Five numbers are selected and deleted from a set of consecutive positive integers beginning with 1. The arithmetic mean of the remaining numbers in the set is 45.1.
Find the largest possible single number that could have been deleted from the original set.
If n is the highest number in the original set, and we use x to represent what the total of the five numbers to be taken away must be, then
(n*(n+1)/2 - x) / (n - 5) = 45.1
n*(n+1)/2 - x = 45.1 * (n - 5) = 45.1 n - 225.5
5*n*(n+1) - 10*x = 451 n - 2255
10*x = 5*n^2 + 5*n - 451*n + 2255
10*x = 5*n^2 - 446*n + 2255
The lhs is even so so must be the right. The rhs is even iff n is odd. The lhs is a multiple of 5, so so must be the right, which happend only if n is a multiple of 5, so n is an odd multiple of 5.
Since x represents the total amount that has to be taken away, it must be positive. To find the demarcaton value(s) of n between positive and negative, solve
5n^2 - 446n + 2255 = 0
n = (446 +/- sqrt(153816)) / 10
One of these zeroes lies between 5 and 6, and the other between 83 and 84. The values are negative between these, but positive before the first and after the second.
The value n=5 doesn't work, as nothing is left to average after removing 5 numbers.
That leaves n = 85, 95, 105, ...
The actual amount needed to be subtracted from the total, as seen above, is
x = n^2/2 - 44.6*n + 225.5
But when n = 95, this means that 501 must be subtracted from the total by the removal of 5 numbers, so the average that would need to be removed is over 100, but none of the numbers even equals 100 when n = 95. This situation gets even worse as higher n's are sought, as x varies quadratically with n.
So n=85 is the only possibility. At that point, x = 47 is the total by which the sum must be reduced by eliminating 5 numbers. To maximize the highest of these, the other four should be as small as possible: 1 thru 4, totaling 10. Then 37 is also removed, which is the answer.
Posted by Charlie
on 2007-05-09 11:06:47