VIII.2. The sum of all natural numbers from 1 to p is equal to a three-digit number with identical digits. How many natural numbers were there in the summation?
Solution: According to the condition of the problem, we have: $1+2+3+\ldots+n=$ $\overline{\text { aaa }}=\mathrm{a} \cdot 111$. It is known that $1+2+3+\ldots+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$. Therefore, $\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\mathrm{a} \cdot 111$; $n(n+1)=\mathrm{a} \cdot 222 ; n(n+1)=\mathrm{a} \cdot 6 \cdot 37$. On the left side, we have the product of two consecutive natural numbers, meaning one number is 37, and the other is 36, as it is divisible by 6, i.e., $\mathrm{n}=\mathrm{a} \cdot 6=36$. Therefore, there were 36 natural numbers in the sum, and their sum is 666.