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== | + | ==Solution 1 (Factor)== |
− | We have <cmath>2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1) | + | We have |
+ | <cmath>\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*}</cmath> | ||
+ | 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 (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> | ||
+ | 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> | ||
+ | |||
+ | ~icecreamrolls8 | ||
+ | |||
+ | ==Solution 3 (Answer Choices)== | ||
+ | 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> | ||
+ | |||
+ | ~[[User:emerald_block|emerald_block]] ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution by Omega Learn== | ||
+ | https://youtu.be/CPYCUnL_Yc0 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== | ||
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~North America Math Contest Go Go Go | ~North America Math Contest Go Go Go | ||
− | ==Video Solution | + | ==Video Solution== |
https://youtu.be/zYIuBXDhJJA | https://youtu.be/zYIuBXDhJJA | ||
Latest revision as of 10:05, 16 June 2023
Contents
Problem
For which of the following integers is the base- number not divisible by ?
Solution 1 (Factor)
We have which is divisible by unless The only choice congruent to modulo is
~MRENTHUSIASM
Solution 2 (Vertical Subtraction)
Vertically subtracting we see that the ones place becomes and so does the place. Then, we perform a carry (make sure the carry is in base ). Let Then, we have our final number as Now, when expanding, we see that this number is simply
Now, notice that the final number will only be congruent to If either or if (because note that would become and would become as well, and therefore the final expression would become Therefore, must be Among the answers, only is and therefore our answer is
~icecreamrolls8
Solution 3 (Answer Choices)
By the definition of bases, we have For values and such that we get Note that answer choices are congruent to modulo respectively. So, and are either both correct or both incorrect. Since there is only one correct answer, and are both incorrect. Similarly, and are both incorrect. This leaves us with the answer choice with a unique residue modulo
~emerald_block ~MRENTHUSIASM
Video Solution by Omega Learn
~ pi_is_3.14
Video Solution (Simple and Quick)
~ 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
~savannahsolver
Video Solution by TheBeautyofMath
~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.