Difference between revisions of "2021 AIME I Problems/Problem 7"
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The <math>sin</math> of <math>(90+360k)</math> is equal to 1. | The <math>sin</math> of <math>(90+360k)</math> is equal to 1. | ||
− | Assuming <math>mx</math> and <math>nm</math> are both positive, m and n could be <math>1,5,9,13,17,21,25,29</math>. There are <math>8</math> ways, so <math> | + | Assuming <math>mx</math> and <math>nm</math> are both positive, m and n could be <math>1,5,9,13,17,21,25,29</math>. There are <math>8</math> ways, so <math>\dbinom{8}{2}</math>. |
− | If bother are negative, m and n could be <math>3,7,11,15,19,23,27</math>. There are <math>7</math> ways, so <math> | + | If bother are negative, m and n could be <math>3,7,11,15,19,23,27</math>. There are <math>7</math> ways, so <math>\dbinom{7}{2}</math>. |
+ | |||
+ | However, the pair <math>(1,5)</math> could also be <math>(2, 10)</math> and so on. The same goes for some other pairs. | ||
+ | |||
+ | In total there are <math>14</math> of these extra pairs. | ||
+ | |||
+ | The answer is <math>28+21+14 = \boxed{063}</math> | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=6|num-a=8}} | {{AIME box|year=2021|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:08, 13 March 2021
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.
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 pairings of that satisfy the conditions.
-KingRavi
Solution 3
We know that the range of is between and .
Thus, the only way for the sum to be is the of and to both be .
The of is equal to 1.
Assuming and are both positive, m and n could be . There are ways, so .
If bother are negative, m and n 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
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 |
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