Difference between revisions of "2003 AMC 10A Problems/Problem 13"
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<math> \mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800 </math> | <math> \mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800 </math> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === |
Revision as of 00:18, 19 October 2020
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.