Problem Solving (PS) | Re: If k = [3/3]*[4/3]*[5/3]*[6/3]*.....*[99/3]*[100/3], in which [a] is t

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
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Statistics : Posted by chetan2u • on 26 Jun 2023, 11:41 • Replies 4 • Views 252

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