Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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− | 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 base <math>b</math>!). Let <math>b-2 = A</math> | + | 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 base <math>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 number is simply <math>b^3 - (b - 2)^2</math> | + | 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^3-(b-2)^2\equiv0\pmod{3}</cmath> if either <math>b\equiv0\pmod{3}</math> | + | Now, notice that the final number will only be congruent to <cmath>b^3-(b-2)^2\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 <math>b^3</math> 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> Among the answers, only 8 is <math>\equiv2\pmod{3},</math> and therefore our answer is <math>\boxed{\textbf{(E)} ~8}.</math> |
~icecreamrolls8 | ~icecreamrolls8 |
Revision as of 15:06, 22 November 2021
Contents
Problem
For which of the following integers is the base- number not divisible by ?
Solution 1 (Factor)
We have which is divisible by unless The only choice congruent to modulo is
~MRENTHUSIASM
Solution 2 (Vertical Subtraction)
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 base !). 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 would become as well, and therefore the final expression would become Therefore, must be Among the answers, only 8 is and therefore our answer is
~icecreamrolls8
Solution 3 (Residues)
By definition of bases, this number is a polynomial in terms of ( to be exact). Thus, for two values of that share the same residue modulo , the two resulting numbers will share the same residue modulo , so either both or neither will be divisible by .
Choices are congruent to modulo , respectively. This means and are either both wrong or both right (and the latter obviously cannot be the case), and likewise with and . This leaves , the only choice with a unique residue.
~MRENTHUSIASM (revised by emerald_block)
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.