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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>.
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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=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 want <math>a = 5</math>. Then we have <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 15:41, 13 March 2020

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

Since $a$, $b$, and $c$ are digits in base eight, they are all 7 or less. Now, from the given equation, $121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c$. 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 want $a = 5$. Then we have $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.

~ JHawk0224

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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