Difference between revisions of "2003 AMC 10A Problems/Problem 13"
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== Solution == | == Solution == | ||
− | Let the numbers be <math>x</math>, <math>y</math>, and <math>z</math> in that order. | + | === Solution 1 === |
+ | Let the numbers be <math>x</math>, <math>y</math>, and <math>z</math> in that order. The given tells us that | ||
− | < | + | <cmath>\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*}</cmath> | ||
− | <math> | + | Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \boxed{\mathrm{(A)}\ 28}</math>. |
− | + | === Solution 2 === | |
+ | Alternatively, we can set up the system in [[equation]] form: | ||
− | < | + | <cmath>\begin{eqnarray*}1x+1y+1z&=&20\\ |
+ | 1x-4y-4z&=&0\\ | ||
+ | 0x+1y-7z&=&0\\ | ||
+ | \end{eqnarray*}</cmath> | ||
− | <math> | + | Or, in [[matrix]] form |
+ | <math> | ||
+ | \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} | ||
+ | </math> | ||
− | + | To solve this matrix equation, we can rearrange it thus: | |
− | + | <math>\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} | ||
+ | </math> | ||
− | + | Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields | |
− | <math>x | + | <math>\begin{bmatrix} |
− | + | x \\ | |
− | + | y \\ | |
− | + | z \\ | |
− | + | \end{bmatrix} = | |
+ | \begin{bmatrix} | ||
+ | \frac{1}{2} \\ | ||
+ | \frac{7}{2} \\ | ||
+ | 16 \\ | ||
+ | \end{bmatrix} | ||
+ | </math> | ||
− | + | Which means that <math>x = \frac{1}{2}</math>, <math>y = \frac{7}{2}</math>, and <math>z = 16</math>. Therefore, <math>xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 \Rightarrow \boxed{\mathrm{(A)}\ 28}</math> | |
− | <math> | + | === Solution 3 === |
+ | Let's denote the 3rd number as <math>x</math>, the 2nd as <math>7x</math>, and the 1st as <math>4(7x+x)</math> according to the information given in the problem. We know that all three numbers add up to <math>20</math>, so <math>4(8x)+7x+x = 20</math>. Solving the equation we get <math>x = \frac{1}{2}</math>. Then substitute this value of x to solve for the other two numbers. Lastly, we obtain <math>28</math> as the product of all three numbers. | ||
− | + | ==Video Solution by WhyMath== | |
+ | https://youtu.be/3hY9YXqn5zg | ||
− | + | ~savannahsolver | |
− | + | ==Video Solution== | |
− | + | https://www.youtube.com/watch?v=7V2k1WxooJ0 ~David | |
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− | == See | + | == See also == |
− | + | {{AMC10 box|year=2003|num-b=12|num-a=14|ab=A}} | |
− | |||
− | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:45, 19 August 2023
Contents
Problem
The sum of three numbers is . 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
Solution 1
Let the numbers be ,
, and
in that order. The given tells us that
Therefore, the product of all three numbers is .
Solution 2
Alternatively, we can set up the system in equation form:
Or, in matrix form
To solve this matrix equation, we can rearrange it thus:
Solving this matrix equation by using inverse matrices and matrix multiplication yields
Which means that ,
, and
. Therefore,
Solution 3
Let's denote the 3rd number as , the 2nd as
, and the 1st as
according to the information given in the problem. We know that all three numbers add up to
, so
. Solving the equation we get
. Then substitute this value of x to solve for the other two numbers. Lastly, we obtain
as the product of all three numbers.
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=7V2k1WxooJ0 ~David
See also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.