[ Pairing and grouping; bijections $]$ [ Decimal number system ]

Find the last digit of the number $1^{2}+2^{2}+\ldots+99^{2}$.
$1^{2}+2^{2}+\ldots+99^{2} \equiv 10\left(1^{2}+2^{2}+\ldots+9^{2}\right)=0(\bmod 10)$.

## Answer

0.