# 1988 AJHSME Problems/Problem 21

## Problem

A fifth number, $n$, is added to the set $\{ 3,6,9,10 \}$ to make the mean of the set of five numbers equal to its median. The number of possible values of $n$ is $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than }4$

## Solution

The possible medians after $n$ is added are $6$, $n$, or $9$. Now we use casework.

Case 1: The median is $6$

In this case, $n<6$ and $$\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2$$ so this case contributes $1$.

Case 2: The median is $n$

We have $6 and $$\frac{3+6+n+9+10}{5}=n \Rightarrow n=7$$ so this case also contributes $1$.

Case 3: The median is $9$

We have $9 and $$\frac{3+6+9+n+10}{5}=9 \Rightarrow 17$$ so this case adds $1$.

In all there are $3\rightarrow \boxed{\text{C}}$ possible values of $n$.

## See Also

 1988 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS