Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>. | In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>. | ||
− | Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math> | + | Thus, our answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>. |
~JHawk0224 ~awesomediabrine | ~JHawk0224 ~awesomediabrine | ||
==Solution 3 (Deeper)== | ==Solution 3 (Deeper)== | ||
− | 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)}}</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 | ||
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==Solution 5== | ==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. | + | 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 <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>.. |
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 15:06, 15 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
- 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 ..
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.