15.3. Boys are dividing nuts. The first one took 3 nuts and a fifth of the remainder; the second took twice 3 nuts and a fifth of the new remainder; the third took thrice 3 nuts and a fifth of the next remainder, and so on. How many boys were there if it turned out that as a result of such division, each received an equal number of nuts?

$$
\text { (6-8 grades) }
$$
15.3. Let the number of all boys be denoted by $n$. Then the last boy received $3 n$ nuts with nothing left over. The second-to-last boy received $3(n-1)$ nuts and one fifth of the remainder. Since the second-to-last boy received as many nuts as the last one, i.e., $3 n$ nuts, the one fifth of the remainder amounts to $3 n-3(n-1)=3$ nuts, and the entire remainder before the second-to-last boy took his one fifth consisted of 15 nuts. Therefore, the last boy received $15-3=12$ nuts. Then $3 n=12$, from which $n=4$. Answer: There were 4 boys in total.