Difference between revisions of "2021 AMC 12A Problems/Problem 18"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Comprehensive, Similar to Solution 3)) |
MRENTHUSIASM (talk | contribs) m (→Solution) |
||
(39 intermediate revisions by 4 users not shown) | |||
Line 6: | Line 6: | ||
<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math> | <math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math> | ||
− | ==Solution 1 | + | ==Solution 1== |
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>. | Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>. | ||
Line 41: | Line 41: | ||
~yofro | ~yofro | ||
− | ==Solution 4 (Comprehensive, Similar to Solution 3)== | + | ==Solution 4 (Extremely Comprehensive, Similar to Solution 3)== |
+ | ===Results=== | ||
We have the following important results: | We have the following important results: | ||
− | <math> | + | <ol style="margin-left: 1.5em;"> |
+ | <li><math>f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)</math> for all positive rational numbers <math>a_k</math> and positive integers <math>n</math></li><p> | ||
+ | <li><math>f\left(a^n\right)=nf(a)</math> for all positive rational numbers <math>a</math> and positive integers <math>n</math></li><p> | ||
+ | <li><math>f(1)=0</math></li><p> | ||
+ | <li><math>f\left({\frac 1a}\right)=-f(a)</math> for all positive rational numbers <math>a</math></li><p> | ||
+ | </ol> | ||
− | + | ~MRENTHUSIASM | |
− | <math>(3) \ f\left({\frac 1a}\right)=-f(a) | + | ===Proofs=== |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>Result 1: We can show Result 1 by induction.</li><p> | ||
+ | <li>Result 2: Since positive powers are just repeated multiplication of the base, we will use Result 1 to prove Result 2: <cmath>f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).</cmath></li><p> | ||
+ | <li>Result 3: For all positive rational numbers <math>a,</math> we have <cmath>f(a)=f(a\cdot1)=f(a)+f(1).</cmath> Therefore, we get <math>f(1)=0,</math> from which Result 3 is true.</li><p> | ||
+ | <li>Result 4: We have <cmath>f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.</cmath> Therefore, we get <math>f\left({\frac 1a}\right)=-f(a),</math> from which Result 4 is true.</li><p> | ||
+ | </ol> | ||
− | + | ~MRENTHUSIASM | |
− | + | ===Solution=== | |
+ | For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{m}p_k^{e_k}</math> and <math>\prod_{k=1}^{n}q_k^{d_k}</math> are their respective prime factorizations, we have | ||
+ | <cmath>\begin{align*} | ||
+ | f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) & \text{by Result 1} \\ | ||
+ | &=f(x)-f(y) & \text{by Result 4} \\ | ||
+ | &=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \\ | ||
+ | &=\left[\sum_{k=1}^{m}f\left(p_k^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] & \text{by Result 1} \\ | ||
+ | &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] &\hspace{10mm} \text{by Result 2} \\ | ||
+ | &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. | ||
+ | \end{align*}</cmath> | ||
− | + | We apply function <math>f</math> to each fraction in the choices: | |
− | + | <cmath>\begin{alignat*}{10} | |
+ | &\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)} \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)} \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)} \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)} \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{alignat*}</cmath> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> | ||
~MRENTHUSIASM | ~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== | ==Video Solution by Hawk Math== | ||
https://www.youtube.com/watch?v=dvlTA8Ncp58 | 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 Punxsutawney Phil== | ||
Line 70: | Line 104: | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/IUJ_A9KiLEE | ||
+ | |||
+ | ~IceMatrix | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}} | ||
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:25, 15 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 from which Result 3 is true.
- Result 4: We have Therefore, we get from which 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.