Determine the sum of all positive integers $N<1000$ for which $N+2^{2015}$ is divisible by 257 .
First, we note that $257=256+1=2^{8}+1$.

Next, we note that $2^{2008}+1$ is divisible by $2^{8}+1$.

This is because since

$1-2^{8}+2^{16}-2^{24}+\cdots-2^{1992}+2^{2000}=1+\left(-2^{8}\right)+\left(-2^{8}\right)^{2}+\left(-2^{8}\right)^{3}+\cdot+\left(-2^{8}\right)^{249}+\left(-2^{8}\right)^{250}$

and the right side is a geometric sequence with first term 1, common ratio $-2^{8}$, and 251 terms, giving

$$
1-2^{8}+2^{16}-2^{24}+\cdots-2^{1992}+2^{2000}=\frac{1\left(1-\left(-2^{8}\right)^{251}\right)}{1-\left(-2^{8}\right)}=\frac{2^{2008}+1}{2^{8}+1}
$$

Since the left side is an integer, then $2^{2008}+1$ is divisible by $2^{8}+1=257$.

Since $2^{2008}+1$ is divisible by 257 , then any multiple of $2^{2008}+1$ is divisible by 257 .

In particular, $2^{7}\left(2^{2008}+1\right)=2^{2015}+128$ is divisible by 257 .

Since multiples of 257 differ by 257 , then the largest multiple of 257 less than $2^{2015}+128$ is $2^{2015}-129$ which is smaller than $2^{2015}$.

The next multiples of 257 larger than $2^{2015}+128$ are $2^{2015}+385,2^{2015}+642,2^{2015}+899$, and $2^{2015}+1156$.

Those between $2^{2015}$ and $2^{2015}+1000$ are $2^{2015}+128,2^{2015}+385,2^{2015}+642,2^{2015}+899$, which give values of $N$ of $128,385,642$, and 899 .

The sum of these values is $128+385+642+899=2054$.

ANSWER: 2054

## Team Problems