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# The set \$\{2, 4, 6, \dots, n\}\$ contains the positive consecutive even integers from 2 through \$n\$. When one of the integers from the set is

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The set \$\{2, 4, 6, \dots, n\}\$ contains the positive consecutive even integers from 2 through \$n\$. When one of the integers from the set is removed, the average of the remaining integers in the set is 28. What is the least possible value of \$n\$ ?

Sep 29, 2017

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Since the set consists of even integers from: 2, 4,  6, 8.........n, and if the average of the set is 28, then 28th term must be=28 x 2 =56=n. The sum of 28 terms from 2+4+ 6+ 8.......56=812. 812/28 =29 average. Therefore: 812 - 56 =756 / 27=28 - the average.

Therefore, the least possible value of n = 56.

Sep 29, 2017
#2
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I get a slightly different answer than the guest does

Notice that  the  greatest possible average from  adding the digits  2 + 4 + 6 + ....+ n

is given by :

[ ( n / 2) (n / 2  + 1 ) - 2 ]  /   [ (n - 2) / 2 ]

Example....   2 , 4 , 6 , 8  10

And the greatest  average possible average is produced when the least number, 2, is omitted

And this average is   [ ( 10 / 2) (10 / 2  + 1 ) - 2 ]  /   [ (10 - 2) / 2 ]  =

[(5)(6) - 2] / (8 / 2 )  =  28 / 4  =  7

So...we want to solve this :

[ ( n / 2) (n / 2  + 1 ) - 2 ]  /   [ (n - 2) / 2 ]  =  28      simplify

2 [ n^2 / 4  +  n/2  - 2 ] =   28 [ n - 2 ]

2 [ ( n^2 + 2n - 8) / 4 ]  =  28 [n - 2]

( n^2 + 2n - 8) / 2  = 28 [n - 2]

( n^2 + 2n - 8) =  56 [n - 2]

n^2 + 2n - 8 =  56n - 112

n^2 - 54n + 104  = 0     factor

( n - 52) ( n - 2)  = 0

And its obvious that  n  = 52  and this is the smallest value of n

Proof :

2  +   4  +   6  +  ....... +  52     can be written as

1   +  2   +  3 +  ........ +  26

+

1   +  2   +  3  + ......... + 26

So  52  is the 26th term

And  the sum of these two series  =  2 (26)(27)/ 2  =  26 * 27

And the average of this series omitting the first term is given by

[Series sum -  2] / [ Number of remaining terms omitting the first one, 2 ]

[ 26 * 27 - 2 ] / 25  =

[ 702 - 2] / 25  =

700 / 25  =

28

Sep 30, 2017
edited by CPhill  Sep 30, 2017
edited by CPhill  Sep 30, 2017