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Re: Product of sequence [#permalink]
17 Sep 2010, 02:29

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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

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.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
16 Nov 2012, 06:58

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I did it in a different way..... since it is multiplication of all 3 multiples.... 3*6*9*..... *99=3(1*2*3*4*5*......33)=3*33! 3 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.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
04 Jun 2013, 04:48

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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

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]
22 Aug 2013, 02:53

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mumbijoh wrote:

Dear Bunuel I came across this question and i really do not understand it.I read the "Everything about factorial " link but i cant seem to apply what i have read there to this question. How did you come up with this?Please help " once in 15; once in 30; once in 45; once in 60; twice in 75 (5*5*3); once in 90;

I understand the basic concept you explained in the mathbook by Gmatclub and various explanations you have given,but I am finding it difficult to apply on hard questions that involve multiple factorilas and questions that do not specifically give any fcatorial but give a complex product of numbers.

In that case, I must say that practice should help. _________________

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
25 Sep 2013, 12:34

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TAL010 wrote:

Finding the powers of a prime number p, in the n! The formula is: Example: What is the power of 2 in 25!?

^^ Taken from the GMAT Club book...what is the logic behind this question? What are they really asking?

It means calculating number of instances of P in n! Consider the simple example ---> what is the power of 3 in 10! We can find four instances of three in 10! -----> 1 * 2 * 3 * 4 * 5 * (2*3) * 7 * 8 * (3*3) * 10

You can see above we can get four 3s in the expression.

Calculating the number of instances in this way could be tedious in the long expressions. but there is a simple formula to calculate the powers of a particular prime.

the powers of Prime P in n! can be given by \frac{n}{p} + \frac{n}{p^2} + \frac{n}{p^3} + ................. till the denominator equal to or less than the numerator. what is the power of 3 in 10! ------> \frac{10}{3} + \frac{10}{3^2} = 3 + 1 = 4

Analyze how the process works........ We first divided 10 by 1st power of 3 i.e. by 3^1 in order to get all red 3s Later we divided 10 by 2nd power of 3 i.e. by 3^2 in order to get the leftover 3 (blue) we can continue in this way by increasing power of P as long as it does not greater than n

Back to the original question.............. What is the power of 2 in 25!? ---------> 25/2 + 25/4 + 25/8 + 25/16 = 12 + 6 + 3 + 1 = 22

Re: Product of sequence [#permalink]
20 Oct 2010, 03:14

Expert's post

nonameee wrote:

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. _________________

Re: Product of sequence [#permalink]
15 Nov 2012, 11:10

Bunuel wrote:

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?

Re: Product of sequence [#permalink]
16 Nov 2012, 02:32

Expert's post

Amateur wrote:

Bunuel wrote:

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).

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
16 Nov 2012, 07:47

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 _________________

Thriving for CHANGE

gmatclubot

Re: If N is the product of all multiples of 3 between 1 and 100
[#permalink]
16 Nov 2012, 07:47