Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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==Solution== | ==Solution== | ||
Because the sum of <math>n</math> positive integers is <math>(n)(n+1)/2</math>, and we want this to not be a divisor of the <math>n!</math>, <math>n+1</math> must be prime. The greatest three-digit integer that is prime is <math>997</math>. Subtract <math>1</math> to get <math>996 => B</math> | Because the sum of <math>n</math> positive integers is <math>(n)(n+1)/2</math>, and we want this to not be a divisor of the <math>n!</math>, <math>n+1</math> must be prime. The greatest three-digit integer that is prime is <math>997</math>. Subtract <math>1</math> to get <math>996 => B</math> | ||
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-Lcz | -Lcz | ||
Revision as of 17:26, 9 February 2019
Problem
What is the greatest three-digit positive integer 
for which the sum of the first positive integers is 
a divisor of the product of the first positive integers?
Solution
Because the sum of positive integers is 
, and we want this to not be a divisor of the 
, 
must be prime. The greatest three-digit integer that is prime is 
. Subtract 
to get 
-Lcz
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
