Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Generalization:)
 
(15 intermediate revisions by 5 users not shown)
Line 6: Line 6:
  
 
== Solution 2 ==
 
== 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!.
+
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:
 
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).
+
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 (n-3) consecutive positive integers are (5, 6, 7…, n+1)
+
So the <math>n-3</math> consecutive positive integers are <math>5, 6, 7…, n+1</math>
  
−
So we have (n+1)! /4! = n!
+
So we have <math>\frac{(n+1)!}{4!} = n!</math>
−
=> n+1 = 24
+
<math>\Longrightarrow  n+1 = 24</math>
−
=> n = 23
+
<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>
  
−
== Generalization: ==
+
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>
−
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 <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>.
  
−
Proof:
+
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>.
−
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!.
+
 
 +
So the <math>n-a</math> consecutive positive integers are <math>a+2, a+3, … n+1</math>
  
−
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 we have <math>\frac{(n+1)!}{(a+1)!} = n!</math>
 +
<math>\Longrightarrow  n+1 = (a+1)!</math>
 +
<math>\Longrightarrow  n = (a+1)! -1</math>
  
−
So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)
+
<math>Kris17</math>
  
−
So we have (n+1)! / (a+1)! = n!
+
== Video Solution!!! ==
−
=> n+1 = (a+1)!
+
https://www.youtube.com/watch?v=H-ZmsYjF-xE
−
=> n = (a+1)! -1
 
  
 
== See also ==
 
== See also ==
Line 38: Line 42:
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 +
{{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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&oldid=175420"