PRMO 2017 Question 15

15. 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? [PRE-RMO – 2017]

Solution:

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Let two numbers moved be 1, 2.

Then average of remaining (n − 2) numbers

$$\bar{x}_1 = \frac{n(n+1)}{2} - 3 \div (n - 2) = \frac{n^2 + n - 6}{2(n - 2)} = \frac{(n + 3)(n - 2)}{2(n - 2)} = \frac{n + 3}{2}$$

If two numbers be $n, n - 1$, then

$$\bar{x}_2 = \frac{n(n + 1)}{2} - (2n - 1) \div (n - 2) = \frac{n^2 - 3n + 2}{2(n - 2)} = \frac{n - 1}{2}$$

Now, the removed numbers are $m, k$, then new
average is 17 i.e.,

$$17 = \frac{n(n + 1)}{2} - (m + k) \div (n - 2)$$

Clearly,

$$\frac{n - 1}{2} < 17 < \frac{n + 3}{2}$$

⇒ $n < 35$ & $n > 31$
∴ $n = 32, 33, 34$

$$17 = \frac{n(n + 1)}{2} - (m + k) \div (n - 2)$$

∴

$$(m + k) = \frac{n(n + 1) - 34(n - 2)}{2}$$

When $n = 32$ ⇒ $m + k = 18$
$n = 33$ ⇒ $m + k = 34$
$n = 34$ ⇒ $m + k = 51$

∴ $(m + k)_{\text{max}} = 51$