Difference between revisions of "1983 AIME Problems/Problem 8"
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First notice that | First notice that | ||
<cmath>{200\choose100}=\frac{200!}{100!100!}=\frac{200\cdot199\cdot198\cdot...\cdot102\cdot101}{100!}=\frac{2\cdot199\cdot2\cdot...\cdot101}{50!}</cmath> | <cmath>{200\choose100}=\frac{200!}{100!100!}=\frac{200\cdot199\cdot198\cdot...\cdot102\cdot101}{100!}=\frac{2\cdot199\cdot2\cdot...\cdot101}{50!}</cmath> | ||
| + | Since we know that <math>n</math> is an integer, all factors of <math>2</math> in the denominator must be cancelled out by the factors of <math>2</math> in the numerator. | ||
== See Also == | == See Also == | ||
Revision as of 10:49, 14 August 2019
Problem
What is the largest 
-digit prime factor of the integer 
?
Solution
Expanding the binomial coefficient, we get 
. Let the required prime be 
; then 
. If 
, then the factor of appears twice in the denominator. Thus, we need to appear as a factor at least three times in the numerator, so 
. The largest such prime is 
, which is our answer.
Solution 2: Clarification of Solution 1
First notice that
![\[{200\choose100}=\frac{200!}{100!100!}=\frac{200\cdot199\cdot198\cdot...\cdot102\cdot101}{100!}=\frac{2\cdot199\cdot2\cdot...\cdot101}{50!}\]](http://latex.artofproblemsolving.com/3/8/e/38ea28fb3e36c1eb7c58620b591a6313dabec6f4.png)
Since we know that 
is an integer, all factors of in the denominator must be cancelled out by the factors of in the numerator.
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 | ||
