Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
− | We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get \begin{align*} | + | We know that <math>f(2)=2</math>. Adding <math>f(1)</math> to both sides, we get <math>\begin{align*} |
f(2)+f(1)&=2+f(1)\\ | f(2)+f(1)&=2+f(1)\\ | ||
f(2)&=2+f(1)\\ | f(2)&=2+f(1)\\ | ||
2&=2+f(1)\\ | 2&=2+f(1)\\ | ||
f(1)&=0 | f(1)&=0 | ||
− | \end{align*} | + | \end{align*}</math> |
Also | Also | ||
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath> | <cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath> |
Revision as of 15:27, 11 February 2021
Problem
Let be a function defined on the set of positive rational numbers with the property that for all positive rational numbers and . Furthermore, suppose that also has the property that for every prime number . For which of the following numbers is ?
Solution 1
Looking through the solutions we can see that can be expressed as so using the prime numbers to piece together what we have we can get , so or .
-Lemonie
Solution 2
We know that . Adding to both sides, we get $\begin{align*} f(2)+f(1)&=2+f(1)\\ f(2)&=2+f(1)\\ 2&=2+f(1)\\ f(1)&=0 \end{align*}$ (Error compiling LaTeX. Unknown error_msg) Also In we have .\\ In we have .\\ In we have .\\ In we have .\\ In we have .\\ Thus, our answer is ~JHawk0224
Video Solution by Punxsutawney Phil
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.