Difference between revisions of "2006 AIME A Problems/Problem 4"
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== Problem == | == Problem == | ||
− | Let <math> | + | 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. |
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== Solution == | == Solution == |
Revision as of 14:03, 25 September 2007
Problem
Let be the number of consecutive 0's at the right end of the decimal representation of the product
Find the remainder when
is divided by 1000.
Solution
Clearly, . Now, consider selecting
of the remaining
values. Sort these values in descending order, and sort the other
values in ascending order. Now, let the
selected values be
through
, and let the remaining
be
through
. It is now clear that there is a bijection between the number of ways to select
values from
and ordered 12-tuples
. Thus, there will be
such ordered 12-tuples.
See also
2006 AIME II (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 |