\section*{Problem 2 - 101012}

If \(n\) is a positive integer, then \(s_{n}\) denotes the sum of all positive integers from 1 to \(n\).

a) For which positive integer \(n\) do we get \(s_{n}=2415\)?

b) For which positive integer \(m\) is \(s_{m}\) exactly 69 times as large as \(m\)?

\section*{a) It holds that \(s_{n}=\frac{n(n+1)}{2}\).}
}
Assuming there is a positive integer \( n \) for which \(\frac{n(n+1)}{2}=2415\) holds, then \(n^{2}+n-4830=0\) follows, and from this either \(n=69\) or \(n=-70\). Since \(-70<0\), only \(n=69\) can have the desired property. Indeed, for the positive integer 69:

\[
s_{69}=\frac{69 \cdot 70}{2}=2415
\]

b) Assuming there is a positive integer \( m \) for which \(\frac{m(m+1)}{2}=69 m\) holds. Because \( m \neq 0 \), it follows that \(\frac{m+1}{2}=69\), so \(m=137\). Therefore, only this number can have the desired property. Indeed, for the positive integer 137

\[
s_{137}=\frac{137 \cdot 138}{2}=137 \cdot 69
\]

\section*{Adopted from \([5]\)}