1. Given 6000 cards, on which natural numbers from 1 to 6000 are written (each card has exactly one number, and the numbers do not repeat). It is required to choose two cards such that the sum of the numbers written on them is divisible by 100. In how many ways can this be done?
Answer: 179940.

Solution. We will take the cards in turn. There are several cases depending on the number written on the first card.

1) The number on the card ends in 00 (there are 60 such cards). For the sum to be divisible by 100, the second card must also have a number ending in 00. We get a total of $C_{60}^{2}=\frac{60 \cdot 59}{2}=1770$ options.
2) Similarly, if the number on the card ends in 50 (there are also 60 such cards), then for the sum to be divisible by 100, the second card must also have a number ending in 50, i.e., here we also have 1770 options.
3) The number on the card ends in a number from 1 to 49 (there are $49 \cdot 60=2940$ such cards). For each of them, a pair can be chosen in 60 ways (if the number ends in 1, then any card with a number ending in 99 will do; if the number ends in 2, any card with a number ending in 98, and so on). Thus, we get $2940 \cdot 60=176400$ options.
4) The number on the card ends in a number from 51 to 99. All such options were accounted for when considering the third case (these cards formed pairs with the cards initially chosen).

In total, we get $1770+1770+176400=179940$ ways.