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

Revision as of 11:55, 21 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 ${200\choose100}=\frac{200!}{100!100!}$ Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$ . (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,)

So basically, $p$ is the largest prime number such that $\lfloor\frac{200}{p}\rfloor>3$ Since $p<\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\boxed{61}$

~ Nafer

@@ Line 15: / Line 15: @@
 <cmath>\lfloor\frac{200}{p}\rfloor>3</cmath>
 Since <math>p<\frac{200}{3}=66.66...</math>, the largest prime value for <math>p</math> is <math>p=\boxed{61}</math>
 ~ Nafer

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

Revision as of 11:55, 21 August 2019

Contents

Problem

Solution

Solution 2: Clarification of Solution 1

See Also