Write two two-digit numbers in which the number 2 is written in the tens place and which are divisible by 3.

Let’s find two-digit natural numbers that are divisible by 3, while in the tens place they must have the digit 2. In order for a number to be divisible by three without a remainder, it is necessary that the sum of its digits be divisible by three. The smallest amount divisible by three, including the digit “2” will be: 2 + 1 = 3, which means that the first required number is 21. The next sum, divisible by three, including the digit “2” will be: 2 + 4 = 6, which means the second the required number is 24. Another option is possible: 2 + 7 = 9, that is, the third number is 27.
Answer: the required numbers are 21; 24; 27.