Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2== | + | ==Solution 2 (Easy)== |
− | Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, the <math>b^1</math> place | + | 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>base 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 is simply <math>b^3 | + | 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^2(b+1)\equiv0\pmod{3}</cmath> if either | + | Now, notice that the final number will only be congruent to <cmath>b^2(b+1)\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 b^3 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>. |
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== |
Revision as of 14:48, 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 (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 !). 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 b^3 would become as well, and therefore the final expression would become . Therefore, must 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.