\section*{Problem 3 - 131013}

Someone wants to pose a calculation problem where from the set of natural numbers from 1 to a given natural number \(n\), exactly one given natural number \(x\) is removed, and the remaining \(n-1\) numbers are added. He wants to specify the numbers \(n\) and \(x\) such that the result of this calculation is 448.

Determine all possibilities for specifying \(x\) and \(n\) in this way!
}

The sum of all natural numbers from 1 to \(\mathrm{n}\) is \(\frac{n(n+1)}{2}\). Therefore, for natural numbers \(n, x\), the conditions of the problem are satisfied if and only if

\[
\frac{n(n+1)}{2}=448+x \quad(1) \quad \text { and } \quad 0 \leq x \leq n
\]
holds. If (1) and (2) are valid, then \(44829\); because if \(0896\)

From both conditions for \(n\), it follows that \(n=30\), and from this and (1)

\[
x=\frac{30 \cdot 31}{2}-448=17
\]

Therefore, only \(n=30, x=17\) can satisfy the requirements of the problem. Since \(\frac{30 \cdot 31}{2}=448+17\) and \(0<17 \leq 30\), they indeed satisfy the conditions.