Difference between revisions of "1983 AIME Problems/Problem 8"

 
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== Problem ==
 
== Problem ==
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What is the largest 2-digit prime factor of the integer <math>\binom{200}{100}</math>?
  
 
== Solution ==
 
== Solution ==
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Expanding the [[binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>.
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Therefore, our two digit [[prime]] <math>p</math> must satisfy <math>3p<200</math>. The largest such prime is <math>61</math>, which is our answer.
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----
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* [[1983 AIME Problems/Problem 7|Previous Problem]]
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* [[1983 AIME Problems/Problem 9|Next Problem]]
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* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
* [[1983 AIME Problems]]
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 00:06, 24 July 2006

Problem

What is the largest 2-digit prime factor of the integer $\binom{200}{100}$?

Solution

Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$.

Therefore, our two digit prime $p$ must satisfy $3p<200$. The largest such prime is $61$, which is our answer.


See also