Difference between revisions of "2006 AIME I Problems/Problem 4"

Latest revision as of 21:56, 27 September 2024

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 $\boxed{124}$ .

Solution 2 (Legendre's Formula )

This problem can be easily solved using Legendre's Formula:

First is to account for all of the factorials that are greater than $5n$ or $5, 10, 15, 20...100$ , then the factorials that are greater than $5^2n$ or $25, 50, 75, 100$ . Since $5^3=125>100$ we do not need to account for it.

This gives $96+91+86+...+1$ and $76+51+26+1$ respectively. Adding all of these terms up gives 1124 or $\boxed{124}$ .

~PeterDoesPhysics

Video Solution by OmegaLearn

https://youtu.be/p5f1u44-pvQ?t=413

~ pi_is_3.14

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math> N </math> be the number of consecutive 0'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 1000.
+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>.
+== 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.
-== Solution ==
+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>.
-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 <math>96 + 91 + 86 + \ldots + 1</math>.  Summing this [[arithmetic series]] of 20 terms, we get 970.  However, we have neglected some powers of 5 -- every n! term for <math>n\geq25</math> has an additional power of 5 dividing it, for 76 extra; every n! for <math>n\geq 50</math> 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 124.
+== Solution 2 (Legendre's Formula )==
+This problem can be easily solved using [[Legendre's Formula]]:
+First is to account for all of the factorials that are greater than <math>5n</math> or <math>5, 10, 15, 20...100</math>, then the factorials that are greater than <math>5^2n</math> or <math>25, 50, 75, 100</math>. Since <math>5^3=125>100</math> we do not need to account for it.
+This gives <math>96+91+86+...+1</math> and <math>76+51+26+1</math> respectively. Adding all of these terms up gives 1124 or <math>\boxed{124}</math>.
+~PeterDoesPhysics
+== Video Solution by OmegaLearn ==
+https://youtu.be/p5f1u44-pvQ?t=413
+~ pi_is_3.14
 == See also ==
-* [[2006 AIME I Problems/Problem 3 | Previous problem]]
+{{AIME box|year=2006|n=I|num-b=3|num-a=5}}
-* [[2006 AIME I Problems/Problem 5 | Next problem]]
-* [[2006 AIME I Problems]]
 [[Category:Intermediate Combinatorics Problems]]
 [[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

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