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 | |
| 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 | ||
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