square of Every number ending in 5, say x5, is equal to x*(x+1) followed by 25. Number ending in 5 can be represented as 10n +5
(10n +5)^2 = 100n^2 + 100n+25= 10n * 10 (n +1) + 25.
A Vedic mathematics trick