One number is removed from the set of the first n integers. The average of the remainder is 40 ¾. Which number was removed?
- IndicaLv 75 years agoFavourite 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
- Barry GLv 75 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.
- Rita the dogLv 75 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
- Gia huy HoLv 55 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.
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- 5 years ago
Details of your reasoning? Or did you just plug the problem into Wolfram Alpha?