Difference between revisions of "2014 AMC 12B Problems/Problem 23"
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===Sidenote=== | ===Sidenote=== | ||
− | Another way to finish, using the fact that <math>\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)</math>: | + | Another way to finish, using the fact that <math>\dbinom{k+2}{2} = 1 + 2 + \dots + (k+1)</math> (Hockey-Stick Identity): |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} | \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} | ||
Line 32: | Line 32: | ||
&\equiv \boxed{1024} \mod 2017 | &\equiv \boxed{1024} \mod 2017 | ||
\end{align*}</cmath> | \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}} |
Latest revision as of 00:58, 19 January 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 (Hockey-Stick Identity):
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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