Difference between revisions of "2020 AIME I Problems/Problem 3"

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== Solution ==
 
== Solution ==
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Since <math>a</math>, <math>b</math>, and <math>c</math> are digits in base eight, they are all 7 or less. Now, from the given equation, <math>121a+11b+c=512+64b+8c+a \implies 120a-53b-7c=512</math>. Since <math>a</math>, <math>b</math>, and <math>c</math> have to be positive, <math>a \geq 5</math>. If <math>a = 5</math>, then by casework we see that <math>b = 1</math> and <math>c = 5</math> is the only solution. Since the question asks for the lowest working value, we can stop here. Finally, <math>515_{11} = 621_{10}</math>, so our answer is <math>\boxed{621}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:03, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

Solution

Since $a$, $b$, and $c$ are digits in base eight, they are all 7 or less. Now, from the given equation, $121a+11b+c=512+64b+8c+a \implies 120a-53b-7c=512$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. If $a = 5$, then by casework we see that $b = 1$ and $c = 5$ is the only solution. Since the question asks for the lowest working value, we can stop here. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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

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