Difference between revisions of "1990 AIME Problems/Problem 11"
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=> n+1 = (a+1)! | => n+1 = (a+1)! | ||
=> n = '''(a+1)! -1''' | => n = '''(a+1)! -1''' | ||
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| + | Kris17 | ||
== See also == | == See also == | ||
Revision as of 22:09, 16 November 2014
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 (n-3) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-3) consecutive positive integers will be less than n!.
Key observation: Now for n to be maximum the smallest number (or starting number) of the (n-3) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).
So the (n-3) consecutive positive integers are (5, 6, 7…, n+1)
So we have (n+1)! /4! = n! => n+1 = 24 => n = 23
Kris17
Generalization:
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
Kris17
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
