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If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

a. 3 b. 6 c. 7 d. 8 e. 10

How do you solve these sort of questions quickly Thanks

We should determine # of trailing zeros of N=3*6*9*12*15*...*99 (a sequence of 0's of a number, after which no other digits follow).

Since there are at least as many factors 2 in N as factors of 5, then we should count the number of factors of 5 in N and this will be equivalent to the number of factors 10, each of which gives one more trailing zero.

Factors of 5 in N: once in 15; once in 30; once in 45; once in 60; twice in 75 (5*5*3); once in 90;

1+1+1+1+2+1=7 --> N has 7 trailing zeros, so greatest integer \(m\) for which \(\frac{N}{10^m}\) is an integer is 7.

Bunuel, is it necessary to count the number of trailing zeros? I have solved the problem by counting the number of 5's in N.

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.
_________________

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.

How did you know that 2 factors and 5 factors in N are same?

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.

How did you know that 2 factors and 5 factors in N are same?

No, that's not what I'm saying (see the red part). The power of 2 in N is at least as high as the power of 5 in N.

We are told that N=3*6*9*12*15*18*21*...*90*93*96*99 --> as you can observe, the power of 2 in N will be higher than the power of 5 (there are more even numbers than multiples of 5).

If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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16 Nov 2012, 07:58

3

This post received KUDOS

I did it in a different way..... since it is multiplication of all 3 multiples.... 3*6*9*..... *99=(3^33)(1*2*3*4*5*......33)=(3^33)*33! (3 power 33 because a 3 can be extracted from each number inside) (3^33) doesn't have any multiples between 1-9 which can contribute a 0..... so number of trailing 0's should be number of trailing 0's of 33! which is 7. So C is the answer... we don't need to count 5's and 2's and complicate things in this case! Let me know if you think this approach of mine has loop holes.

Last edited by Amateur on 23 Jan 2015, 09:04, edited 2 times in total.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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16 Nov 2012, 08:47

i got answer as '6' bunuel am i missing something? I may be completely wrong. Below is my approach :-

N= 3*6*9*12*.......*99

Then total multiple of 3 will be 33

then every alternate number have factor of 2 in it so total factor of 2 will be 6 Since highest number which is multiple of 2 is 96 which have total 6 factors of 2
_________________

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

A. 3 B. 6 C. 7 D. 8 E. 10

How do you solve these sort of questions quickly Thanks

Responding to a pm:

First, check out this post. It is an application of a concept that discusses the maximum power of a number in a factorial. This post discusses how and why we find the maximum power. http://www.veritasprep.com/blog/2011/06 ... actorials/

Once you are done, note that this question can be easily broken down into the factorial form.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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17 Aug 2013, 12:37

rafi wrote:

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

A. 3 B. 6 C. 7 D. 8 E. 10

How do you solve these sort of questions quickly Thanks

stunning math.

my solution: 3.6.9.12..............................99 = 3^33 . (1.2.3.4.5...............................33) factors that can bring zeros are = 2,5,10,12,15,20,22,25,30,32 (All from 2 , 0 and 5 ) But only 25 can produce two 5.

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