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Manager  Joined: 04 Jun 2010
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Concentration: General Management, Technology
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GMAT 1: 670 Q47 V35 GMAT 2: 730 Q49 V41 If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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Question Stats: 47% (02:00) correct 53% (02:16) wrong based on 2249 sessions

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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

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

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55
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

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.

Check this for more:
http://gmatclub.com/forum/everything-ab ... 85592.html

Hope it helps.
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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42
1
10
N = The product of the sequence of 3*6*9*12....*99

N therefore is also equal to 3* (1*2*3*.....*33)

Therefore N = 3* 33!

From here we want to find the exponent number of prime factors, specifically the factors of 10.

10 = 5*2 so we want to find which factors is the restrictive factor

We can ignore the 3, since a factor that is not divisible by 5 or 2 is still not divisible if that number is multiplied by 3.

Therefore:

33/ 2 + 33/4 + 33/8 = 16+8+4 = 28

33/ 5 + 33/25 = 6 + 1 = 7

5 is the restrictive factor.

Here is a similar problem: number-properties-from-gmatprep-84770.html
##### General Discussion
Director  Joined: 23 Apr 2010
Posts: 504
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58988
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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

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11
Manager  Joined: 07 Jun 2010
Posts: 76
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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7
2
N = The product of the sequence of 3*6*9*12....*99

N therefore is also equal to 3* (1*2*3*.....*33)

Therefore N = 3* 33!

From here we want to find the exponent number of prime factors, specifically the factors of 10.

10 = 5*2 so we want to find which factors is the restrictive factor

We can ignore the 3, since a factor that is not divisible by 5 or 2 is still not divisible if that number is multiplied by 3.

Therefore:

33/ 2 + 33/4 + 33/8 = 16+8+4 = 28

33/ 5 + 33/25 = 6 + 1 = 7

5 is the restrictive factor.

Here is a similar problem: number-properties-from-gmatprep-84770.html
Manager  Joined: 05 Nov 2012
Posts: 138
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58988
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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1
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).

Hope it's clear.
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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3
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.

Originally posted by Amateur on 16 Nov 2012, 07:58.
Last edited by Amateur on 23 Jan 2015, 09:04, edited 2 times in total.
Intern  Joined: 22 Sep 2012
Posts: 9
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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I am not convinced by the answer of Bunuel, so I used excel to calculate the product.

The answer is 48,271,088,561,614,000,000,000,000,000,000,000,000,000,000,000,000,000, which means the maximum of m will be 39.

This is not a good question
Math Expert V
Joined: 02 Sep 2009
Posts: 58988
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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7
lunar255 wrote:
I am not convinced by the answer of Bunuel, so I used excel to calculate the product.

The answer is 48,271,088,561,614,000,000,000,000,000,000,000,000,000,000,000,000,000, which means the maximum of m will be 39.

This is not a good question

1. There is nothing wrong with the question.

2. Solution is correct, answer is C.

3. Excel rounds big numbers. Actual result is 48,271,088,561,613,960,642,858,365,853,327,381,832,862,269,440,000,000.
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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18
2
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.

$$3*6*9*...*99 = 3^{33} * (1*2*3*4*...*32*33) = 3^{33} * 33!$$

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.

Note that we ignore $$3^{33}$$ because it has no 5s in it.
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Intern  Joined: 03 Apr 2012
Posts: 22
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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1
I think the easiest way to do it is to count the number of 5's from 1 to 33.
3^ 33 ( 1 x 2x 3...... 33)

5 factors

5 - 5x1
10- 5x2
15- 5x3
20 - 5x4
25 - 5x5
30 - 5x6

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GMAT Date: 01-17-2014
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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1
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.
"
once in 15;
once in 30;
once in 45;
once in 60;
twice in 75 (5*5*3);
once in 90;
Math Expert V
Joined: 02 Sep 2009
Posts: 58988
Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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1
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.
"
once in 15;
once in 30;
once in 45;
once in 60;
twice in 75 (5*5*3);
once in 90;

15=5*3
30=5*6
45=5*9
60=5*12
75=5^2*3
90=5*18

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

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

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1
1
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

Hope that helps! _________________
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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1
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?

Check out this post: http://www.veritasprep.com/blog/2011/06 ... actorials/
It answers this question in detail explaining the logic behind it.
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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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2
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.\ Re: If N is the product of all multiples of 3 between 1 and 100, what is   [#permalink] 25 Sep 2013, 23:35

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