Problem 2. The sum of 15 consecutive natural numbers is a number with different digits, including the digits $0,1,2$ and 4. What is the smallest possible number among the 15 numbers?
Solution. If $n$ is the smallest number sought, then the 15 consecutive numbers mentioned in the statement are $n, n+1, \ldots, n+14$, whose sum $S$ is equal to $15n + 105$, i.e., $S = 3 \cdot 5 \cdot (n + 7) \ldots \ldots . .3$ p

Since the sum $S$ is divisible by 3, the sum of the digits of $S$ is, in turn, divisible by 3. Since $0, 1, 2$ and 4 are digits of $S$, and $0 + 1 + 2 + 4 = 7 \text{ does not}$ divide by 3, it follows that $S$ has at least one more digit, different from $0, 1, 2$ and 4. The conditions of minimality and divisibility by 3 lead to the fact that $S$ has only one more digit, equal to 5

$2 p$

The smallest number of the form $15(n + 7)$ that is written with the digits $0, 1, 2, 4$ and 5 is 10245, for which $n = 676$ $2 \text{p}$