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Difference between revisions of "2020 AIME I Problems/Problem 3"
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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:
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==Video Solution==
  
 
https://youtu.be/hZSBUXCX5hI
 
https://youtu.be/hZSBUXCX5hI
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Minor edits by TryhardMathlete
 
Minor edits by TryhardMathlete
  
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== Video Solution ==
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== Video Solution by OmegaLearn==
 
https://youtu.be/mgEZOXgIZXs?t=1204
 
https://youtu.be/mgEZOXgIZXs?t=1204
  

Latest revision as of 03:48, 21 January 2023

Problem

A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.

Solution 1

From the given information, $121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. Since we need to minimize the value of $n$, we want to minimize $a$, so we have $a = 5$. Then we know $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.

~ JHawk0224

Solution 2 (Official MAA)

The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $600=512 + 53b+7c$, which implies that $88 = 53b+7c$. Then $b$ must be $0$ or $1.$ If $b = 0$, then $c$ is not an integer, and if $b = 1$, then $7c = 35$, so $c = 5$. Thus $N = 515_{11}$, and $N=5\cdot 121 + 1\cdot 11 + 5 = 621$. The number $637_{11} =1376_{8} = 766$ also satisfies the conditions of the problem, but $621$ is the least such number.

Video Solution

https://youtu.be/hZSBUXCX5hI

Minor edits by TryhardMathlete

Video Solution by OmegaLearn

https://youtu.be/mgEZOXgIZXs?t=1204

~ pi_is_3.14

Video Solution

https://youtu.be/SuVsBIz8pZ8

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

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