Difference between revisions of "2019 AMC 10A Problems/Problem 9"
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− | The sum of <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this not to be a divisor of <math>n!</math> (the product of the first <math>n</math> positive integers). Notice that if and only if <math>n+1</math> were composite, all of its factors would be less than or equal to <math>n</math>, | + | The sum of <math>n</math> positive integers is <math>\frac{(n)(n+1)}{2}</math>, and we want this not to be a divisor of <math>n!</math> (the product of the first <math>n</math> positive integers). Notice that if and only if <math>n+1</math> were composite, all of its factors would be less than or equal to <math>n</math>, which means they would be able to cancel with the factors in <math>n!</math>. Thus, the sum of <math>n</math> positive integers would be a divisor of <math>n!</math> when <math>n+1</math> is composite. Hence in this case, <math>n+1</math> must instead be prime. The greatest three-digit integer that is prime is <math>997</math>, so we subtract <math>1</math> to get <math>n=\boxed{\textbf{(B) } 996}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 17:28, 29 December 2019
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?
Solution 1
The sum of 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. 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 .
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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