3. On the table, there are 200 different cards with numbers $201, 203, 205, \ldots, 597, 599$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3?
Answer: 437844.

Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next number is $3k+2$ - and it gives a remainder of 2 when divided by 3, followed by $3k+4 = 3(k+1)+1$, which gives a remainder of 1 when divided by 3, and then $3k+6$, which is again divisible by 3. Thus, the remainders of the given numbers when divided by 3 go in the order ... $0 ; 2 ; 1 ; 0 ; 2 ; 1 ; 0 \ldots$.

Among the given numbers, there are 66 that give a remainder of 1 when divided by 3 (they form the set $A=\{205 ; 211 ; 217 ; \ldots ; 595\}$), 67 numbers that are divisible by 3 ($B=\{201 ; 207 ; 213 ; \ldots ; 597\}$), and 67 numbers that give a remainder of 2 when divided by 3 ($C=\{203 ; 209 ; 215 ; \ldots ; 599\}$).

The sum of three numbers can be divisible by 3 in the following cases.

1) All three numbers give the same remainder when divided by 3. There are $C_{66}^{3}$ ways to choose 3 numbers from set $A$ and $C_{67}^{3}$ ways to choose 3 numbers from sets $B$ and $C$. In total, we get $\frac{66 \cdot 65 \cdot 64}{6} + 2 \cdot \frac{67 \cdot 66 \cdot 65}{6} = 45760 + 2 \cdot 47905 = 141570$ ways.
2) If not all remainders are the same, then only the case where all three remainders are different fits, i.e., we must choose one number from each of the sets $A, B, C$. We get $66 \cdot 67 \cdot 67 = 296274$ ways.

In total, we get 437844 ways.