Difference between revisions of "2020 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
| + | 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 
, 
, and 
are digits in base eight, they are all 7 or less. Now, from the given equation, 
. Since , , and have to be positive, 
. If 
, then by casework we see that 
and 
is the only solution. Since the question asks for the lowest working value, we can stop here. Finally, 
, so our answer is 
.
See Also
| 2020 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. 
