Difference between revisions of "2014 AMC 12B Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | + | 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> | |
− | 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? | ||
<math>\textbf{(A) }32\qquad | <math>\textbf{(A) }32\qquad | ||
Line 13: | Line 12: | ||
<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= | + | <cmath>\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017</cmath> |
This is simply an alternating series of triangular numbers that goes like this: <math>1-3+6-10+15-21....</math> | This is simply an alternating series of triangular numbers that goes like this: <math>1-3+6-10+15-21....</math> | ||
After finding the first few sums of the series, it becomes apparent that | After finding the first few sums of the series, it becomes apparent that | ||
Line 24: | Line 22: | ||
<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 + 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 == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=22|num-a=24}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:31, 31 October 2021
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 |
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