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

(Fixed the problem statement)
(Cleaned up the solution (and deleted Solution 2, which, as far as I can tell, is simply wrong))
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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>?
  
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== Solution 1 ==
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== Solution ==
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Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the 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 three times in the numerator, or <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.
 
  
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== Solution 2: Parity and Logic ==
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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.
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Expanding <math>{200 \choose 100}</math>,  we find that the remaining product is all the odd number from 199 to 101. We want a prime number to be a factor of one such odd number. If that is the case, the prime number must end in an odd number as well, for the last digit of the multiple must be odd. Therefore the prime number must end in one, and going down the list of two digit multiples we try 71, then 61. We find that 61 does indeed work, as <math>61 \cdot 3 = 183.</math>
 
  
 
== See Also ==
 
== See Also ==

Revision as of 18:37, 15 February 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.

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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