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 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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