# 1955 AHSME Problems/Problem 38

## Problem 38

Four positive integers are given. Select any three of these integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers $29, 23, 21$, and $17$ are obtained. One of the original integers is: $\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17$

## Solution

Define numbers $a, b, c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.) $$\frac{a+b+c}{3}+d=29$$ $$\frac{a+b+d}{3}+c=23$$ $$\frac{a+c+d}{3}+b=21$$ $$\frac{b+c+d}{3}+a=17$$ Adding all of the equations together, we get: $2(a+b+c+d)=90$. This means that $a+b+c+d=45$.

We can determine that $\frac{a+b+c}{3}+d+16=a+b+c+d$. This, with some algebra, means that $\frac{1}{3}(a+b+c)=8$. $d$ must be $\boxed{\textbf{(B)} 21}$.

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