5. We consider all possible $100-$-digit numbers, where each digit is equal to 1 or 2. For each, the remainder when divided by 1024 is calculated. How many different remainders are there among these?
5. Solution: Subtract the number $11 \ldots 1$ (one hundred ones) from all the numbers. The numbers under consideration have turned into numbers consisting of zeros and ones from 0 to $11 \ldots 1$. We will show that if two of these numbers have differences in the last 10 digits, then they have different remainders when divided by 1024. If this is not the case, then consider the difference between these numbers. It should be divisible by 1024, but it ends with fewer than 10 zeros, preceded by a 1 or 9. According to the divisibility rule for powers of two, such a number is not divisible by $2^{10}$.

## Instructions for checking:

Most olympiad participants know the divisibility rule for powers of two:

A number is divisible by $2^{\mathrm{n}}$ if and only if the last $\mathrm{n}$ digits of this number form a number that is divisible by $2^{\text {n }}$.

Proving it is simple, but of course, it is not necessary.