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

15

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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.

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=(3^33)(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, 08:04, edited 2 times in total.

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

3

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Expert's post

1

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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

1

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Expert's post

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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Expert's post

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 _________________

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

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