Difference between revisions of "2006 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
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| + | 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 | ||
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| + | <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> | ||
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| + | 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 == | ||
Revision as of 13:03, 25 September 2007
Problem
Let 
be a permutation of 
for which
An example of such a permutation is 
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.
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 
. Summing this arithmetic series of 20 terms, we get 970. However, we have neglected some powers of 5 -- every n! term for 
has an additional power of 5 dividing it, for 76 extra; every n! for 
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 | ||
| All AIME Problems and Solutions | ||
