Bětka wrote down thirty consecutive integers and added all these numbers together. Then she erased the second, fifth, eighth, and every subsequent third number. She added all the numbers that remained after erasing and found that the new sum was 265 less than the original sum. Determine the number that Bětka erased first.

(E. Novotná)
Let the original thirty numbers be denoted as follows:

$$
a, a+1, a+2, a+3, a+4, a+5, \ldots, a+27, a+28, a+29 .
$$

The erased numbers were ten, specifically

$$
a+1, a+4, a+7, a+10, a+13, a+16, a+19, a+22, a+25, a+28 .
$$

The sum of these numbers is $10a + 145$, which corresponds to the given difference of 265. From this, we get $10a = 120$, hence $a = 12$.

The first number that Betka erased was $a+1 = 13$.

Evaluation. 3 points for writing down the relationships and auxiliary calculations; 3 points for the result and quality of the commentary.

Note. The intermediate sum above can be calculated efficiently as follows:

$$
1+4+\cdots+25+28=(1+28)+(4+25)+\cdots=29 \cdot 5=145 .
$$

Similarly, the consecutive terms of any arithmetic sequence can be summed in this manner.