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

m (See also: c)
 
(16 intermediate revisions by 5 users not shown)
Line 2: Line 2:
 
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>\displaystyle (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>\displaystyle ^{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.
+
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 <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>.
 +
 
 +
Key observation:
 +
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>.
 +
 
 +
So the <math>n-3</math> consecutive positive integers are <math>5, 6, 7…, n+1</math>
 +
 
 +
So we have <math>\frac{(n+1)!}{4!} = n!</math>
 +
<math>\Longrightarrow  n+1 = 24</math>
 +
<math>\Longrightarrow  n = 23</math>
 +
 
 +
== 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>
 +
 
 +
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:
 +
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>.
 +
 
 +
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>
 +
<math>\Longrightarrow  n+1 = (a+1)!</math>
 +
<math>\Longrightarrow  n = (a+1)! -1</math>
 +
 
 +
<math>Kris17</math>
  
 
== See also ==
 
== See also ==
Line 9: Line 39:
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:53, 4 December 2018

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

$Kris17$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS