Difference between revisions of "2003 AMC 10A Problems/Problem 13"
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
| − | Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \mathrm{(A)}</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 === | === Solution 2 === | ||
Revision as of 21:00, 3 January 2014
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 matrix 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, 
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. 
