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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.

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.

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

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16 Nov 2012, 07:58

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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, 09:04, edited 2 times in total.

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

Dear Reto, My friend, before you post anything else, please familiarize yourself with the protocols. This question has been posted many times before, for example, here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html where there's already a long discussion. Always search for a question before you start a new thread from scratch. Presumably, Bunuel, the math genius moderator, will merge this post into one of the larger previous posts on the same topic. If you have any questions that are not already answered there, you are more than welcome to ask me. Best of luck, Mike
_________________

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?

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).

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.
_________________

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

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25 Sep 2013, 23:35

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We know that for a number to be divisible by 10 must have at least one zero. Let's break the 10 into its prime factors, ie. 5 and 2. Now, we need to find pairs of 2 and 5 in the numerator. Here, 5 is our limiting factor, as it appears less than 2 does. therefore two cont the number of 5s, we must count the 5s in all multiples of 3 between 1 and 100.

15= One 5 30= One 5 45= One 5 60= One 5 75 = Two 5s (5 x 5 x3=75) 90= One 5.\

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.

We need to find the number of 5s in 33! because you need a 2 and a 5 to make a 10. The number of 5s will certainly be fewer than the number of 2s.

33/5 = 6 6/5 = 1

So you will have a total of 6+1 = 7 5s and hence can make 7 10s. So maximum power of 10 must be 7.

Answer C

Note that we ignore \(3^{33}\) because it has no 5s in it.

Dear Karishma

Could you explain step by step how to arrive at \(3^{33}*33!\) It's logical for me that we have to illustrate the product of all multiples of 3 between 1-100. The following is however not quite clear for me:

1. Did you count all the multiples of 3 between 1 and 100 "manually" or is there a smart way? 2. Why do you multiply by 33! ?

Could you help me here? Thank you!

You don't have to count the multiples of 3. Just look at the pattern.

Now from each term, separate out the 3 and put all 3s together in the front. You have 33 terms so you will get 33 3s. Also you will be left with all second terms 1, 2, 3, 4 etc

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