Difference between revisions of "2022 AIME I Problems/Problem 2"
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Of course, <math>99</math> can't be made of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>28</math>, which can't be made of <math>8</math>'s either. So <math>99</math> doesn't work. | Of course, <math>99</math> can't be made of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>28</math>, which can't be made of <math>8</math>'s either. So <math>99</math> doesn't work. | ||
<math>198</math> can't be made up of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>127</math>, which can't be made of <math>8</math>'s. If we use two <math>71</math>'s, we get a remainder of <math>56</math>, which can be made of <math>8</math>'s. | <math>198</math> can't be made up of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>127</math>, which can't be made of <math>8</math>'s. If we use two <math>71</math>'s, we get a remainder of <math>56</math>, which can be made of <math>8</math>'s. | ||
− | Therefore we get <math>99\cdot2=71\cdot2+8\cdot7</math> so <math>a=2,b=2,</math> and <math>c=7</math>. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, <math>\underline{a}\underline{b}\underline{c}=227.</math> | + | Therefore we get <math>99\cdot2=71\cdot2+8\cdot7</math> so <math>a=2,b=2,</math> and <math>c=7</math>. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, <math>\underline{a}\underline{b}\underline{c}=\boxed {227}.</math> |
~Technodoggo | ~Technodoggo | ||
Revision as of 22:31, 18 February 2022
Contents
Problem
Find the three-digit positive integer whose representation in base nine is where and are (not necessarily distinct) digits.
Solution 1
We are given that which rearranges to Taking both sides modulo we have The only solution occurs at from which
Therefore, the requested three-digit positive integer is
~MRENTHUSIASM
Solution 2
As shown in Solution 1, we get . Note that and are big numbers comparatively to , so we hypothesize that and are equal and fills the gap between them. The difference between and is , which is a multiple of . So, if we multiply this by , it will be a multiple of and thus the gap can be filled. Therefore, a viable solution is , and the answer is .
~KingRavi
Solution 3
As shown in Solution 1, we get We list a few multiples of out: \begin{align*} 99\\198\\297\\396\\ \end{align*} Of course, can't be made of just 's. If we use one , we get a remainder of , which can't be made of 's either. So doesn't work. can't be made up of just 's. If we use one , we get a remainder of , which can't be made of 's. If we use two 's, we get a remainder of , which can be made of 's. Therefore we get so and . Plugging this back into the original problem shows that this answer is indeed correct. Therefore, ~Technodoggo
Solution 4
As shown in Solution 1, the equation simplifies to .
We can see that is larger than , and we have an . We can clearly see that is a multiple of , and any larger than would result in being larger than . Therefore, our only solution is . Our answer is .
~Arcticturn
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=z5Y4bT5rL-s
Video Solution
https://www.youtube.com/watch?v=CwSkAHR3AcM
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.