I Need A Concrete Information

Probability Level 3

The largest of 21 distinct integers is 2015, and one of the other number is 101. The sum of any 11 of them is greater than the sum of the other 10.

Let those 21 integers are arranged in ascending order, find the number in the middle. In other words, find the median of these numbers.

1995 2003 2005

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Charuka Bandara by Charuka Bandara , 21, Sri Lanka

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1 solution

Otto Bretscher
Apr 16, 2016

Let the numbers besides 101 be a 1 < . . < a 19 < a 20 = 2015 a_1<..<a_{19}<a_{20}=2015 , with a 10 a_{10} being the number in the middle. Since the sum of the bottom 11 exceeds the sum of the top 10, we find 101 + ∑ k = 1 10 a k > ∑ k = 11 19 a k + 2015 101+\sum_{k=1}^{10}a_k>\sum_{k=11}^{19}a_k+2015 or a 10 > ∑ k = 11 19 a k − ∑ k = 1 9 a k + 1914 ≥ 9 × 10 + 1914 = 2004 a_{10}>\sum_{k=11}^{19}a_k-\sum_{k=1}^{9}a_k+1914\geq 9\times 10+1914=2004 since a k + 10 ≥ a k + 10 a_{k+10}\geq a_k+10 .

On the other hand we have a 10 ≤ a 20 − 10 = 2005 a_{10}\leq a_{20}-10 =2005 so a 10 = 2005 a_{10}=\boxed{2005}

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