What is $\nu_{2}\left(2^{100}+2^{200}\right)$?
On a $2^{100}+2^{200}=2^{100} \times\left(1+2^{100}\right)$, and $1+2^{100}$ is odd, so $2^{101}$ does not divide $2^{100}+2^{200}$, but $2^{100}$ does divide it. Therefore, $\nu_{2}\left(2^{100}+2^{200}\right)=100$

## Principle of infinite descent