1983 AIME Problems/Problem 8: Difference between revisions

Revision as of 23:06, 23 July 2006

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

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.

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