2019 AMC 10A Problems/Problem 9
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 1
The sum of the first positive integers is , and we want this not to be a divisor of (the product of the first positive integers). Notice that if and only if were composite, all of its factors would be less than or equal to , which means they would be able to cancel with the factors in . Thus, the sum of positive integers would be a divisor of when is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, must instead be prime. The greatest three-digit integer that is prime is , so we subtract to get .
Solution 2
As in Solution 1, we deduce that must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of . Choices , , and don't work because is even, and all even numbers are divisible by two, which makes choices , , and composite and not prime. Choice also does not work since is divisible by , which means it's a composite number and not prime. Thus, the correct answer must be .
Video Solution
~savannahsolver
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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All AMC 10 Problems and Solutions |
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