1. Given 2414 cards, on which natural numbers from 1 to 2414 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: 29112.

Solution. We will take 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 24 such cards). For the sum to be divisible by 100, the second card must also end in 00. We get a total of $C_{2} 4^{2}=\frac{24 \cdot 23}{2}=276$ options.
2) Similarly, if the number on the card ends in 50 (there are also 24 such cards), then for the sum to be divisible by 100, the second card must also end in 50, i.e., here we also have 276 options.
3) The number on the card ends in a number from 1 to 14 (there are $14 \cdot 25=350$ such cards). For each of them, a pair can be chosen in 24 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 $350 \cdot 24=8400$ options.
4) The number on the card ends in a number from 15 to 49 (there are $35 \cdot 24=840$ such cards). For each of them, a pair can be chosen in 24 ways (similarly to the previous case). Thus, we get $840 \cdot 24=20160$ options.
5) The number on the card ends in a number from 51 to 99. All such options were considered when examining the third and fourth cases (these cards formed pairs with the cards initially chosen).

In total, we get $276+276+8400+20160=29112$ ways.