Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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<u><b>Remark</b></u> | <u><b>Remark</b></u> | ||
− | Similarly, we can | + | Similarly, we can find the outputs of <math>f</math> at the inputs of the other answer choices: |
<cmath>\begin{alignat*}{10} | <cmath>\begin{alignat*}{10} | ||
&\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\ | &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\ | ||
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&\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 | &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 | ||
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
− | Alternatively, refer to Solutions 2 and 4 for the full | + | Alternatively, refer to Solutions 2 and 4 for the full processes. |
~Lemonie ~awesomediabrine ~MRENTHUSIASM | ~Lemonie ~awesomediabrine ~MRENTHUSIASM | ||
− | ==Solution 2== | + | ==Solution 2 (Specific)== |
We know that <math>f(p) = f(p \cdot 1) = f(p) + f(1)</math>. By transitive, we have <cmath>f(p) = f(p) + f(1).</cmath> | We know that <math>f(p) = f(p \cdot 1) = f(p) + f(1)</math>. By transitive, we have <cmath>f(p) = f(p) + f(1).</cmath> | ||
Subtracting <math>f(p)</math> from both sides gives <math>0 = f(1).</math> | Subtracting <math>f(p)</math> from both sides gives <math>0 = f(1).</math> | ||
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~JHawk0224 ~awesomediabrine | ~JHawk0224 ~awesomediabrine | ||
− | ==Solution 3 ( | + | ==Solution 3 (Generalized)== |
Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>. | Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>. | ||
~yofro | ~yofro | ||
− | ==Solution 4 (Extremely Comprehensive | + | ==Solution 4 (Generalized: Extremely Comprehensive)== |
===Results=== | ===Results=== | ||
We have the following important results: | We have the following important results: | ||
− | |||
<ol style="margin-left: 1.5em;"> | <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(\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> | ||
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<li><math>f\left({\frac 1a}\right)=-f(a)</math> for all positive rational numbers <math>a</math></li><p> | <li><math>f\left({\frac 1a}\right)=-f(a)</math> for all positive rational numbers <math>a</math></li><p> | ||
</ol> | </ol> | ||
− | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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<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> | <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> | </ol> | ||
− | |||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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&=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. | &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | We apply function <math>f</math> to each fraction in the answer choices: | |
− | We apply function <math>f</math> to each fraction in the choices: | ||
− | |||
<cmath>\begin{alignat*}{10} | <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{(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 \\ | ||
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==Solution 5== | ==Solution 5== | ||
− | The problem gives us that f(p)=p. | + | The problem gives us that <math>f(p)=p.</math> If we let <math>a=p</math> and <math>b=1,</math> we get <math>f(p)=f(p)+f(1),</math> which implies <math>f(1)=0.</math> 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 <math>a=p</math> and <math>b=1/p,</math> we get <math>f(1)=f(p)+f(1/p).</math> We can solve for <math>f(1/p)</math> as <math>-f(p)!</math> This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> |
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Latest revision as of 08:30, 16 June 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 (Intuitive)
- 3 Solution 2 (Specific)
- 4 Solution 3 (Generalized)
- 5 Solution 4 (Generalized: Extremely Comprehensive)
- 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 (Intuitive)
From the answer choices, note that On the other hand, we have Equating the expressions for produces from which Therefore, the answer is
Remark
Similarly, we can find the outputs of at the inputs of the other answer choices: Alternatively, refer to Solutions 2 and 4 for the full processes.
~Lemonie ~awesomediabrine ~MRENTHUSIASM
Solution 2 (Specific)
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 (Generalized)
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 (Generalized: Extremely Comprehensive)
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 answer choices: Therefore, the answer is
~MRENTHUSIASM
Solution 5
The problem gives us that If we let and we get which implies 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 and we get We can solve for as This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of
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.