Difference between revisions of "1990 AIME Problems/Problem 11"
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Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest [[positive]] [[integer]] <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the [[product]] of <math>n - 3_{}^{}</math> [[consecutive]] positive integers. | Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest [[positive]] [[integer]] <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the [[product]] of <math>n - 3_{}^{}</math> [[consecutive]] positive integers. | ||
| − | == Solution == | + | == Solution 1 == |
| − | The product of <math>n - 3</math> consecutive integers can be written as <math>\frac{(n - 3 + a)!}{a!}</math> for some integer <math>a</math>. Thus, <math>n! = \frac{(n - 3 + a)!}{a!}</math>, from which it becomes evident that <math>a \ge 3</math>. Since <math> | + | The product of <math>n - 3</math> consecutive integers can be written as <math>\frac{(n - 3 + a)!}{a!}</math> for some integer <math>a</math>. Thus, <math>n! = \frac{(n - 3 + a)!}{a!}</math>, from which it becomes evident that <math>a \ge 3</math>. Since <math>(n - 3 + a)! > n!</math>, we can rewrite this as <math>\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!</math>. For <math>a = 4</math>, we get <math>n + 1 = 4!</math> so <math>n = 23</math>. For greater values of <math>a</math>, we need to find the product of <math>a-3</math> consecutive integers that equals <math>a!</math>. <math>n</math> can be approximated as <math>^{a-3}\sqrt{a!}</math>, which decreases as <math>a</math> increases. Thus, <math>n = 23</math> 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 | ||
| + | |||
| + | |||
| + | == 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 | ||
| + | |||
== See also == | == See also == | ||
Revision as of 21:38, 28 October 2009
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
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
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 | ||
