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

Revision as of 19:26, 25 September 2007

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 124.

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 == 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> (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 ==
+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:  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.
 == 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
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Difference between revisions of "2006 AIME I Problems/Problem 4"

Revision as of 19:26, 25 September 2007

Problem

Solution

See also