Difference between revisions of "2022 AIME I Problems/Problem 2"
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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}=227.</math> | ||
~Technodoggo | ~Technodoggo | ||
| + | |||
| + | == Solution 4 == | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
Revision as of 23:28, 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 ![\[100a + 10b + c = 81b + 9c + a,\]](http://latex.artofproblemsolving.com/9/8/4/98419c4a18d61bbd8b2c652976c8427a082c8167.png)
which rearranges to ![\[99a = 71b + 8c.\]](http://latex.artofproblemsolving.com/0/2/a/02a0693b866a5a516e33b7d14ea36547a9b2e122.png)
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
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. 
