Lv 7
JOHN asked in Science & MathematicsMathematics · 5 years ago

One number is removed from the set of the first n integers. The average of the remainder is 40 ¾. Which number was removed?

5 Answers

  • Indica
    Lv 7
    5 years ago
    Favourite answer

    If the number removed is k then n(n+1)/2 – k = 40¾*(n−1) with 1 ≤ k ≤ n

    Fiddling around a bit gives 4(n−k) = (163−2n)(n−1) so 2|(n−1) and n is odd

    Now 0 ≤ (n−k)/(n−1) ≤ 1 hence 0 ≤ (163−2n)/4 ≤ 1 or 79½ ≤ n ≤ 81½

    ∴ only solution is n=81 and k = 81*41−40¾*80 = 61

  • 5 years ago

    The sum of the 1st n integers is (1/2)n(n+1).

    If 1 is removed the sum will be (1/2)(n-1)(n+2) and the average of the remainder will be (1/2)(n+2).

    If n is removed the sum will be (1/2)(n-1)(n-1+1)=(1/2)(n-1)n and the average of the remainder will be (1/2)n.

    The actual average of the remainder must lie between these two extremes :

    (1/2)n < 40.75 < (1/2)(n+2)

    n < 81.5 < n+2.

    The only possible values for n are 80 and 81.

    The total after one integer is removed is either 79*40.75=3219.25 or 80*40.75=3260.

    This total must be an integer, so it must be 3260 with n=81.

    The sum of the 1st 81 integers is (1/2)(81)(82) = 81*41 = 1+40+80+3200 = 61+3260.

    So the number removed was 61.

  • 5 years ago

    The number 61 was removed.

    n = 81.

    @JOHN: I'm sure there is a way to reason to figure it out, but being lazy I wrote this totally dumb maple program:

    for n from 2 to 200 do:

    for k from 1 to n do:

    if (n*(n+1)/2 - k)/(n-1) = 163/4 then print(n,k) end if:



    And it spit out

    81, 61

  • 5 years ago

    I think that the removed number was supposed to be number, but my calculation contains many algebraic works...

    Check my answer :) I am not sure if it is right.

    Attachment image
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  • JOHN
    Lv 7
    5 years ago

    Details of your reasoning? Or did you just plug the problem into Wolfram Alpha?

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