AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
Personal tools

1990 AIME Problems/Problem 11

From AoPSWiki

Problem

Someone observed that . Find the largest positive integer for which can be expressed as the product of consecutive positive integers.

Solution

The product of consecutive integers can be written as for some integer . Thus, , from which it becomes evident that . Since \displaystyle (n - 3 + a)! > n!, we can rewrite this as \frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!. For , we get so . For greater values of , we need to find the product of consecutive integers that equals . can be approximated as , which decreases as increases. Thus, is the greatest possible value to satisfy the given conditions.

See also

1990 AIME (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Support local problem solving programs by contributing to the Art of Problem Solving Foundation.
Click here for more information about the Foundation.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us