3. There are 207 different cards with numbers $1,2,3,2^{2}, 3^{2}, \ldots, 2^{103}, 3^{103}$ (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 6?
Answer: 267903.

Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of two and three must be recorded. There are 51 ways to choose an even power of two and 51 ways to choose an even power of three, and since this choice is made independently, the total number of ways in this case is $51^{2}=2601$.

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

Suppose we take two cards with powers of two. Since the power of two in the product of the selected numbers must be even, the powers of two on the selected cards must have the same parity. We have a total of 52 cards with odd powers and 51 cards with even powers. Thus, there are $C_{52}^{2}=\frac{52 \cdot 51}{2}=1326$ ways to choose cards with odd powers and $C_{51}^{2}=\frac{51 \cdot 50}{2}=1275$ ways to choose two cards with even powers, totaling 2601 ways. After selecting the cards with powers of two, we can choose any card with an even power of three (51 ways). Since the choice of the card with the power of three is independent of the previous choice, in total, we get $2601 \cdot 51=132651$ ways.

It is easy to see that if we take two cards with powers of three and one with a power of two, the number of ways will be the same as in the previous case. Therefore, we get $2601 + 132651 \cdot 2 = 267903$ ways.