1983 AIME Problems/Problem 8: Difference between revisions

Revision as of 22:25, 11 May 2024

Problem

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

Solution

Solution 1

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 $\left \lfloor\frac{200}{p}\right \rfloor>3$ Since $p<\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\boxed{61}$

~ Nafer

Note

Similar to 2023 MATHCOUNTS State Sprint #25

@@ Line 20: / Line 20: @@
 ~ Nafer
+==Note==
+Similar to 2023 MATHCOUNTS State Sprint #25
 == See Also ==

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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