Difference between revisions of "2019 AMC 10A Problems/Problem 9"
(→Solution) |
(→Solution) |
||
Line 9: | Line 9: | ||
-Lcz | -Lcz | ||
+ | |||
+ | Following from the fact that <math>n+1</math> must be prime, we can use to answer choices as possible solutions for n. <math>A</math>, <math>C</math>, and <math>E</math> don't work because <math>n+1</math> is even, and <math>D</math> does not work since <math>999</math> is divisible by <math>9</math>. Thus, the only correct answer is <math>996 => B</math>. | ||
==See Also== | ==See Also== |
Revision as of 19:39, 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
Following from the fact that must be prime, we can use to answer choices as possible solutions for n. , , and don't work because is even, and does not work since is divisible by . Thus, the only correct answer is .
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.