Difference between revisions of "1983 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | What is the largest 2-digit prime factor of the integer <math>{200\choose 100}</math>? | + | What is the largest 2-digit [[prime]] factor of the integer <math>{200\choose 100}</math>? |
== Solution == | == Solution == | ||
| − | Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. | + | 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. |
| − | + | == See also == | |
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{{AIME box|year=1983|num-b=7|num-a=9}} | {{AIME box|year=1983|num-b=7|num-a=9}} | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 11:04, 25 April 2008
Problem
What is the largest 2-digit prime factor of the integer 
?
Solution
Expanding the binomial coefficient, we get 
. Let the prime be 
; then 
. If 
, then the factor of appears twice in the denominator. Thus, we need to appear as a factor three times in the numerator, or 
. The largest such prime is 
, 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 | ||
