Difference between revisions of "2006 AIME I Problems/Problem 4"
(→Solution) |
|||
(11 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == 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. | ||
− | + | 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>. | |
− | + | ||
− | One way to do this is as follows: | + | == 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+61+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 == | == See also == | ||
− | + | {{AIME box|year=2006|n=I|num-b=3|num-a=5}} | |
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:55, 2 September 2024
Contents
Problem
Let be the number of consecutive
's at the right end of the decimal representation of the product
Find the remainder when
is divided by
.
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: of the numbers
have a factor of
.
have a factor of
.
have a factor of
. And so on. This gives us an initial count of
. Summing this arithmetic series of
terms, we get
. However, we have neglected some powers of
- every
term for
has an additional power of
dividing it, for
extra; every n! for
has one more in addition to that, for a total of
extra; and similarly there are
extra from those larger than
and
extra from
. Thus, our final total is
, and the answer is
.
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 or
, then the factorials that are greater than
or
. Since
we do not need to account for it.
This gives and
respectively. Adding all of these terms up gives 1124 or
.
~PeterDoesPhysics
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=413
~ pi_is_3.14
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.