# 1952 AHSME Problems/Problem 47

## Problem

In the set of equations $z^x = y^{2x},\quad 2^z = 2\cdot4^x, \quad x + y + z = 16$, the integral roots in the order $x,y,z$ are: $\textbf{(A) } 3,4,9 \qquad \textbf{(B) } 9,-5,-12 \qquad \textbf{(C) } 12,-5,9 \qquad \textbf{(D) } 4,3,9 \qquad \textbf{(E) } 4,9,3$

## Solution #1

The easiest method, which in this case is not very time consuming, is to guess and check and use process of elimination. We can immediately rule out (B) since it does not satisfy the third equation. The first equation allows us to eliminate (A) and (C) because the bases are different and aren't powers of each other. Finally, we manually check (D) and (E) and see that our answer is $\fbox{D}$

## Solution #2

Let us actually solve the problem through algebra. We start by simplifying each of the equations given to us.

Equation #1 $$z^x = y^{2x}$$ $$z = y^2$$

Equation #2 $$2^z = 2 * 4^x$$ $$2^z = 2^{2x + 1}$$ $$z = 2x + 1$$

Equation #3 $$x + y + z = 16$$

Equations 1 & 2 tell us that $z$ is an odd perfect square. They also tell us that $z$ is the largest of the three integers. We see that if we assume $z = 9$, then $y = 3$ and $x = 4$, which satisfies all three equations.

Therefore, the answer is $\fbox{(D) 4,3,9}$

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