Difference between revisions of "2021 AMC 10A Problems/Problem 11"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2 ( | + | ==Solution 2 (Vertical Subtraction)== |
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> | 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> | ||
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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> | 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 | |
==Solution 3 (Residues)== | ==Solution 3 (Residues)== |
Revision as of 15:56, 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.