1990 AIME Problems/Problem 11
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.
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 k needs to be minimum. But the least is
So the consecutive positive integers are
So we have
Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = (a+1)! – 1
For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039
Proof: Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.
Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).
So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)
So we have (n+1)! / (a+1)! = n! => n+1 = (a+1)! => n = (a+1)! -1
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