Difference between revisions of "2022 AIME I Problems/Problem 4"
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<li><math>s+3k\equiv0\pmod{3}</math></li><p> | <li><math>s+3k\equiv0\pmod{3}</math></li><p> | ||
− | There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv0\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> | + | There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv0\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p> |
+ | There are <math>8\cdot33=264</math> ordered pairs in this case. | ||
<li><math>s+3k\equiv1\pmod{3}</math></li><p> | <li><math>s+3k\equiv1\pmod{3}</math></li><p> | ||
+ | There are <math>9</math> values for <math>s+3k,</math> so there are <math>9</math> values for <math>r.</math> It follows that <math>s\equiv1\pmod{3},</math> so there are <math>34</math> values for <math>s.</math> <p> | ||
+ | There are <math>9\cdot34=306</math> ordered pairs in this case. | ||
<li><math>s+3k\equiv2\pmod{3}</math></li><p> | <li><math>s+3k\equiv2\pmod{3}</math></li><p> | ||
+ | There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv2\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p> | ||
+ | There are <math>8\cdot33=264</math> ordered pairs in this case. | ||
</ol> | </ol> | ||
+ | Together, the answer is <math>264+306+264=\boxed{834}.</math> | ||
+ | |||
~MRENTHUSIASM ~bluesoul | ~MRENTHUSIASM ~bluesoul | ||
Revision as of 17:07, 17 February 2022
Problem
Let and where Find the number of ordered pairs of positive integers not exceeding that satisfy the equation
Solution
We rewrite and in polar form: The equation becomes for some integer
Since and we conclude that Note that the values for and the values for have one-to-one correspondence.
We apply casework to the values for
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
Together, the answer is
~MRENTHUSIASM ~bluesoul
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.