Difference between revisions of "2021 AMC 10A Problems/Problem 11"

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<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8</math>
 
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8</math>
  
==Solution==
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==Solution 1==
 
We have <cmath>2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).</cmath> This expression is divisible by <math>3</math> <b><i>unless</i></b> <math>b\equiv2\pmod{3}.</math> The only choice congruent to <math>2</math> modulo <math>3</math> is <math>\boxed{\textbf{(E)} ~8}.</math>
 
We have <cmath>2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).</cmath> This expression is divisible by <math>3</math> <b><i>unless</i></b> <math>b\equiv2\pmod{3}.</math> The only choice congruent to <math>2</math> modulo <math>3</math> is <math>\boxed{\textbf{(E)} ~8}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
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==Solution 2==
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Vertically subtracting 2021 base b - 221 base b, we see that the ones place becomes 0, the b^1 place becomes 0 as well. Now, at the b^2 place, we must perform a carry, but instead of incrementing the place's value by 10 like we normally would in base 10, we do so by b, and make the b^3 place in 2021 base b equal to 1. Thus, we have our final number as 1100 base b.
  
 
==Video Solution (Simple and Quick)==
 
==Video Solution (Simple and Quick)==

Revision as of 14:26, 5 May 2021

Problem

For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

Solution 1

We have \[2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).\] This expression is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$

~MRENTHUSIASM


Solution 2

Vertically subtracting 2021 base b - 221 base b, we see that the ones place becomes 0, the b^1 place becomes 0 as well. Now, at the b^2 place, we must perform a carry, but instead of incrementing the place's value by 10 like we normally would in base 10, we do so by b, and make the b^3 place in 2021 base b equal to 1. Thus, we have our final number as 1100 base b.

Video Solution (Simple and Quick)

https://youtu.be/1TZ1uI9z8fU

~ Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=XBfRVYx64dA&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=10

~North America Math Contest Go Go Go

Video Solution 3

https://youtu.be/zYIuBXDhJJA

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE

~IceMatrix

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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