## Task $7 / 68$

We are looking for all natural numbers $n$ for which $\sum_{k=1}^{n} k$ is a three-digit number with identical digits.
It is

$$
\sum_{k=1}^{n} k=\frac{n(n+1)}{2}
$$

according to the problem statement, this expression should be a three-digit number with identical digits.

Such a number can be represented by $111a$ with $1 \leq a \leq 9$ ($a$ being an integer). Therefore, the equation

$$
\frac{n(n+1)}{2}=111a
$$

must hold. Rearranged, this gives $n(n+1)=222a=2 \cdot 3 \cdot 37 \cdot a$.

The left side of this equation is a product of two consecutive numbers. Consequently, the right side must also be such a product. From this, one concludes that

$$
2 \cdot 3 \cdot a=6a=36 \text{ or } 2 \cdot 3 \cdot a=6a=38
$$

must hold (one can easily verify that other factor combinations do not satisfy the condition for $a$). Only the first of these two equations has an integer solution for $a$, namely $a=6$. Thus, $n=36$, and the three-digit number is 666.