2021 AMC 12A Problems/Problem 18
Contents
[hide]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 
Also
![\[f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2\]](http://latex.artofproblemsolving.com/6/e/c/6ecc6b0d5ad8e58eb436e0df1f88a8c3d4e5f1ac.png)
![\[f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3\]](http://latex.artofproblemsolving.com/4/2/d/42d125cac45ee3bfe2b5c9740a154268363d6289.png)
![\[f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11\]](http://latex.artofproblemsolving.com/5/f/e/5feb78c9665b9d473e56fc8bc309458f1c0f6f4e.png)
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 | |
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