1. $[20]$ Find the sum of the positive integer divisors of $2^{2007}$.
Answer: $2^{\mathbf{2 0 0 8}}-\mathbf{1}$. The divisors are the powers of two not exceeding $2^{2007}$. So the sum is
$$
1+2+2^{2}+\cdots+2^{2007}=-1+2+2+2^{2}+\cdots+2^{2007}=-1+2^{2}+2^{2}+\cdots+2^{2007}=\cdots=-1+2^{2008}
$$