Difference between revisions of "2021 AIME I Problems/Problem 7"
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From the first set of congruences, we find that <math>a</math> and <math>b</math> can be two of | From the first set of congruences, we find that <math>a</math> and <math>b</math> can be two of | ||
− | <math>\{1, 5, 9, \ | + | <math>\{1, 5, 9, \ldots, 29\}</math>. |
From the second set of congruences, we find that <math>a</math> and <math>b</math> can be two of | From the second set of congruences, we find that <math>a</math> and <math>b</math> can be two of | ||
− | <math>\{3, 7, 11, \ | + | <math>\{3, 7, 11, \ldots, 27\}</math>. |
Now all we have to do is multiply by <math>2^p</math> to get back to <math>m</math> and <math>n</math>. | Now all we have to do is multiply by <math>2^p</math> to get back to <math>m</math> and <math>n</math>. | ||
Let’s organize the solutions in order of increasing values of <math>p</math>, keeping in mind that <math>m</math> and <math>n</math> are bounded between 1 and 30. | Let’s organize the solutions in order of increasing values of <math>p</math>, keeping in mind that <math>m</math> and <math>n</math> are bounded between 1 and 30. | ||
− | For <math>p = 0</math> we get <math>\{1, 5, 9, \ | + | For <math>p = 0</math> we get <math>\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}</math>. |
For <math>p = 1</math> we get <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}</math> | For <math>p = 1</math> we get <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}</math> |
Revision as of 16:37, 8 September 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 pairs 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 for 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 both 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
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
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.