Difference between revisions of "1983 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | Expanding the [[binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. | + | Expanding the [[combination|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. | 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. | ||
Revision as of 00:09, 24 July 2006
Problem
What is the largest 2-digit prime factor of the integer 
?
Solution
Expanding the binomial coefficient, we get 
.
Therefore, our two digit prime 
must satisfy 
. The largest such prime is 
, which is our answer.
