Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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We apply function <math>f</math> to each fraction in the choices: | We apply function <math>f</math> to each fraction in the choices: | ||
− | <cmath>\begin{ | + | <cmath>\begin{alignat*}{10} |
− | \textbf{(A) } & f\left(\frac{17}{32}\right) & = & f\left(\frac{17^1}{2^5}\right) & = & [1(17)]-[5(2)] & = & 7 \\ | + | &&\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && f\left(\frac{17^1}{2^5}\right) \quad && = \quad && [1(17)]-[5(2)] \quad && = \quad && 7 \\ |
− | \textbf{(B) } & f\left(\frac{11}{16}\right) & = & f\left(\frac{11^1}{2^4}\right) & = & [1(11)]-[4(2)] & = & 3 \\ | + | &&\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && f\left(\frac{11^1}{2^4}\right) \quad && = \quad && [1(11)]-[4(2)] \quad && = \quad && 3 \\ |
− | \textbf{(C) } & f\left(\frac{7}{9}\right) | + | &&\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && f\left(\frac{7^1}{3^2}\right) \quad && = \quad && [1(7)]-[2(3)] \quad && = \quad && 1 \\ |
− | \textbf{(D) } & f\left(\frac{7}{6}\right) | + | &&\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && f\left(\frac{7^1}{2^1\cdot3^1}\right) \quad && = \quad && [1(7)]-[1(2)+1(3)] \quad && = \quad && 2 \\ |
− | \textbf{(E) } & f\left(\frac{25}{11}\right) & = & f\left(\frac{5^2}{11^1}\right) & = & [2(5)]-[1(11)] & = & -1 | + | &&\textbf{(E)} \qquad && f\left(\frac{25}{11}\right) \quad && = \quad && f\left(\frac{5^2}{11^1}\right) \quad && = \quad && [2(5)]-[1(11)] \quad && = \quad && -1 |
− | \end{ | + | \end{alignat*}</cmath> |
Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> | Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> | ||
Revision as of 21:49, 14 April 2021
- The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Deeper)
- 5 Solution 4 (Extremely Comprehensive, Similar to Solution 3)
- 6 Solution 5
- 7 Video Solution by Hawk Math
- 8 Video Solution by North America Math Contest Go Go Go Through Induction
- 9 Video Solution by Punxsutawney Phil
- 10 Video Solution by OmegaLearn (Using Functions and manipulations)
- 11 Video Solution by TheBeautyofMath
- 12 See also
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
- awesomediabrine
Solution 2
We know that . By transitive, we have Subtracting from both sides gives Also In we have .
In we have .
In we have .
In we have .
In we have .
Thus, our answer is
~JHawk0224 ~awesomediabrine
Solution 3 (Deeper)
Consider the rational , for integers. We have . So . Let be a prime. Notice that . And . So if , . We simply need this to be greater than what we have for . Notice that for answer choices and , the numerator has less prime factors than the denominator, and so they are less likely to work. We check first, and it works, therefore the answer is .
~yofro
Solution 4 (Extremely Comprehensive, Similar to Solution 3)
Results
We have the following important results:
- for all positive rational numbers and positive integers
- for all positive rational numbers and positive integers
- for all positive rational numbers
~MRENTHUSIASM
Proofs
- Result 1: We can show Result 1 by induction.
- Result 2: Since positive powers are just repeated multiplication of the base, we will use Result 1 to prove Result 2:
- Result 3: For all positive rational numbers we have Therefore, we get and Result 3 is true.
- Result 4: For all positive rational numbers we have Therefore, we get and Result 4 is true.
~MRENTHUSIASM
Solution
For all positive integers and suppose and are their respective prime factorizations, we have
We apply function to each fraction in the choices:
Therefore, the answer is
~MRENTHUSIASM
Solution 5
The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.
Video Solution by Hawk Math
https://www.youtube.com/watch?v=dvlTA8Ncp58
Video Solution by North America Math Contest Go Go Go Through Induction
https://www.youtube.com/watch?v=ffX0fTgJN0w&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=12
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Using Functions and manipulations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
See also
2021 AMC 10A (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 10 Problems and Solutions |
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.