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Algebra Difficulty 3.8 AMC 10/12 Find the answer Canada

Consider the five distinct integers $16$ , $x$ , $8$ , $17$ , and $11$ . Their mean (average) and their median
are equal. What is the sum of all possible values of $x$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The mean of the five integers is $\dfrac{16+x+8+17+11}{5}=\dfrac{52+x}{5}$ .

The median must be smaller than two other integers in the list, so
$17$ cannot be the median since there
are at least three integers in the list that are less than $17$ . Similarly, $8$ cannot be the median since there are at
least three integers in the list that are greater than $8$ .

The possible values for the median are $x$ , $11$ , and $16$ .

Since the median and mean are equal, we must have $x=\dfrac{52+x}{5}$ , which implies $x=13$ , or $11=\dfrac{52+x}{5}$ which implies $x=3$ , or $16=\dfrac{52+x}{5}$ which implies $x=28$ .

Therefore, the possibilities are $x=3$ , $x=13$ , and $x=28$ . Their sum is $3+13+28=44$ .

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.