3. There are 183 different cards with numbers $1,2,11,2^{2}, 11^{2}, \ldots, 2^{91}, 11^{91}$ (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 selected cards is a square of an integer divisible by 22?
Answer: 184275.

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 two and eleven must be recorded. There are 45 ways to choose an even power of two and 45 ways to choose an even power of eleven, and since this choice is made independently, the total number of ways in this case is $45^{2}=2025$.

Now let's move on to the case when the card with the number one is not used. There are two possibilities: take two cards with powers of two and one with a power of eleven, or vice versa.

Suppose we take two cards with powers of two. Since the number two must be present in an even power in the product of the selected numbers, the powers of two on the selected cards must have the same parity. We have a total of 46 cards with odd powers and 45 cards with even powers. Thus, there are $C_{46}^{2}=\frac{46 \cdot 45}{2}=1035$ ways to choose cards with odd powers and $C_{45}^{2}=\frac{45 \cdot 44}{2}=990$ ways to choose two cards with even powers - in total, we get 2025 ways. After selecting the cards with powers of two, we can choose any card with an even power of eleven (45 ways). Since the choice of the card with the power of eleven is independent of the previous choice, in total, we get $2025 \cdot 45=91125$ ways.

It is easy to see that if we take two cards with powers of eleven and one with a power of two, the number of ways will be the same as in the previous case. Therefore, we get $2025+91125 \cdot 2=184275$ ways.