Difference between revisions of "2014 AMC 12B Problems/Problem 23"
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<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 == | == 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}} |
Revision as of 15:30, 1 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 :
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.