Difference between revisions of "2021 AIME I Problems/Problem 7"
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==Solution 4== | ==Solution 4== | ||
− | The equation implies that <math>\sin(mx)=\sin(nx)=1</math>. Therefore, we can write <math>mx</math> as <math>2{\pi}k_1+\frac{\pi}{2}</math> and <math>nx</math> as <math>2{\pi}k_2+\frac{\pi}{2}</math> for integers <math>k_1</math> and <math>k_2</math>. Then, <math>\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}</math> Cross multiplying we get | + | The equation implies that <math>\sin(mx)=\sin(nx)=1</math>. Therefore, we can write <math>mx</math> as <math>2{\pi}k_1+\frac{\pi}{2}</math> and <math>nx</math> as <math>2{\pi}k_2+\frac{\pi}{2}</math> for integers <math>k_1</math> and <math>k_2</math>. Then, <math>\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}</math>. Cross multiplying, we get <math>m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m</math>. Let <math>n-m=a</math> so the equation becomes <math>4(m(k_2-k_1)+k_1a)=a</math>. Let <math>k_2-k_1=X</math> and <math>k_1=Y</math>, then the equation becomes <math>a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y</math> Note that <math>X</math> and <math>Y</math> are not relevant so they can vary accordingly, and <math>a\mid{4}</math>. Next, we do casework on <math>m\pmod{4}</math>: |
If <math>m\equiv 1\pmod{4}</math>: | If <math>m\equiv 1\pmod{4}</math>: | ||
− | Once <math>a</math> and <math>m</math> are determined, <math>n</math> is determined, so <math>a+m\leq30</math>. <math>a\in {4,8,12,\dots,28}</math> and <math>m\in {1,5,9,\dots,29}</math>. Therefore, there are <math>\sum_{i=1}^{7}{i}=28</math> ways for this case such that <math>a+m\leq30</math> | + | Once <math>a</math> and <math>m</math> are determined, <math>n</math> is determined, so <math>a+m\leq30</math>. <math>a\in {4,8,12,\dots,28}</math> and <math>m\in {1,5,9,\dots,29}</math>. Therefore, there are <math>\sum_{i=1}^{7}{i}=28</math> ways for this case such that <math>a+m\leq30</math>. |
+ | |||
+ | If <math>m\equiv 3\pmod{4}</math>: | ||
+ | |||
+ | <math>a\in {4,8,12,\dots,28}</math> and <math>m\in {3,7,11,\dots,27}</math>. Therefore, there are <math>\sum_{i=1}^{6}{i}=21</math> ways such that <math>a+m\leq30</math>. | ||
+ | |||
+ | If <math>m\equiv 2\pmod{4}</math>: | ||
+ | |||
+ | Note that <math>a\mid8</math> since <math>m</math> in this case will have a factor of <math>2</math>, which will cancel out a factor of <math>2</math> in <math>a</math>, and we need the left hand side to divide <math>4</math>. Also, <math>1-4X\equiv 1\pmod{4}</math> so it is odd and will therefore never contribute a factor of <math>2</math>. <math>a\in {8,16,24}</math> and <math>m\in {2,6,10,\dots,30}</math>. Following the condition <math>a+m\leq30</math>, there are <math>6+4+2=12</math> ways for this case. | ||
+ | |||
+ | If <math>m\equiv 0\pmod{4}</math>: | ||
+ | |||
+ | Note that <math>a\mid16</math> since <math>m</math> will cancel out a factor of <math>4</math> from <math>a</math>, and <math>\frac{a}{m}</math> must contain a factor of <math>4</math>. We see two feasible values for <math>(a,m)</math> such that <math>a+m\leq30</math>. There are <math>2</math> ways for this case. | ||
+ | |||
+ | Adding all the cases up, we obtain <math>28+21+12+2=\boxed{63}</math> | ||
==Remark== | ==Remark== |
Revision as of 18:51, 22 May 2023
Contents
Problem
Find the number of pairs of positive integers with
such that there exists a real number
satisfying
Solution 1
The maximum value of is
, which is achieved at
for some integer
. This is left as an exercise to the reader.
This implies that , and that
and
, for integers
.
Taking their ratio, we have
It remains to find all
that satisfy this equation.
If , then
. This corresponds to choosing two elements from the set
. There are
ways to do so.
If , by multiplying
and
by the same constant
, we have that
. Then either
, or
. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set
. There are
ways here. (This argument seems to have a logical flaw)
Finally, if , note that
must be an integer. This means that
belong to the set
, or
. Taking casework on
, we get the sets
. Some sets have been omitted; this is because they were counted in the other cases already. This sums to
.
In total, there are pairs of
.
This solution was brought to you by ~Leonard_my_dude~
Solution 2
In order for ,
.
This happens when
mod
This means that and
for any integers
and
.
As in Solution 1, take the ratio of the two equations:
Now notice that the numerator and denominator of are both odd, which means that
and
have the same power of two (the powers of 2 cancel out).
Let the common power be : then
, and
where
and
are integers between 1 and 30.
We can now rewrite the equation:
Now it is easy to tell that mod
and
mod
. However, there is another case: that
mod
and
mod
. This is because multiplying both
and
by
will not change the fraction, but each congruence will be changed to
mod
mod
.
From the first set of congruences, we find that and
can be two of
.
From the second set of congruences, we find that and
can be two of
.
Now all we have to do is multiply by to get back to
and
.
Let’s organize the solutions in order of increasing values of
, keeping in mind that
and
are bounded between 1 and 30.
For we get
.
For we get
For we get
If we increase the value of more, there will be less than two integers in our sets, so we are done there.
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
In each of these sets we can choose 2 numbers to be and
and then assign them in increasing order. Thus there are:
possible pairs
that satisfy the conditions.
-KingRavi
Solution 3
We know that the range of sine is between and
, inclusive.
Thus, the only way for the sum to be is for
.
Note that .
Assuming and
are both positive,
and
could be
. There are
ways, so
.
If both are negative, and
could be
. There are
ways, so
.
However, the pair could also be
and so on. The same goes for some other pairs.
In total there are of these extra pairs.
The answer is .
Solution 4
The equation implies that . Therefore, we can write
as
and
as
for integers
and
. Then,
. Cross multiplying, we get
. Let
so the equation becomes
. Let
and
, then the equation becomes
Note that
and
are not relevant so they can vary accordingly, and
. Next, we do casework on
:
If :
Once and
are determined,
is determined, so
.
and
. Therefore, there are
ways for this case such that
.
If :
and
. Therefore, there are
ways such that
.
If :
Note that since
in this case will have a factor of
, which will cancel out a factor of
in
, and we need the left hand side to divide
. Also,
so it is odd and will therefore never contribute a factor of
.
and
. Following the condition
, there are
ways for this case.
If :
Note that since
will cancel out a factor of
from
, and
must contain a factor of
. We see two feasible values for
such that
. There are
ways for this case.
Adding all the cases up, we obtain
Remark
The graphs of and
are shown here in Desmos: https://www.desmos.com/calculator/busxadywja
Move the sliders around for and
to observe the geometric representation generated by each pair
~MRENTHUSIASM (inspired by TheAMCHub)
Video Solution
~mathproblemsolvingskills
Video Solution
https://www.youtube.com/watch?v=LUkQ7R1DqKo
~Mathematical Dexterity
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.