How many five-digit numbers are there that end in six and are divisible by three?
I. solution: The numbers ending in 6 and divisible by 3 form an arithmetic sequence whose difference is the least common multiple of 3 and 10, which is 30. The first term of the sequence is 10026, and the last term is 99996. Thus, the number of terms, according to the formula $a_{n}=a_{1}+(n-1) d$,

$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{99996-10026}{30}+1=3000
$$

Tibor Farkas (Veszprém, László Lovassy Grammar School)

II. solution: If we write 6 at the end of a 4-digit number divisible by 3, the divisibility by 3 remains, and all the sought numbers can be formed this way without exception. Therefore, our task is clearly the following: how many 4-digit numbers are divisible by 3. $9999-999=9000$ four-digit numbers exist, and out of these, every third, that is, 3000 are divisible by 3.

Pál Bártfai (Budapest, Petőfi Grammar School)