Difference between revisions of "2020 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
− | + | From the given information, <math>121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c</math>. Since <math>a</math>, <math>b</math>, and <math>c</math> have to be positive, <math>a \geq 5</math>. Since we need to minimize the value of <math>n</math>, we want to minimize <math>a</math>, so we have <math>a = 5</math>. Then we know <math>88=53b+7c</math>, and we can see the only solution is <math>b=1</math>, <math>c=5</math>. Finally, <math>515_{11} = 621_{10}</math>, so our answer is <math>\boxed{621}</math>. | |
~ JHawk0224 | ~ JHawk0224 |
Revision as of 15:42, 13 March 2020
Problem
A positive integer has base-eleven representation and base-eight representation where and represent (not necessarily distinct) digits. Find the least such expressed in base ten.
Solution
From the given information, . Since , , and have to be positive, . Since we need to minimize the value of , we want to minimize , so we have . Then we know , and we can see the only solution is , . Finally, , so our answer is .
~ JHawk0224
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.