3. There are 167 different cards with numbers $1, 3, 11, 3^{2}, 11^{2}, \ldots, 3^{83}, 11^{83}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 33?
Answer: 139523.

Solution. Consider the case when one of the selected cards has the number one written on it. Then, on the other two cards, even powers of the numbers three and eleven must be recorded. There are 41 ways to choose an even power of three and 41 ways to choose an even power of eleven, and since this choice is made independently, the total number of ways in this case is $41^{2}=1681$.

Now let's move on to the case when the card with the number one is not used. There are two possibilities: to take two cards with powers of three and one with a power of eleven, or vice versa. Suppose we take two cards with powers of three. Since the number three must be present in an even power in the product of the selected numbers, the powers of three on the selected cards must have the same parity. We have a total of 42 cards with odd powers and 41 cards with even powers. Thus, there are $C_{42}^{2}=\frac{42 \cdot 41}{2}=861$ ways to choose cards with odd powers and $C_{41}^{2}=\frac{41 \cdot 40}{2}=820$ ways to choose two cards with even powers - in total, we get 1681 ways. After selecting the cards with powers of three, we can choose any card with an even power of eleven (41 ways). Since the choice of the card with the power of eleven is independent of the previous choice, in total, we get $1681 \cdot 41=68921$ ways.

It is easy to see that if we take two cards with powers of eleven and one with a power of three, the number of ways will be the same as in the previous case. Therefore, we get $1681+68921 \cdot 2=139523$ ways.