2019 AMC 10A Problems/Problem 9
Contents
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?
Solutions
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 .
Solution 3 (Elimination)
The sum of the first positive integers is and the product of the positive integers upto is . The quotient of the two is -
which simplifies to . Thus, must be odd for the remainder to not be 0 (as will multiply with some number in , cancelling out if it is even, which leaves us with the answer choices and . Notice that must also be prime as otherwise there will be a factor of in x somewhere. So either or must be prime - is obviously not prime as it is divisible by 9, so our answer should be where , and so our answer is or .
- youtube.com/indianmathguy
Video Solutions
Video Solution 1
~savannahsolver
Video Solution 2 by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=33
~ pi_is_3.14
Video Solution 3
Education, the Study of Everything
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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