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

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== Problem ==
 
== Problem ==
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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.
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== Solution 1 ==
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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.
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== Solution 2 ==
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Let the largest of the <math>n-3</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-3</math> consecutive positive integers will be less than <math>n!</math>.
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Key observation:
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Now for <math>n</math> to be maximum the smallest number (or starting number) of the <math>n-3</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>.
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So the <math>n-3</math> consecutive positive integers are <math>5, 6, 7…, n+1</math>
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So we have <math>\frac{(n+1)!}{4!} = n!</math>
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<math>\Longrightarrow  n+1 = 24</math>
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<math>\Longrightarrow  n = 23</math>
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== Generalization ==
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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>
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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>
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Proof:
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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>.
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So the <math>n-a</math> consecutive positive integers are <math>a+2, a+3, … n+1</math>
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So we have <math>\frac{(n+1)!}{(a+1)!} = n!</math>
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<math>\Longrightarrow  n+1 = (a+1)!</math>
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<math>\Longrightarrow  n = (a+1)! -1</math>
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<math>Kris17</math>
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== Video Solution!!! ==
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https://www.youtube.com/watch?v=H-ZmsYjF-xE
  
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== Solution ==
 
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{{solution}}
 
 
== See also ==
 
== See also ==
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* [[1990 AIME Problems/Problem 10 | Previous problem]]
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{{AIME box|year=1990|num-b=10|num-a=12}}
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* [[1990 AIME Problems/Problem 12 | Next problem]]
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* [[1990 AIME Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 08:28, 28 June 2022

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 $\frac{(n+1)!}{4!} = n!$ $\Longrightarrow n+1 = 24$ $\Longrightarrow n = 23$

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

$Kris17$

Video Solution!!!

https://www.youtube.com/watch?v=H-ZmsYjF-xE

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AIME Problems and Solutions

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