Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 25 Mar 2017, 06:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Jun 2010
Posts: 113
Concentration: General Management, Technology
Schools: Chicago (Booth) - Class of 2013
GMAT 1: 670 Q47 V35
GMAT 2: 730 Q49 V41
Followers: 15

Kudos [?]: 260 [7] , given: 43

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

### Show Tags

17 Sep 2010, 03:14
7
KUDOS
98
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

44% (02:20) correct 56% (01:41) wrong based on 2160 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

Consider Kudos if my post helped you. Thanks!
--------------------------------------------------------------------
My TOEFL Debrief: http://gmatclub.com/forum/my-toefl-experience-99884.html
My GMAT Debrief: http://gmatclub.com/forum/670-730-10-luck-20-skill-15-concentrated-power-of-will-104473.html

Math Expert
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [25] , given: 11022

### Show Tags

17 Sep 2010, 03:29
25
KUDOS
Expert's post
33
This post was
BOOKMARKED
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.

Check this for more:

Hope it helps.
_________________
Director
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 82 [0], given: 7

### Show Tags

20 Oct 2010, 04:09
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
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [1] , given: 11022

### Show Tags

20 Oct 2010, 04:14
1
KUDOS
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.
_________________
Intern
Joined: 06 Feb 2011
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 4

### Show Tags

14 Feb 2011, 06:11
I am only able to get ans B (6) and not the OA: 7(C).

Allow me to share my take on this,

Since N which is the product of all the multiples of 3 between 1 and 100 i.e.
N = 3X6X9X12X.....99,

for N to be divided by 10 and remain an integer, I need to find out the number of factors with "0" in the ones digit

N contains factors 30,60 and 90 so m will be at least 3 since (30X60X90)/1000 is an integer

and since 10 = 5X2, any multiple of 3 with a "5" as a ones digit when multiplied by an even number will yield a number with "0" in the ones digit.

So 15,45 and 75 (all multiples of 3) come to mind. Since there are plenty of even number factors in N, I get another 3 for the value of m

So m = 3+3 = 6. I dont get how m can be 7 though.. so am i missing something?
Manager
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 107
Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16
Followers: 1

Kudos [?]: 28 [0], given: 22

### Show Tags

14 Feb 2011, 06:17
IrinaTyan wrote:
If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which is N/(10^m) is an integer?

1. 3
2. 6
3. 7
4. 8
5. 10

Add up the terms that can lead to a zero that are multiples of 3

30
60
90
15*12
45*42
75*72

cant think of the 7th and gota run to work, but that is how you do it!
Math Expert
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [1] , given: 11022

### Show Tags

14 Feb 2011, 06:34
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
Intern
Joined: 06 Feb 2011
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 4

### Show Tags

14 Feb 2011, 06:59
Thanks Brunel! I didnt consider that 75 has 2 factors of 5 thus adding 1 more to m.
Manager
Joined: 07 Jun 2010
Posts: 86
Followers: 1

Kudos [?]: 33 [18] , given: 0

### Show Tags

15 Feb 2011, 20:36
18
KUDOS
2
This post was
BOOKMARKED
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: 07 Jun 2010
Posts: 86
Followers: 1

Kudos [?]: 33 [5] , given: 0

### Show Tags

15 Feb 2011, 20:36
5
KUDOS
1
This post was
BOOKMARKED
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: 171
Followers: 1

Kudos [?]: 33 [0], given: 57

### Show Tags

15 Nov 2012, 12:10
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?
Math Expert
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [1] , given: 11022

### Show Tags

16 Nov 2012, 03:32
1
KUDOS
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).

Hope it's clear.
_________________
Manager
Joined: 05 Nov 2012
Posts: 171
Followers: 1

Kudos [?]: 33 [3] , given: 57

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

### Show Tags

16 Nov 2012, 07:58
3
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.
Manager
Joined: 29 Jul 2012
Posts: 189
GMAT Date: 11-18-2012
Followers: 0

Kudos [?]: 95 [0], given: 23

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

### Show Tags

16 Nov 2012, 08:47
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

Intern
Joined: 22 Sep 2012
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 4

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

### Show Tags

30 Dec 2012, 19:21
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
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [4] , given: 11022

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

### Show Tags

31 Dec 2012, 03:34
4
KUDOS
Expert's post
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.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7246
Location: Pune, India
Followers: 2204

Kudos [?]: 14326 [9] , given: 222

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

### Show Tags

04 Jun 2013, 05:48
9
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 37581
Followers: 7394

Kudos [?]: 99440 [0], given: 11022

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

### Show Tags

17 Jul 2013, 00:35
From 100 hardest questions
Bumping for review and further discussion.
_________________
Intern
Joined: 03 Apr 2012
Posts: 27
Followers: 0

Kudos [?]: 12 [1] , given: 10

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

### Show Tags

17 Aug 2013, 00:58
1
KUDOS
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

Senior Manager
Joined: 10 Jul 2013
Posts: 335
Followers: 3

Kudos [?]: 336 [0], given: 102

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

### Show Tags

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.

so, 2*5 , 10, 12*15, 20, 22*5, 30, 32*5 = 7 zeros.

so m=7 we can afford at most. Answer (C)
_________________

Asif vai.....

Re: If N is the product of all multiples of 3 between 1 and 100   [#permalink] 17 Aug 2013, 12:37

Go to page    1   2   3   4    Next  [ 63 posts ]

Similar topics Replies Last post
Similar
Topics:
3 If 7^n is a factor of the product of the integers from 1 to 100, inclu 4 26 Apr 2016, 01:01
13 If 3^k is a divisor of the product of all even integers between 2 5 07 Feb 2016, 18:08
23 If M is the set of all consecutive multiples of 9 between 100 and 500, 15 17 Sep 2015, 01:27
10 If N is the product of all multiples of 10 between 199 and 301, what i 7 26 Aug 2015, 12:23
4 D is the set of all the multiples of 3 between 20 and 100. E is the se 3 09 Jul 2015, 23:58
Display posts from previous: Sort by