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

(Solution 2: Clarification of Solution 1)
(Solution 2: Clarification of Solution 1)
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We know that  
 
We know that  
 
<cmath>{200\choose100}=\frac{200!}{100!100!}</cmath>
 
<cmath>{200\choose100}=\frac{200!}{100!100!}</cmath>
 +
Since <math>p<100</math>, there is a factor of <math>p</math> in each of the <math>100!</math> in the denominator. Thus there must be at least <math>3</math> factors of <math>p</math> in the numerator <math>200!</math> for <math>p</math> to be a factor of <math>n=\frac{200!}{100!100!}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 11:17, 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 \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$, there is a 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 $n=\frac{200!}{100!100!}$.

See Also

1983 AIME (ProblemsAnswer KeyResources)
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