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

Revision as of 10:50, 14 August 2019

Problem

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

Solution

Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$ . Let the required prime be $p$ ; then $10 \le p < 100$ . If $p > 50$ , then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$ . The largest such prime is $\boxed{061}$ , which is our answer.

Solution 2: Clarification of Solution 1

First notice that ${200\choose100}=\frac{200!}{100!100!}=\frac{200\cdot199\cdot198\cdot...\cdot102\cdot101}{100!}=\frac{2\cdot199\cdot2\cdot...\cdot101}{50!}$

@@ Line 10: / Line 10: @@
 First notice that
 <cmath>{200\choose100}=\frac{200!}{100!100!}=\frac{200\cdot199\cdot198\cdot...\cdot102\cdot101}{100!}=\frac{2\cdot199\cdot2\cdot...\cdot101}{50!}</cmath>
-Since we know that <math>n</math> is an integer, all factors of <math>2</math> in the denominator must be cancelled out by the factors of <math>2</math> in the numerator.
 == See Also ==

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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Difference between revisions of "1983 AIME Problems/Problem 8"

Revision as of 10:50, 14 August 2019

Contents

Problem

Solution

Solution 2: Clarification of Solution 1

See Also