Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
| − | 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 <math>b^2</math> 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 <math>b^3</math> place in <math>2021_b</math> equal to 1. Thus, we have our final number as <cmath>1100_b</cmath> | + | Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, the <math>b^1</math> place becomes 0 as well. Now, at the <math>b^2</math> 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 <math>b^3</math> place in <math>2021_b</math> equal to 1. Thus, we have our final number as <cmath>1100_b</cmath> |
Now, when expanding, we see that this is simply <math>b^3 + b^2</math>, which factors into <cmath>b^2(b+1)</cmath> | Now, when expanding, we see that this is simply <math>b^3 + b^2</math>, which factors into <cmath>b^2(b+1)</cmath> | ||
| − | Now, notice that the final number will only be | + | Now, notice that the final number will only be congruent to <cmath>b^2(b+1)\equiv0\pmod{3}</cmath> if either $b\equiv2\pmod{3}, or |
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== | ||
Revision as of 13:37, 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 
place becomes 0 as well. Now, at the 
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 
place in 
equal to 1. Thus, we have our final number as ![\[1100_b\]](http://latex.artofproblemsolving.com/5/f/c/5fc3474f12f05da456dda6c13d62064742edd578.png)
Now, when expanding, we see that this is simply 
, which factors into ![\[b^2(b+1)\]](http://latex.artofproblemsolving.com/6/a/b/6aba90c0578ce07482f887b2b8f13c8a58dd5a06.png)
Now, notice that the final number will only be congruent to ![\[b^2(b+1)\equiv0\pmod{3}\]](http://latex.artofproblemsolving.com/3/a/a/3aa51f010460922f50482e4a6f3ddbc93f441f6a.png)
if either $b\equiv2\pmod{3}, or
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. 
