Difference between revisions of "2022 AIME I Problems/Problem 2"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) (→Solution) |
||
| Line 9: | Line 9: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
28a &\equiv 8c \pmod{71} \\ | 28a &\equiv 8c \pmod{71} \\ | ||
| − | 7a &\equiv 2c \ | + | 7a &\equiv 2c \pmod{71}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
The only solution occurs at <math>(a,c)=(2,7),</math> from which <math>b=2.</math> | The only solution occurs at <math>(a,c)=(2,7),</math> from which <math>b=2.</math> | ||
Revision as of 16:18, 17 February 2022
Problem
Find the three-digit positive integer 
whose representation in base nine is 
where 
and 
are (not necessarily distinct) digits.
Solution
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
See Also
| 2022 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
