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

(Solution 2: Clarification of Solution 1)
m
 
(8 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
__TOC__
 +
 
== Problem ==
 
== Problem ==
 
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>?
 
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>?
  
== Solution ==
+
==Solution==
 +
=== Solution 1 ===
  
 
Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the required prime be <math>p</math>; then <math>10 \le p < 100</math>. If <math>p > 50</math>, then the factor of <math>p</math> appears twice in the denominator. Thus, we need <math>p</math> to appear as a factor at least three times in the numerator, so <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.
 
Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the required prime be <math>p</math>; then <math>10 \le p < 100</math>. If <math>p > 50</math>, then the factor of <math>p</math> appears twice in the denominator. Thus, we need <math>p</math> to appear as a factor at least three times in the numerator, so <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.
  
 
+
=== Solution 2: Clarification of Solution 1 ===
== Solution 2: Clarification of Solution 1 ==
 
  
 
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 at least <math>1</math> 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>.
+
Since <math>p<100</math>, there is at least <math>1</math> 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>. (Note that here we assume the minimum because as <math>p</math> goes larger in value, the number of factors of <math>p</math> in a number decreases,)
  
 
So basically, <math>p</math> is the largest prime number such that
 
So basically, <math>p</math> is the largest prime number such that
<cmath>\lfloor\frac{200}{3}\rfloor>3</cmath>
+
<cmath>\left \lfloor\frac{200}{p}\right \rfloor>3</cmath>
Since <math>p<\frac{200}{3}\approx66.66...</math>, the largest prime value for <math>p</math> is <math>p=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>
  
 
~ Nafer
 
~ Nafer

Latest revision as of 20:23, 14 January 2023

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

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