## Task A-4.5.

Determine the sum of all natural numbers $n$ less than 1000 for which $2^{n}+1$ is divisible by 11.
## Solution.

We need to determine the sum of all natural numbers $n<1000$ such that the number $2^{n}$ gives a remainder of 10 when divided by 11.

Let's consider the remainders that the numbers $2^{n}$ give when divided by 11:

| $2^{n}$ | $2^{1}$ | $2^{2}$ | $2^{3}$ | $2^{4}$ | $2^{5}$ | $2^{6}$ | $2^{7}$ | $2^{8}$ | $2^{9}$ | $2^{10}$ | $2^{11}$ | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| remainder | 2 | 4 | 8 | 5 | 10 | 9 | 7 | 3 | 6 | 1 | 2 | $\ldots$ |

Thus, we conclude that the natural number $2^{n}$ gives a remainder of 10 when divided by 11 if and only if $n=10k+5$, for some integer $k \geqslant 0$.

We need to determine the sum

$$
5 \cdot 1 + 5 \cdot 3 + 5 \cdot 5 + \cdots + 5 \cdot 197 + 5 \cdot 199
$$

However, this sum is equal to

$$
5 \cdot (1 + 3 + 5 + \cdots + 197 + 199) = 5 \cdot 100^{2} = 50000
$$