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# If k = [3/3]*[4/3]*[5/3]*[6/3]*.....*[99/3]*[100/3], in which [a] is t

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Re: If k = [3/3]*[4/3]*[5/3]*[6/3]*.....*[99/3]*[100/3], in which [a] is t [#permalink]
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gmatophobia wrote:
If $$k = [\frac{3}{3}]*[\frac{4}{3}]*[\frac{5}{3}]*[\frac{6}{3}]*.....*[\frac{98}{3}]*[\frac{99}{3}]*[\frac{100}{3}]$$, in which $$[a]$$ is the greatest integer less than or equal to $$a$$.

How many trailing zeros will appear in $$k$$?

A. 14
B. 18
C. 21
D. 49
E. 343

When you divide consecutive numbers by n and discard the decimal part, we are looking at set of n numbers giving same value and each subsequent set giving consecutive values, that is 1,1,1,2,2,2,3… as here n is 3.
The last number will be 100/3 or 33.
Thus the pattern we get is 1,1,1,2,2,2……33,33.
For trailing zeroes, we have to concentrate on number of 5s we get. => 5,5,5,10,10,10….30,30,30
Thus, the product of these number will give $$(5*10*15*20*25*30)^3$$
Number of 5s = $$(5*5*5*5*5^2*5)^3=(5^7)^3=5^{21}$$

So, 21 trailing zeroes.

C
Re: If k = [3/3]*[4/3]*[5/3]*[6/3]*.....*[99/3]*[100/3], in which [a] is t [#permalink]
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