Difference between revisions of "2020 AIME I Problems/Problem 3"

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A positive integer <math>N</math> has base-eleven representation <math>\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}</math> and base-eight representation <math>\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},</math> where <math>a,b,</math> and <math>c</math> represent (not necessarily distinct) digits. Find the least such <math>N</math> expressed in base ten.
 
A positive integer <math>N</math> has base-eleven representation <math>\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}</math> and base-eight representation <math>\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},</math> where <math>a,b,</math> and <math>c</math> represent (not necessarily distinct) digits. Find the least such <math>N</math> expressed in base ten.
  
==video solution==
+
==Video Solution==
 
https://youtu.be/SuVsBIz8pZ8
 
https://youtu.be/SuVsBIz8pZ8
 +
 
== 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>.
 
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>.

Revision as of 02:09, 22 January 2021

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.

Video Solution

https://youtu.be/SuVsBIz8pZ8

Solution

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

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

~ pi_is_3.14

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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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