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== | + | ==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 (Easy)== | ==Solution 2 (Easy)== | ||
− | 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 <math> | + | 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> |
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^ | + | 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> |
+ | |||
+ | - icecreamrolls8 | ||
+ | |||
+ | ==Solution 3 (Educated Guess)== | ||
+ | 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. | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== |
Latest revision as of 03:02, 12 June 2021
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 (Easy)
Vertically subtracting we see that the ones place becomes 0, 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 8 is , and therefore our answer is
- icecreamrolls8
Solution 3 (Educated Guess)
Note that choices are congruent to modulo respectively. Since only one of these choices is correct, we pick due to its uniqueness.
~MRENTHUSIASM
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 3
~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.