Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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| − | 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>unless</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 | ||
Revision as of 11:21, 12 March 2021
Contents
Problem
For which of the following integers 
is the base- number 
not divisible by 
?
Solution
We have ![\[2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).\]](http://latex.artofproblemsolving.com/d/a/2/da2e0428c3dd55e006ffbfdc6cc33faa719312fd.png)
This expression is divisible by unless 
The only choice congruent to 
modulo is 
~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. 
