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

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

Revision as of 10:55, 4 September 2020

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

~ Nafer

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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