Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (Vertical Subtraction))
 
(11 intermediate revisions by 3 users not shown)
Line 17: Line 17:
  
 
==Solution 2 (Vertical Subtraction)==
 
==Solution 2 (Vertical Subtraction)==
−
Vertically subtracting <math>2021_b - 221_b.</math> we see that the ones place becomes <math>0,</math> 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>
+
Vertically subtracting <math>2021_b - 221_b,</math> we see that the ones place becomes <math>0,</math> 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>
 
Now, when expanding, we see that this number is simply <math>b^3 - (b - 2)^2.</math>
 
Now, when expanding, we see that this number is simply <math>b^3 - (b - 2)^2.</math>
  
−
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 <math>8</math> is <math>\equiv2\pmod{3},</math> and therefore our answer is <math>\boxed{\textbf{(E)} ~8}.</math>
+
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 <math>8</math> is <math>\equiv2\pmod{3},</math> and therefore our answer is <math>\boxed{\textbf{(E)} ~8}.</math>
  
 
~icecreamrolls8
 
~icecreamrolls8
  
−
==Solution 3 (Residues)==
+
==Solution 3 (Answer Choices)==
−
By definition of bases, this number is a polynomial in terms of <math>b</math> (<math>(2b^3+0b^2+2b^1+1b^0)-(2b^2+2b^1+1b^0)</math> to be exact). Thus, for two values of <math>b</math> that share the same residue modulo <math>3</math>, the two resulting numbers will share the same residue modulo <math>3</math>, so either both or neither will be divisible by <math>3</math>.
+
By the definition of bases, we have <cmath>2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).</cmath>
 +
For values <math>b_1</math> and <math>b_2</math> such that <math>b_1\equiv b_2\pmod{3},</math> we get <cmath>\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.</cmath>
 +
Note that answer 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. So, <math>\textbf{(A)}</math> and <math>\textbf{(C)}</math> are either both correct or both incorrect. Since there is only one correct answer, <math>\textbf{(A)}</math> and <math>\textbf{(C)}</math> are both incorrect. Similarly, <math>\textbf{(B)}</math> and <math>\textbf{(D)}</math> are both incorrect. This leaves us with <math>\boxed{\textbf{(E)} ~8},</math> the answer choice with a unique residue modulo <math>3.</math>
  
−
Choices <math>\textbf{(A)} ~3,\textbf{(B)} ~4,\textbf{(C)} ~6,\textbf{(D)} ~7,\textbf{(E)} ~8</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3</math>, respectively. This means <math>\textbf{(A)} ~3</math> and <math>\textbf{(C)} ~6</math> are either both wrong or both right (and the latter obviously cannot be the case), and likewise with <math>\textbf{(B)} ~4</math> and <math>\textbf{(D)} ~7</math>. This leaves <math>\boxed{\textbf{(E)} ~8}</math>, the only choice with a unique residue.
+
~[[User:emerald_block|emerald_block]] ~MRENTHUSIASM
  
−
~MRENTHUSIASM (revised by [[User:emerald_block|emerald_block]])
+
==Video Solution by Omega Learn==
 +
https://youtu.be/CPYCUnL_Yc0
 +
 
 +
~ pi_is_3.14
  
 
==Video Solution (Simple and Quick)==
 
==Video Solution (Simple and Quick)==
Line 41: Line 46:
 
~North America Math Contest Go Go Go
 
~North America Math Contest Go Go Go
  
−
==Video Solution 3==
+
==Video Solution==
 
https://youtu.be/zYIuBXDhJJA
 
https://youtu.be/zYIuBXDhJJA
  

Latest revision as of 10:05, 16 June 2023

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 (Vertical Subtraction)

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 (Answer Choices)

By the definition of bases, we have \[2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).\] For values $b_1$ and $b_2$ such that $b_1\equiv b_2\pmod{3},$ we get \[\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.\] Note that answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. So, $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves us with $\boxed{\textbf{(E)} ~8},$ the answer choice with a unique residue modulo $3.$

~emerald_block ~MRENTHUSIASM

Video Solution by Omega Learn

https://youtu.be/CPYCUnL_Yc0

~ pi_is_3.14

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

https://youtu.be/zYIuBXDhJJA

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE

~IceMatrix

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_11&oldid=194525"