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 1==
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==Solution 1 (Factor)==
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>
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We have  
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<cmath>\begin{align*}
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2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\
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&= 2000_b - 200_b \\
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&= 2b^3 - 2b^2 \\
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&= 2b^2(b-1),
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\end{align*}</cmath>
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which 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
  
==Solution 2==
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==Solution 2 (Easy)==
Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, the b^1 place becomes 0 as well. Now, at the <cmath>b^2</cmath> 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 <math>b</math>, and make the <cmath>b^3</cmath> place in <cmath>2021_b</cmath> equal to 1. Thus, we have our final number as <math></math>1100_b.
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Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, and so does the <math>b^1</math> place. Then, we perform a carry (make sure the carry is in base <math>b</math>!). Let <math>b-2 = A</math>. Then, we have our final number as <cmath>1A00_b</cmath>
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Now, when expanding, we see that this number is simply <math>b^3 - (b - 2)^2</math>.
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Now, notice that the final number will only be congruent to <cmath>b^3-(b-2)^2\equiv0\pmod{3}</cmath> if either <math>b\equiv0\pmod{3}</math>, or if <math>b\equiv1\pmod{3}</math> (because note that <math>(b - 2)^2</math> would become <math>\equiv1\pmod{3}</math>, and <math>b^3</math> would become <math>\equiv1\pmod{3}</math> as well, and therefore the final expression would become <math>1-1\equiv0\pmod{3}</math>. Therefore, <math>b</math> must be <math>\equiv2\pmod{3}</math>. Among the answers, only 8 is <math>\equiv2\pmod{3}</math>, and therefore our answer is <math>\boxed{\textbf{(E)} ~8}.</math>
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- icecreamrolls8
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==Solution 3 (Educated Guess)==
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Note that choices <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3,</math> respectively. Since only one of these choices is correct, we pick <math>\boxed{\textbf{(E)} ~8}</math> due to its uniqueness.
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~MRENTHUSIASM
  
 
==Video Solution (Simple and Quick)==
 
==Video Solution (Simple and Quick)==

Latest revision as of 03:02, 12 June 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 (Factor)

We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which 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 (Easy)

Vertically subtracting \[2021_b - 221_b\] we see that the ones place becomes 0, and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$!). Let $b-2 = A$. Then, we have our final number as \[1A00_b\]

Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2$.

Now, notice that the final number will only be congruent to \[b^3-(b-2)^2\equiv0\pmod{3}\] if either $b\equiv0\pmod{3}$, or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3}$, and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}$. Therefore, $b$ must be $\equiv2\pmod{3}$. Among the answers, only 8 is $\equiv2\pmod{3}$, and therefore our answer is $\boxed{\textbf{(E)} ~8}.$

- icecreamrolls8

Solution 3 (Educated Guess)

Note that choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. Since only one of these choices is correct, we pick $\boxed{\textbf{(E)} ~8}$ due to its uniqueness.

~MRENTHUSIASM

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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