Difference between revisions of "2020 AIME I Problems/Problem 3"
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==Solution 2 (Official MAA)== | ==Solution 2 (Official MAA)== | ||
The conditions of the problem imply that <math>121a + 11b + c = 512 + 64b + 8 c + a</math>, so <math>120 a = 512+ 53b+7c</math>. The maximum digit in base eight is <math>7,</math> and because <math>120a \ge 512</math>, it must be that <math>a</math> is <math>5, 6,</math> or <math>7.</math> When <math>a = 5</math>, it follows that <math>600=512 + 53b+7c</math>, which implies that <math>88 = 53b+7c</math>. Then <math>b</math> must be <math>0</math> or <math>1.</math> If <math>b = 0</math>, then <math>c</math> is not an integer, and if <math>b = 1</math>, then <math>7c = 35</math>, so <math>c = 5</math>. Thus <math>N = 515_{11}</math>, and <math>N=5\cdot 121 + 1\cdot 11 + 5 = 621</math>. The number <math>637_{11} =1376_{8} = 766</math> also satisfies the conditions of the problem, but <math>621</math> is the least such number. | The conditions of the problem imply that <math>121a + 11b + c = 512 + 64b + 8 c + a</math>, so <math>120 a = 512+ 53b+7c</math>. The maximum digit in base eight is <math>7,</math> and because <math>120a \ge 512</math>, it must be that <math>a</math> is <math>5, 6,</math> or <math>7.</math> When <math>a = 5</math>, it follows that <math>600=512 + 53b+7c</math>, which implies that <math>88 = 53b+7c</math>. Then <math>b</math> must be <math>0</math> or <math>1.</math> If <math>b = 0</math>, then <math>c</math> is not an integer, and if <math>b = 1</math>, then <math>7c = 35</math>, so <math>c = 5</math>. Thus <math>N = 515_{11}</math>, and <math>N=5\cdot 121 + 1\cdot 11 + 5 = 621</math>. The number <math>637_{11} =1376_{8} = 766</math> also satisfies the conditions of the problem, but <math>621</math> is the least such number. | ||
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Video Solution: | Video Solution: | ||
https://youtu.be/hZSBUXCX5hI | https://youtu.be/hZSBUXCX5hI | ||
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+ | Minor edits by TryhardMathlete | ||
==See Also== | ==See Also== |
Revision as of 09:15, 8 January 2021
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.
video solution
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
Solution 2 (Official MAA)
The conditions of the problem imply that , so . The maximum digit in base eight is and because , it must be that is or When , it follows that , which implies that . Then must be or If , then is not an integer, and if , then , so . Thus , and . The number also satisfies the conditions of the problem, but is the least such number.
Video Solution:
Minor edits by TryhardMathlete
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.