Difference between revisions of "2021 AMC 10A Problems/Problem 11"
(→Solution 2) |
(→Solution 2) |
||
Line 10: | Line 10: | ||
==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 < | + | 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> |
+ | |||
+ | 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 | ||
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== |
Revision as of 14:32, 5 May 2021
Contents
Problem
For which of the following integers is the base- number not divisible by ?
Solution 1
We have This expression is divisible by unless The only choice congruent to modulo is
~MRENTHUSIASM
Solution 2
Vertically subtracting we see that the ones place becomes 0, the b^1 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
Now, when expanding, we see that this is simply , which factors into
Now, notice that the final number will only be
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.