Question 494976: Let n!!! denote the product n*(n-3)*(n-6)*...*x where x is either 1, 2, or 3, if n is 1, 2, or 0 more than a multiple of 3, respectively. For example, 10!!! is 10*7*4*1 = 280. Let k be the largest positive integer with the property that 2011!!! is a multiple of 10^k. What is k?
choose the answer and explain the reason to choose your response
166
134
167
168
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! 2011!!! is equal to 2011*2008*2005*...*4*1. The factors that are multiples of 5 are {2005, 1990, 1975, ..., 10}, and the factors that are multiples of 2 are {2008, 2002, ..., 4}. Hence, we can compute the largest power of 5 that divides 2005*1990*1975*...*10 (there will be sufficient 2's).
There are 134 elements in the set {10, 25, ..., 2005}. However, we need to account for multiples of 25, 125, and 625. The multiples of 25 in that set are {25, 100, 175, ..., 1975} (simply all multiples of 25 that are 1 mod 3; 27 numbers); the multiples of 125 are {250, 625, ..., 1750} (5 numbers), and the multiples of 625 are {625} (1 number). It suffices to simply add the sizes of each set, so that we count multiples of 25 twice, multiples of 125 three times, and multiples of 625 four times. Hence, 134+27+5+1 = 167, so k = 167.
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