2003 AMC 10A Problems/Problem 13

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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

Solution 1

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 $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \mathrm{(A)}$ .

Solution 2

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

$\begin{eqnarray*}1x+1y+1z&=&20\\1x-4y-4z&=&0\\0x+1y-7z&=&0\\\end{eqnarray*}$

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, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28$

2003 AMC 10A (Problems)
Preceded by Problem 12	Followed by Problem 14
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

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