Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"
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<math>\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11</math> | <math>\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11</math> | ||
− | ==Solution== | + | ==Solution 1 (Complete Residue System)== |
For a fixed value of <math>m,</math> there is a total of <math>m(m-1)(m-2)(m-3)</math> possible ordered quadruples <math>(a_1, a_2, a_3, a_4).</math> | For a fixed value of <math>m,</math> there is a total of <math>m(m-1)(m-2)(m-3)</math> possible ordered quadruples <math>(a_1, a_2, a_3, a_4).</math> | ||
Let <math>S=a_1+a_2+a_3+a_4.</math> We claim that exactly <math>\frac1m</math> of these <math>m(m-1)(m-2)(m-3)</math> ordered quadruples satisfy that <math>m</math> divides <math>S:</math> | Let <math>S=a_1+a_2+a_3+a_4.</math> We claim that exactly <math>\frac1m</math> of these <math>m(m-1)(m-2)(m-3)</math> ordered quadruples satisfy that <math>m</math> divides <math>S:</math> | ||
− | Since <math>\gcd(m,4)=1,</math> we conclude that <cmath>\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}</cmath> is the complete system | + | Since <math>\gcd(m,4)=1,</math> we conclude that <cmath>\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}</cmath> is the complete residue system modulo <math>m</math> for all integers <math>k.</math> |
Given any ordered quadruple <math>(a'_1, a'_2, a'_3, a'_4)</math> in modulo <math>m,</math> it follows that exactly one of these <math>m</math> ordered quadruples satisfy that <math>m</math> divides <math>S:</math> | Given any ordered quadruple <math>(a'_1, a'_2, a'_3, a'_4)</math> in modulo <math>m,</math> it follows that exactly one of these <math>m</math> ordered quadruples satisfy that <math>m</math> divides <math>S:</math> |
Revision as of 11:47, 24 November 2021
Contents
Problem
Let be an odd integer, and let
denote the number of quadruples
of distinct integers with
for all
such that
divides
. There is a polynomial
such that
for all odd integers
. What is
Solution 1 (Complete Residue System)
For a fixed value of there is a total of
possible ordered quadruples
Let We claim that exactly
of these
ordered quadruples satisfy that
divides
Since we conclude that
is the complete residue system modulo
for all integers
Given any ordered quadruple in modulo
it follows that exactly one of these
ordered quadruples satisfy that
divides
We conclude that
so
By Vieta's Formulas, we get
~MRENTHUSIASM
Solution 2 (if you're running out of time)
Note that you see numbers with absolute value ,
, and
in the answer choices. What is special about those numbers? Well, you should notice that they are the coefficients of the polynomial
when expanded (if you've already memmed this). Then, you can probably guess the polynomial is some form of
whether negative or positive. Since
is asked, the answer should be reasoned out as
you can gain further confidence in your guess since that is the only answer choice with absolute value
-fidgetboss_4000
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.