Difference between revisions of "2024 AIME II Problems/Problem 4"
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− | Adding all three equations, <math>\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}</math>. Subtracting this from every equation, we have: < | + | Adding all three equations, <math>\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}</math>. Subtracting this from every equation, we have: <cmath>2\log_2x = -\frac{7}{12},</cmath> <cmath>2\log_2y = -\frac{3}{4},</cmath> <cmath>2\log_2z = -\frac{5}{6}</cmath> Our desired quantity is the absolute value of <math>4\log_2x+3\log_2y+2\log_2z = 2(\frac{7}{12})+3/2(\frac{3}{4})+\frac{5}{6} = \frac{25}{8}</math>, so our answer is <math>25+8 = \boxed{033}</math>. |
~Spoirvfimidf | ~Spoirvfimidf | ||
Revision as of 09:16, 12 February 2024
Problem
Let and be positive real numbers that satisfy the following system of equations: Then the value of is where and are relatively prime positive integers. Find .
Solution 1
Denote , , and .
Then, we have:
Now, we can solve to get . Plugging these values in, we obtain . ~akliu
Solution 2
~Callisto531
Solution 3
Adding all three equations, . Subtracting this from every equation, we have: Our desired quantity is the absolute value of , so our answer is . ~Spoirvfimidf
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.