2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 9 Easy

What is the greatest three-digit positive integer n for which the sum of the first n positive integers is not a divisor of the product of the first n positive integers? (2019 AMC 10A Problem, Question#9)

  • A.

    995

  • B.

    996

  • C.

    997

  • D.

    998

  • E.

    999

Answer:B

The sum of n positive integers is \frac{(n)(n+1)}{2}, and we want this not to be a divisor of n! (the product of the first n positive integers). Notice that if and only if n+1 were composite, all of its factors would be less than or equal to n , so would be able to cancel with these factors in n!, and thus the sum would be a divisor. Hence in this case, n+1 must instead be prime. The greatest three-digit integer that is prime is 997 , so we subtract 1 to get n=996.

As in Solution 1, we deduce that n+1 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 n.Choices A, C, and E don't work because n+1 is even, and choice D does not work since 999 is divisible by 9. Thus, the correct answer must be 996.