feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A. 150
B. 200
C. 250
D. 245
E. 225
Responding to a pm:
The method discussed in my post is useful while finding the maximum power of a number in a factorial. The given product is not in factorial form and hence the method needs to be suitably modified. That said, it should not be a big problem to modify the method if you understand the basics. Zeroes are produced by multiplying a 2 and a 5. So number of 0s in this product will depend on how many matching 2s and 5s we have here. In factorials, we have more 2s than 5s because we have consecutive numbers so we usually don't bother about finding the number of 2s. Here we have handpicked numbers so we need to ensure that we have both.
From where do we get 2s? From even multiples of 5.
10^10, 20^20, 30^30, 40^40, 50^50
\((2*5)^{10}, (2^2*5)^{20}, (2*15)^{30}, (2^3*5)^{40}, (2*25)^{50}\)
So number of 2s is 10 + 40 + 30 + 120 + 50 = 250
Now, let's see the number of 5s. Each term of the product has a 5. So the number of 5s is at least 5 + 10 + 15 + 20 + 25 + ... + 50
Then we also need to account for terms that have multiple 5s such as 25 and 50 but let's get to that later.
5 + 10 + 15 + 20 + 25 + ... + 50 = 5(1 + 2 + 3 + ..10) = 5*10*11/2 = 275
Note that number of 5s will be even more than 275 while the number of 2s is only 250. So there will be 250 trailing zeroes.
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