Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 10"

(Solution)
(the current start is incorrect, but I can't find any non-brute force way to calculate this ....)
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==Problem==
 
==Problem==
 
Compute the [[remainder]] when <center><p><math>{2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}</math></p></center> is divided by 1000.
 
Compute the [[remainder]] when <center><p><math>{2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}</math></p></center> is divided by 1000.
 +
 
==Solution==
 
==Solution==
Since
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{{solution}}
 
 
 
 
<math>\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n</math>
 
 
 
 
 
that sum equals <math>2^{2007}</math>. So we just need to find n such that
 
  
<math>n\equiv 2^{2007} \pmod {1000}</math>
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== See also ==
 
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{{Mock AIME box|year=2006-2007|n=4|num-b=9|num-a=11|source=125025}}
 
 
 
 
{{solution}}
 
----
 
  
*[[Mock AIME 4 2006-2007 Problems/Problem 11| Next Problem]]
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*[[Binomial theorem]]
*[[Mock AIME 4 2006-2007 Problems/Problem 9| Previous Problem]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 
* [[Binomial theorem]]
 
 
*[[Modular arithmetic]]
 
*[[Modular arithmetic]]

Revision as of 16:38, 8 October 2007

Problem

Compute the remainder when

${2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}$

is divided by 1000.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15