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Peter

#1 h May 25, 2007, 12:26 AM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Four consecutive even numbers are removed from the set $A=\{ 1, 2, 3, \cdots, n \}.$ If the arithmetic mean of the remaining numbers is $51.5625$ , which four numbers were removed?

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rem

#2 h Oct 11, 2007, 6:56 PM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Let the sum of the remaining numbers be $A$ . Then, because there are
$n - 4$ numbers remaining after $4$ numbers are removed, have
$\frac {A}{n - 4} = 51.5625 = \frac {825}{16}$
Now, the sum of the $4$ even numbers removed is greater than
$10 = 1 + 2 + 3 + 4$ but less than $n + (n - 1) + (n - 2) + (n - 3)$ because the numbers removed are all even and belong to the set $\{1,2,...,n\}$ .
The sum of all the numbers in set $\{1,2,...,n\}$ is $\frac {n(n + 1)}{2}$ .
Then, $\frac {n(n + 1)}{2} - (n + (n - 1) + (n - 2) + (n - 3)) < A < \frac {n(n + 1)}{2} - 10$
Simplifying, get:
$\frac {(n - 3)(n - 4)}{2} < A < \frac {(n - 4)(n + 5)}{2}$
Then:
$\frac {n - 3}{2} < \frac {A}{n - 4} < \frac {n + 5}{2}$
But $\frac {A}{n - 4} = \frac {825}{16}$ so:
$\frac {n - 3}{2} < \frac {825}{16} < \frac {n + 5}{2}$
Clearing denominators, we get $98.125 < n < 106.125$ , and as $n$ is a
positive integer, $99 < = n < = 106$ .
On the other hand, if the four consecutive numbers are
$2(k - 1),2(k),2(k + 1),2(k + 2)$ ( $k$ is an integer) their sum is $2(4k + 2)$ the sum of all the numbers from $1$ to $n$ is:
$A + 2(4k + 2) = \frac {n(n + 1)}{2}$
But $A = \frac {825}{16}(n - 4)$ , so:
$\frac {825}{16}(n - 4) + 2(4k + 2) = \frac {n(n + 1)}{2}$ (*)
Because $2(4k + 2)$ , $\frac {n(n + 1)}{2}$ are integers, $\frac {825}{16}(n - 4)$ is an integer, so $16|(n - 4)$ so $n\equiv 4\pmod {16}$ . Then, the only such value of $n$ such that $99 \le n \le 106$ is $n = 100$ .
Then, substituting $n = 100$ into (*) have:
$2(4k + 2) = 100$ then $k = 12$ and so the four consecutive even integers removed are $22,24,26,28$ .

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