Question

Integers 1, 2, 3, ..., n, where n > 2 are written on a board. Two numbers m, k such that 1 < m < n, 1 < k < n are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers?

Answer 1

51

Answer 2

Let the sum of numbers be $S_n = \frac{n(n+1)}{2}$. Let the two removed numbers be m and k. Constraints: $2 \le m, k \le n-1$ and $m \ne k$. The sum of remaining numbers is $S_n - (m+k)$. The average of remaining numbers is 17: $\frac{S_n - (m+k)}{n-2} = 17$. Substitute $S_n$: $\frac{n(n+1)/2 - (m+k)}{n-2} = 17$. Let $X = m+k$. $X = \frac{n(n+1)}{2} - 17(n-2) = \frac{n^2+n-34n+68}{2} = \frac{n^2-33n+68}{2}$. The minimum possible value for $X$ is $2+3=5$. The maximum possible value for $X$ is $(n-1)+(n-2)=2n-3$. So, $5 \le \frac{n^2-33n+68}{2} \le 2n-3$.

1. $10 \le n^2-33n+68 \implies n^2-33n+58 \ge 0$. Roots are $\frac{33 \pm \sqrt{857}}{2} \approx 1.865, 31.135$. Since $n>2$, $n \ge 32$.
2. $n^2-33n+68 \le 4n-6 \implies n^2-37n+74 \le 0$. Roots are $\frac{37 \pm \sqrt{1073}}{2} \approx 2.125, 34.875$. So $3 \le n \le 34$.

Combining $n \ge 32$ and $3 \le n \le 34$, we get $32 \le n \le 34$. The function $X(n) = \frac{n^2-33n+68}{2}$ is an upward-opening parabola with axis of symmetry at $n=16.5$. For $n \in \{32, 33, 34\}$, the function is increasing. Therefore, the maximum value of X occurs at $n=34$. For $n=34$: $X = \frac{34^2 - 33(34) + 68}{2} = \frac{34(34-33) + 68}{2} = \frac{34+68}{2} = \frac{102}{2} = 51$. For $n=34$, the allowed range for m, k is $2 \le m, k \le 33$. A sum of 51 is achievable (e.g., $m=25, k=26$).

The final answer is 51.