Difference between revisions of "1990 AIME Problems/Problem 11"

m (Generalization:)
m (Generalization:)
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== Generalization: ==
 
== Generalization: ==
Largest positive integer <math>n</math> for which <math>n!</math> can be expressed as the product of <math>n-a</math> consecutive positive integers is <math>(a+1)! 1</math>
+
Largest positive integer <math>n</math> for which <math>n!</math> can be expressed as the product of <math>n-a</math> consecutive positive integers is <math>(a+1)! - 1</math>
  
For ex. largest 4n<math> such that product of </math>n-6<math> consecutive positive integers is equal to </math>n!<math> is </math>7!-1 = 5039<math>
+
For ex. largest <math>n</math> such that product of <math>n-6</math> consecutive positive integers is equal to <math>n!</math> is <math>7!-1 = 5039</math>
  
 
Proof:
 
Proof:
Reasoning the same way as above, let the largest of the </math>n-a<math> consecutive positive integers be </math>k<math>. Clearly </math>k<math> cannot be less than or equal to </math>n<math>, else the product of </math>n-a<math> consecutive positive integers will be less than </math>n!<math>.
+
Reasoning the same way as above, let the largest of the <math>n-a</math> consecutive positive integers be <math>k</math>. Clearly <math>k</math> cannot be less than or equal to <math>n</math>, else the product of <math>n-a</math> consecutive positive integers will be less than <math>n!</math>.
  
Now, observe that for </math>n<math> to be maximum the smallest number (or starting number) of the </math>n-a<math> consecutive positive integers must be minimum, implying that </math>k<math> needs to be minimum. But the least </math>k > n<math> is </math>n+1<math>.
+
Now, observe that for <math>n</math> to be maximum the smallest number (or starting number) of the <math>n-a</math> consecutive positive integers must be minimum, implying that <math>k</math> needs to be minimum. But the least <math>k > n</math> is <math>n+1</math>.
  
So the </math>n-a<math> consecutive positive integers are </math>a+2, a+3, … n+1<math>
+
So the <math>n-a</math> consecutive positive integers are <math>a+2, a+3, … n+1</math>
  
So we have </math>\frac{(n+1)!}/{(a+1)!} = n!<math>
+
So we have <math>\frac{(n+1)!}{(a+1)!} = n!</math>
</math>\Longrightarrow  n+1 = (a+1)!<math>
+
<math>\Longrightarrow  n+1 = (a+1)!</math>
</math>\Longrightarrow  n = (a+1)! -1$
+
<math>\Longrightarrow  n = (a+1)! -1</math>
  
 
Kris17
 
Kris17

Revision as of 17:53, 15 March 2017

Problem

Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers.

Solution 1

The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!$. For $a = 4$, we get $n + 1 = 4!$ so $n = 23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as $^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n = 23$ 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 $(n+1)! /4! = n!$ $\Longrightarrow  n+1 = 24$ $\Longrightarrow  n = 23$

Kris17

Generalization:

Largest positive integer $n$ for which $n!$ can be expressed as the product of $n-a$ consecutive positive integers is $(a+1)! - 1$

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 $\frac{(n+1)!}{(a+1)!} = n!$ $\Longrightarrow  n+1 = (a+1)!$ $\Longrightarrow  n = (a+1)! -1$

Kris17

See also

1990 AIME (ProblemsAnswer KeyResources)
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

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