AMC10 2003 10A, Problem 13

The sum of three numbers is $20$. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

Solution: Let the numbers be $x$, $y$, and $z$ in that order. The given tells us that \begin{eqnarray}y&=&7z\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\ x+y+z&=&32z+7z+z=40z=20\\ z&=&\frac{20}{40}=\frac{1}{2}\\ y&=&7z=7\cdot\frac{1}{2}=\frac{7}{2}\\ x&=&32z=32\cdot\frac{1}{2}=16 \end{eqnarray} Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \fbox{28}$.

Answer: 28