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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.


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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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