8. 8.1. How many increasing arithmetic progressions of 22 different natural numbers exist, in which all numbers are no greater than 1000?
Answer: 23312.

## Solution.

Consider the 22nd term of each such progression, it will have the form $a_{22}=a_{1}+21d$. This means that $a_{1}$ and $a_{22}$ will have the same remainders when divided by 21. Each pair of numbers not exceeding 1000, giving the same remainders when divided by 21, defines one of the required progressions, since to construct a progression it is sufficient to divide the segment between $a_{1}$ and $a_{22}$ on the number line into 21 equal parts. Let's find the number of such pairs. Since $1000=21 \cdot 47+13$, 13 remainders occur 48 times and 8 remainders 47 times. Then the number of the required progressions will be equal to $13 \cdot C_{48}^{2}+8 \cdot C_{47}^{2}=$ $\frac{13 \cdot 48 \cdot 47}{2}+\frac{8 \cdot 47 \cdot 46}{2}=23312$.