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Integers-1-2-3-n-where-n-gt-2-are-written-on-a-board-Two-numbers-m-k-such-that-1-lt-m-lt-n-1-lt-k-lt-n-are-removed-and-the-average-of-the-remaining-numbers-is-found-to-be-17-W




Question Number 20693 by Tinkutara last updated on 31/Aug/17
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?
Integers1,2,3,.,n,wheren>2,arewrittenonaboard.Twonumbersm,ksuchthat1<m<n,1<k<nareremovedandtheaverageoftheremainingnumbersisfoundtobe17.Whatisthemaximumsumofthetworemovednumbers?
Answered by ajfour last updated on 31/Aug/17
(m+k)_(maximum) = 51  ((n(n+1))/2)−(m+k)=17(n−2)  ⇒   m+k=((n(n+1))/2)−17(n−2)≤ 2n−2        first let us see      (n^2 /2)+(n/2)−17n+34−2n+2 ≤ 0  ⇒  n^2 −37n+72 ≤ 0      (n−((37)/2))^2  ≤ (((37)/2))^2 −72     ∣n−((37)/2)∣ ≤ ((1369−288)/4)   ∣n−((37)/2)∣ ≤ (√(270.25))     (≈16.48)  ⇒    n−18.5 ≤ 16.48  or   n=34      m+k = ((n(n+1))/2)−17(n−2)                     =17×35−17×32                     =51 .
(m+k)maximum=51n(n+1)2(m+k)=17(n2)m+k=n(n+1)217(n2)2n2firstletusseen22+n217n+342n+20n237n+720(n372)2(372)272n37213692884n372270.25(16.48)n18.516.48orn=34m+k=n(n+1)217(n2)=17×3517×32=51.
Commented by Tinkutara last updated on 01/Sep/17
Thank you very much Sir!
ThankyouverymuchSir!

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