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 "2006 AIME I Problems/Problem 4"

(→‎Solution)
m (→‎Solution)
(5 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
 
+
Let <math> N </math> be the number of consecutive <math>0</math>'s at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>.
−
 
 
−
Let <math> (a_1,a_2,a_3,\ldots,a_{12}) </math> be a permutation of <math> (1,2,3,\ldots,12) </math> for which
 
−
 
 
−
<center><math> a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\  and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}. </math></center>
 
−
 
 
−
An example of such a permutation is <math> (6,5,4,3,2,1,7,8,9,10,11,12). </math> Find the number of such permutations.
 
  
 
== Solution ==
 
== Solution ==
 +
A number in decimal notation ends in a zero for each power of ten which divides it.  Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression.  Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
  
−
Clearly, <math>a_6=1</math>. Now, consider selecting <math>5</math> of the remaining <math>11</math> values.  Sort these values in descending order, and sort the other <math>6</math> values in ascending order.  Now, let the <math>5</math> selected values be <math>a_1</math> through <math>a_5</math>, and let the remaining <math>6</math> be <math>a_7</math> through <math>{a_{12}}</math>.  It is now clear that there is a [[bijection]] between the number of ways to select <math>5</math> values from <math>11</math> and ordered 12-tuples <math>(a_1,\ldots,a_{12})</math>.  Thus, there will be <math>{11 \choose 5}=462</math> such ordered 12-tuples.
+
One way to do this is as follows: <math>96</math> of the numbers <math>1!,\ 2!,\ 3!,\ 100!</math> have a factor of <math>5</math>. <math>91</math> have a factor of <math>10</math>. <math>86</math> have a factor of <math>15</math>. And so on. This gives us an initial count of <math>96 + 91 + 86 + \ldots + 1</math>.  Summing this [[arithmetic series]] of <math>20</math> terms, we get <math>970</math>.  However, we have neglected some powers of <math>5</math> - every <math>n!</math> term for <math>n\geq25</math> has an additional power of <math>5</math> dividing it, for <math>76</math> extra; every n! for <math>n\geq 50</math> has one more in addition to that, for a total of <math>51</math> extra; and similarly there are <math>26</math> extra from those larger than <math>75</math> and <math>1</math> extra from <math>100</math>.  Thus, our final total is <math>970 + 76 + 51 + 26 + 1 = 1124</math>, and the answer is <math>\boxed{124}</math>.
  
 
== See also ==
 
== See also ==
Line 17: Line 12:
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 18:49, 4 August 2020

Problem

Let $N$ be the number of consecutive $0$'s at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$.

Solution

A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.

One way to do this is as follows: $96$ of the numbers $1!,\ 2!,\ 3!,\ 100!$ have a factor of $5$. $91$ have a factor of $10$. $86$ have a factor of $15$. And so on. This gives us an initial count of $96 + 91 + 86 + \ldots + 1$. Summing this arithmetic series of $20$ terms, we get $970$. However, we have neglected some powers of $5$ - every $n!$ term for $n\geq25$ has an additional power of $5$ dividing it, for $76$ extra; every n! for $n\geq 50$ has one more in addition to that, for a total of $51$ extra; and similarly there are $26$ extra from those larger than $75$ and $1$ extra from $100$. Thus, our final total is $970 + 76 + 51 + 26 + 1 = 1124$, and the answer is $\boxed{124}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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=2006_AIME_I_Problems/Problem_4&oldid=130618"