10. Let $P=\left(3^{1}+1\right)\left(3^{2}+1\right)\left(3^{3}+1\right) \ldots\left(3^{2020}+1\right)$. Find the largest value of the integer $n$ such that $2^{n}$ divides $P$.
Answer: 3030
Solution: If $k$ is even, then note that $3^{k}+1 \equiv 2(\bmod 4)$ and so $2 \| 3^{k}+1$, i.e., $4 \nmid 3^{k}+1$. On the other hand, if $k$ is odd, note that $3^{k}+1 \equiv 4(\bmod 8)$ so $4 \| 3^{k}+1$, i.e., $4 \mid 3^{k}+1$ but $8 \nmid 3^{k}+1$. Thus the greatest value of $m$ for which $2^{m}$ divides $3^{k}+1$ is 2 if $k$ is odd, and 1 if $k$ is even. Summing up over $1 \leq k \leq 2020$ gives us $2+1+\cdots+2+1=1010(2+1)=3030$