5. For how many two-digit natural numbers n are exactly two of these three statements true: (A) n is odd; (B) n is not divisible by $3 ;$ (C) n is divisible by 5?
Solution. We can consider the first 30 two-digit numbers (from 10 to 39), and then multiply the result by 3, since the remainders when dividing by 2, 3, and 5 do not change when shifted by 30 or 60. There are three mutually exclusive cases.

1) (A) and (B) are satisfied, and (C) is not. From (A) and (B), it follows that \( n \) is divisible by 6 with a remainder of 1 or 5, and \( n \) should not be divisible by 5. There are eight such numbers: 11, 13, 17, 19, 23, 29, 31, 37.
2) (A) is satisfied, (B) is not, and (C) is satisfied. Then \( n \) is divisible by 6 with a remainder of 3, and \( n \) is divisible by 5. There is one such number: 15.
3) (A) is not satisfied, (B) and (C) are satisfied. Then \( n \) is divisible by 6 with a remainder of 2 or 4, and \( n \) is divisible by 5. There are two such numbers: 10 and 20.

In total: \( 3 * (8 + 1 + 2) = 33 \) numbers.

Answer: 33.