1983 AIME Problems/Problem 8: Difference between revisions

Revision as of 11:13, 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

We know that ${200\choose100}=\frac{200!}{100!100!}$

@@ Line 9: / Line 9: @@
 We know that
-<math></math>{200\choose100}=\frac{200!}{100!100!}
+<cmath>{200\choose100}=\frac{200!}{100!100!}</cmath>
 == See Also ==

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1983 AIME Problems/Problem 8: Difference between revisions

Revision as of 11:13, 14 August 2019

Contents

Problem

Solution

Solution 2: Clarification of Solution 1

See Also