Difference between revisions of "2014 AMC 12B Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | The number 2017 is prime. Let <math>S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}</math>. What is the remainder when <math>S</math> is divided by 2017? | + | The number <math>2017</math> is prime. Let <math>S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}</math>. What is the remainder when <math>S</math> is divided by <math>2017?</math> |
<math>\textbf{(A) }32\qquad | <math>\textbf{(A) }32\qquad | ||
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<cmath>\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017</cmath> | <cmath>\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017</cmath> | ||
<cmath>\equiv (-1)^k\dbinom{k+2}{k} \mod 2017</cmath> | <cmath>\equiv (-1)^k\dbinom{k+2}{k} \mod 2017</cmath> | ||
− | |||
Therefore | Therefore | ||
<cmath>\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017</cmath> | <cmath>\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017</cmath> | ||
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Obviously, <math>62</math> falls in the second category, so our desired value is | Obviously, <math>62</math> falls in the second category, so our desired value is | ||
<cmath>\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}</cmath> | <cmath>\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{\textbf{(C)}\ 1024}</cmath> | ||
+ | |||
+ | ===Sidenote=== | ||
+ | Another way to finish, using the fact that <math>\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)</math>: | ||
+ | <cmath>\begin{align*} | ||
+ | \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} | ||
+ | &\equiv \sum \limits_{k=1}^{63}(-1)^{k-1} (1 + 2 + \dots + k) \\ | ||
+ | &\equiv 1 - (1+2) + (1+2+3) - (1+2+3+4) + \dots + (1 + \dots + 63) \\ | ||
+ | &\equiv 1 + 3 + 5 + \dots + 63 \\ | ||
+ | &\equiv \boxed{1024} \mod 2017 | ||
+ | \end{align*}</cmath> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2014|ab=B|num-b=22|num-a=24}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:18, 11 October 2022
Contents
Problem
The number is prime. Let
. What is the remainder when
is divided by
Solution
Note that . We have for
Therefore
This is simply an alternating series of triangular numbers that goes like this:
After finding the first few sums of the series, it becomes apparent that
and
Obviously,
falls in the second category, so our desired value is
Sidenote
Another way to finish, using the fact that :
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.