[ Residue arithmetic (other).]

How many zeros does the number $9^{999}+1$ end with? 

#
$9^{999}+1=(9+1)\left(9^{998}-9^{997}+\ldots-9+1\right), 9^{998}-9^{997}+\ldots-9+1 \equiv(-1)^{998}-(-1)^{997}+\ldots-(-1)+1=999(\bmod 10)$. Therefore, $9^{999}+1$ is divisible by 10, but not by 100.

## Answer

By one.