Difference between revisions of "1983 AIME Problems/Problem 8"
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== Solution 2 == | == Solution 2 == | ||
| − | Expanding | + | Expanding <math>{200 \choose 100}=\frac{200!}{100!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 15:03, 1 July 2018
Contents
Problem
What is the largest 2-digit prime factor of the integer 
?
Solution 1
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.
Solution 2
Expanding , 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 
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 | ||
