2003 AMC 10A Problems/Problem 13

Problem

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?

$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$

Solution

Let the numbers be $x$ , $y$ , and $z$ in that order.

$y=7z$

$x=4(y+z)=4(7z+z)=4(8z)=32z$

$x+y+z=32z+7z+z=40z=20$

$z=\frac{20}{40}=\frac{1}{2}$

$y=7z=7\cdot\frac{1}{2}=\frac{7}{2}$

$x=32z=32\cdot\frac{1}{2}=16$

Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A$

Alternatively, we can set up the system in matrix form:

$x+y+z=20$

$x=4(y+z)=4y+4z$

$y=7z$

is equivalent to

$1x+1y+1z=20$

$1x-4y-4z=0$

$0x+1y-7z=0$

Or, in matrix form $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$ To solve this matrix equation, we can rearrange it thus: $\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} ^{-1} \begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$ Solving this matrix equation by using inverse matrices and matrix multiplication yields $\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{7}{2} \\ 16 \\ \end{bmatrix}$ Which means that $x = \frac{1}{2}$ , $y = \frac{7}{2}$ , and $z = 16$ . Therefore, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28$

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2003 AMC 10A Problems/Problem 13

Problem

Solution

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