Difference between revisions of "2001 AMC 10 Problems/Problem 10"
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These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288</math>. We divide <math>xyz = 288</math> by each of the given equations, which yields <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = 22</math>, so the answer is <math>\boxed{\textbf{(D) } 22}</math>. | These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288</math>. We divide <math>xyz = 288</math> by each of the given equations, which yields <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = 22</math>, so the answer is <math>\boxed{\textbf{(D) } 22}</math>. | ||
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+ | ==Solution 3(strategic guess and check)== | ||
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+ | Seeing the equations, we notice that they are all multiples of 12. Trying in factors of 12, we find that <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math> work. <math>4 + 6+ 12 = \boxed{\textbf{(D) } 22} </math> | ||
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+ | ~idk12345678 | ||
== See Also == | == See Also == |
Revision as of 22:42, 9 April 2024
Problem
If , , and are positive with , , and , then is
Solution 1
The first two equations in the problem are and . Since , we have . We can substitute into the third equation to obtain and . We replace into the first equation to obtain .
Since we know every variable's value, we can substitute them in to find .
Solution 2
These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives . We divide by each of the given equations, which yields , , and . The desired sum is , so the answer is .
Solution 3(strategic guess and check)
Seeing the equations, we notice that they are all multiples of 12. Trying in factors of 12, we find that , , and work.
~idk12345678
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.