1990 AIME Problems/Problem 11
Problem
Someone observed that . Find the largest positive integer for which can be expressed as the product of consecutive positive integers.
Solution 1
The product of consecutive integers can be written as for some integer . Thus, , from which it becomes evident that . Since , we can rewrite this as . 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.
Solution 2
Let the largest of the consecutive positive integers be . Clearly cannot be less than or equal to , else the product of consecutive positive integers will be less than .
Key observation: Now for to be maximum the smallest number (or starting number) of the consecutive positive integers must be minimum, implying that needs to be minimum. But the least is .
So the consecutive positive integers are
So we have
Generalization
Largest positive integer for which can be expressed as the product of consecutive positive integers is
For ex. largest such that product of consecutive positive integers is equal to is
Proof: Reasoning the same way as above, let the largest of the consecutive positive integers be . Clearly cannot be less than or equal to , else the product of consecutive positive integers will be less than .
Now, observe that for to be maximum the smallest number (or starting number) of the consecutive positive integers must be minimum, implying that needs to be minimum. But the least is .
So the consecutive positive integers are
So we have
Video Solution!!!
https://www.youtube.com/watch?v=H-ZmsYjF-xE
See also
1990 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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