Combinatorics Problems and Solutions
Spring 2008 - Republished July 6, 2014
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2. The Number of Zeros at the End of 1000! (1000 Factorial)
We wish to find the number of zeros at the end of 1000! = 1000*999*998*997*....*4*3*2*1
We examine the factors of 1000! to see which can be used to produce multiples of 10.
For 1000!, each multiple of 5 - taken together with some even number - contributes a multiple of 10. Each multiple of 25 - taken together with some multiple of 4 - contributes two multiples of 10. Each multiple of 125 - taken together with some multiple of 8 - contributes three multiples of 10. Each multiple of 625 - taken together with some multiple of 16 - contributes four multiples of 10.
We can count each multiple of 5 in 1000! once. Then we can count each multiple of 25 in 1000! once in order to altogether count each multiple of 25 twice. Then we can count each multiple of 125 in 1000! once in order to altogether count each multiple of 125 thrice. There is also one multiple of 625 - i.e., 625 itself, in 1000!, which we count once in order to count the multiples of 625 altogether four times.
There are 1000/5 = 200 multiples of 5 in 1000!
There are 1000/25 = 40 multiples of 25 in 1000!
There are 1000/125 = 8 multiples of 125 in 1000!
There is also one multiple of 625 - i.e., 625 itself, in 1000!
Adding them together, we have 200+40+8+1 = 249 zeros at the end of 1000!.
4. Recurrence Relations and Generating Functions: Problem and Solution I
5. Partitions of Integers: Practice Problems and Solutions - Part I
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