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How many consecutive 0's are there

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Re: How many consecutive 0's are there

by Vemuri » Sun Mar 15, 2009 3:15 am
Is the question correct? Please verify.
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by goelmohit2002 » Sun Mar 15, 2009 5:21 am
Why do you think it is wrong ?

Basically it is asking how many consecutive "0's" towards the end are there.

e.g. 5700 have two consecutive zeros.
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by DanaJ » Sun Mar 15, 2009 5:26 am
goelmohit2002: the way you've written the question is: how many zeros are there in 3! * 4!.
Let's break that down: 3! = 2*3 - no 5's
4! = 2*3*4 - again, no 5's

In order to have trailing zeros, you need to count your 5's, since 10 = 2*5 - and this is the way you're supposed to find out the number of trailing zeros: by counting the 5's.
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by DanaJ » Sun Mar 15, 2009 5:36 am
Now that I think of it, you might be looking for the number of zeros in (3! * 4!)!

This would mean the number of trailing zeros in (6*24)! = (144)!.

I've seen many questions like this one on this forum and people need to understand that: by counting the zeros, you're actually looking for the 5's, since 10 = 2*5. In 144!, there are plenty of 2's - at least one in every other number. This makes the 5's (which are indeed more rare) important. You have my strategy for this below:


Start by counting the number of multiples of 5 up to 144: 5, 10, 15, 20, ... 140. You get 140/5 = 28 multiples of 5 up to 144.

BUT (A REALLY BIG BUT) you have extra 5's, which happens because some of the multiples leading up to 144 are multiples of 25 = 5^2, meaning that they contain 1 or more 5's.
You need:
25 = 5*5 - one extra 5 (since you've already counted one of the two in the list above).
50 = 5*5*2 - one extra 5
75 = 5*5*3 - one extra 5
100 = 5*5*4 - one extra 5
125 = 5*5*5 - TWO extra 5's.

So you get 1 + 1 + 1 + 1 + 2 = 6 extra 5's to add up to the 28 you already counted.

And it seems I was right: you do get the OA: 28 + 6 = 34.
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by goelmohit2002 » Sun Mar 15, 2009 5:39 am
sorry sorry there was a typo...

please read it as (3! * 4!)!.

I have corrected the same in the original post too.
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by DanaJ » Sun Mar 15, 2009 5:46 am
Heh... It seems you made the correction to your post immediately after I realized what the mistake was... :D
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by goelmohit2002 » Sun Mar 15, 2009 5:51 am
Thanks.

on searching on this forum I found another way people have solved this problem

144/5 + (144/(5*5)) + ( 144/(5*5*5)) = 28 + 5 + 1 = 34

In the above only quotients are taken and remainder is ignored.

Can somebody please help me understand what is the logic behind above calculation. This seems to be pretty fast way to solve this question.

Thanks
Mohit
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by goelmohit2002 » Sun Mar 15, 2009 5:54 am
DanaJ wrote:Heh... It seems you made the correction to your post immediately after I realized what the mistake was... :D
Actually this is a contention problem :-)....both working simultaneously....i realised my mistake after your first post that there are no zeros in 144 = 3! * 4!....
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by Ian Stewart » Sun Mar 15, 2009 6:14 am
goelmohit2002 wrote:Thanks.

on searching on this forum I found another way people have solved this problem

144/5 + (144/(5*5)) + ( 144/(5*5*5)) = 28 + 5 + 1 = 34

In the above only quotients are taken and remainder is ignored.

Can somebody please help me understand what is the logic behind above calculation. This seems to be pretty fast way to solve this question.

Thanks
Mohit
The logic is as follows:

-from 1 to 144, there are 144/5 ~ 28 multiples of 5 (we round the decimal down, since 145 doesn't count). Each of these multiples of 5 will add one to the power on the 5 in the prime factorization of 144!

-from 1 to 144, there are also 144/25 ~ 5 multiples of 25 (we again round down, since we don't want to count 150). Each of these multiples of 25 will contribute an additional 5 to the prime factorization of 144!

-from 1 to 144, there is 144/125 ~ 1 multiple of 125, which adds yet another 5 to the prime factorization of 144!

So, in total, the power on the 5 in the prime factorization of 144! will be 28+5+1 = 34.
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by goelmohit2002 » Sun Mar 15, 2009 6:40 am
Thanks Stewart for the fantastic post.

Let's say the question would have been like
y = 12^k.
What is the maximum value of k in (144!)

Then can the above method be applied in calculating the same or there is some other way to solve the same?
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by DanaJ » Sun Mar 15, 2009 7:35 am
Well, if you use the same line of thought as outlined by Ian, you would get:
144/12 + 144/(12^2) = 12 + 1 = 13 for your k.
However, I don't think this is correct, mainly because 12 is not prime (like 5 was). You can easily figure out that 12 = 3*4 - and I guarantee there are more than 13 pairs of 3*4 in 144!. To check that, just consider the number of 3's in 144!:
144/3 + 144/9 + 144/27 + 144/81 - this is obviously going to be greater than 13, since only 144/3 is 48.
There are also plenty of 4's to go around - as I've mentioned, you get at least one 2 every other number and 4 = 2*2, so you get at least one 4 every four numbers.
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by goelmohit2002 » Sun Mar 15, 2009 7:57 am
Thanks.

Basically my doubt is that we have decided to use "5"...when we wanted to calculate 10^p....i.e. maximum value of "p"( as consecutive number of last zeros are same as max. value of p in 10^p). Since 10 = 5 * 2 and 144! has more number of 2's as you mentioned...so we need to calculate using 5's.

Why can't this method be applied to 12 = 3 * 4 = 3 * 2 * 2.

i.e. either calculate the same using no. of 3's or number of 4's or number of 2's.

Is the solution for 10^k was a specific case or can it be generalized for all the numbers ?

Thanks
Mohit
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by cramya » Sun Mar 15, 2009 9:02 am
Is the solution for 10^k was a specific case or can it be generalized for all the numbers ?

Can be applied to all but we just need to know the limiting power.

Eg: For 10 = 2*5

More 2's than 5's and hence 5 would be your limiting number. Hence u divided by 5,5^2 so on...



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Cramya
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by goelmohit2002 » Sun Mar 15, 2009 9:11 am
Can you please tell how to apply the same with 12^k ?

Thanks
Mohit
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