Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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==Solution 4 (Generalized)== | ==Solution 4 (Generalized)== | ||
− | + | We derive the following properties of <math>f:</math> | |
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
<li>By induction, we have <cmath>f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)</cmath> for all positive rational numbers <math>a_k</math> and positive integers <math>n.</math> <p> | <li>By induction, we have <cmath>f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)</cmath> for all positive rational numbers <math>a_k</math> and positive integers <math>n.</math> <p> | ||
Line 78: | Line 78: | ||
&=\left[\sum_{k=1}^{m}d_k p_k \right]-\left[\sum_{k=1}^{n}e_k q_k \right]. | &=\left[\sum_{k=1}^{m}d_k p_k \right]-\left[\sum_{k=1}^{n}e_k q_k \right]. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We apply | + | We apply <math>f</math> to each fraction in the answer 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 \\ |
Revision as of 12:03, 22 September 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)
- 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)
We derive the following properties of
- By induction, we have
for all positive rational numbers
and positive integers
Since positive powers are just repeated multiplication of the base, it follows that
for all positive rational numbers
and positive integers
- For all positive rational numbers
we have
Therefore, we get
- For all positive rational numbers
we have
Therefore, we get
For all positive integers and
suppose
and
are their respective prime factorizations. We have
We apply
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.