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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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If k = [3/3]*[4/3]*[5/3]*[6/3]*.....*[99/3]*[100/3], in which [a] is t [#permalink] New post   26 Jun 2023, 11:41
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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
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rollinbusiness
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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] New post   26 Jun 2023, 12:53
I'm kinda stuck.

So the trailing 0s would be the number of factors of 10 in the series.

so 30/3, 31/3, 32/3 all give 10
60,61,62 all give 20, which has a 10
90,91,92 all give 30 which has a 10.

So I see 9 so far.

Now I'm thinking we also need factors of 5 since 5 times any of the even numbers will give us 10, and I think we have enough even numbers to not worry. 15, 30 (exclude, already counted), 45, 60 (already coutned), 75, 90 (already counted)

so 3 more (15,45,75)

So 13? I see 9 10s, and 3 more 10s formed by 5s with 10s...

I'm definitely missing something...
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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] New post   27 Jun 2023, 05:31
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To find the number of trailing zeros of k, we need to find the number of multiples of 10 in k.
We can re-write k as k = 1 * 1 * 1 * 2 * .... * 33 * 33, therefore we can see that k is a multiple of factorials, i.e.,
k = 32! * 33! * 33!

To find all multiples of 10 in k, we need to find the multiples of 10 in each of the factorials. Since 10 = 5*2 and we have many more 2s than 5s, we need to find the number of 5s in each.

In 32!:
32/5=6 or 5 shows up once 6 times
32/25=1 or 5 shows up twice once.
Therefore, in 32! there are 6+1=7 fives. Since 33! = 32!*33, it will not have additional fives. Thus, in total we have 7*3=21 fives in k and k has 21 trailing zeros, which means the answer is C
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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] New post   28 Jun 2023, 20:41
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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]
28 Jun 2023, 20:41
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