Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
| − | Vertically subtracting <cmath>2021_b - </cmath> | + | 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. |
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== | ||
Revision as of 14:30, 5 May 2021
Contents
Problem
For which of the following integers 
is the base- number 
not divisible by 
?
Solution 1
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
Solution 2
Vertically subtracting ![\[2021_b - 221_b\]](http://latex.artofproblemsolving.com/d/9/d/d9df25ee4f1eade49041469d56d18478b1afdc13.png)
we see that the ones place becomes 0, the b^1 place becomes 0 as well. Now, at the ![\[b^2\]](http://latex.artofproblemsolving.com/8/0/a/80afe638f2ffef249ea1b36687b30d150a734f03.png)
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 , and make the ![\[b^3\]](http://latex.artofproblemsolving.com/8/c/1/8c1c3d192716337a23a2f856e82683047f35548f.png)
place in ![\[2021_b\]](http://latex.artofproblemsolving.com/a/e/8/ae83e9d460032bb5291418ffd22effa7d7b965c1.png)
equal to 1. Thus, we have our final number as $$ (Error compiling LaTeX. Unknown error_msg)1100_b.
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. 
