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

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

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New post 18 Aug 2014, 11:14
apsForGmat wrote:
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 :!:


Hi Bunuel,

I used following approach to solve this question. Please advise if the approach and my assumption are correct :

Product of all multiples of 3 between 1 and 100 = 3*6*9*12...*99
This can be reduced to 3^33 * 33!
As 3^33 will never contribute a 0 to the result (I think. Please confirm), We can just go ahead by calculating the number of trailing zeroes in 33!
# of trailing 0s in 33! = 7 and hence 7 is the answer.
Please confirm if this is the right approach.


That's correct. Check here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1383974
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If N is the product of all multiples of 3 between 1 and 100, what is t [#permalink]

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New post 18 May 2015, 16:46
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reto 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

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

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New post 19 May 2015, 14:14
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 :!:


By Understanding the question well.

This question basically asks number of zeroes in N. Now N is a product of multiples of 3 upto 100---> 3.6.9...99
Now zeroes come from a combination of 2 and 5. In this series we will have more 2's (every second term in the series). So first we need to count combination of 3 with 5's in the series. Since 3 and 5 are mutually co-prime, simply count the multiples of 15 first (=3*5) from 1 to 100 which are 6
Now consider 5^2 = 25 which will yield extra 0's with 3. We get just a single multiple of 3 and 25 in the given range---> 75. 1 more 0.
Total we have 6+1 = 7 zeroes. Option C.

So actually the question was just asking how many multiples of 3*5, 3*25 ... and so on exist from 1 to 100

ALSO, you could have done the question using prime factorisation and applying a simple formula for counting the number of zeroes in factorials.

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

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New post 20 May 2015, 11:27
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 :!:


I found an explaination from The Economist GMAT Tutor. Just for you to read, if interested.

1) First step: understand what the question is asking:

Break down N: N=3⋅6⋅9⋅...30⋅33..60⋅63..90⋅93....⋅96⋅99 - all the multiples of 3 within the range.

[[continue]]

What is the greatest integer m? m is the powers of 10 in the denominator: the greater the m, the more powers of 10 we "stick" in the denominator. So why is m even limited? why can't m be 1,000, for example?

[[continue]]

The limitation is that N/10m needs to be an integer. So that all the powers of 10 in the denominator must be reduced by "10s" in the numerator N. The question is effectively asking "N can be divided by how many 10s?"

2) Don't make the mistake of focusing only on the multiples of 10 in N (30, 60, 90). Remember that a 10 can also be 'constructed' from the building blocks of 2·5 - for example, 15 and 6 (both included in N) multiplied together equal 90, which is another power of 10 than can eliminate a 10 on the denominator. You could theoretically count all of the different ways of making a 10 in N, but that will take too long, and you might miss something.

Try a more systematic approach:

[[continue]]

Break down 10 to 2⋅5. So the question now becomes: "how many pairs of 2s and 5s are there in N?"

[[continue]]

3) Further simplify the question. Which do you have more, the 5s or the 2s? Note that there are many more 2-factors than 5-factors (every even number in N will have a 2), so there is no need to count all the 2s - the 5s (that are also multiples of 3) are scarce, so count only them. You can now count how many multiples of 5 are there in N, and know that for every 5 you will have a 2 to pair up and make a 10. The question now becomes: "how many 5s are there in N?"

[[continue]]

Since N is composed of multiples of 3, you are effectively looking for multiples of both 5 and 3: The multiples of 5 that are also multiples of 3 are basically multiples of 5·3=15. There are 6 multiples of 15 between 1 and a 100: 15, 30, 45, 60, 75, 90.

Focus on those, and count how many powers of 5 are in them.
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Last edited by reto on 23 May 2015, 02:21, edited 1 time in total.

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

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New post 23 May 2015, 02:20
VeritasPrepKarishma wrote:
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.

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

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New post 23 May 2015, 03:24
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reto wrote:
VeritasPrepKarishma wrote:
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.

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.

Multiples of 3:

3 * 6 * 9 * 12 * ... * 96 * 99

3 = 3*1
6 = 3*2
9 = 3*3
...
96 = 3*32
99 = 3*33

So in all, we have 33 multiples of 3.

(3*1) * (3*2) * (3*3) * (3*4) * ... * (3*32) * (3*33)

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

= (3*3*3..*3) * (1 * 2 * 3 * 4 * ... * 32 * 33)

= 3^(33) * (1 * 2 * 3 * 4 * ... * 32 * 33)

But 33! = (1 * 2 * 3 * 4 * ... * 32 * 33)

So you get 3^(33) * 33!
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If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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New post 08 Aug 2015, 02:48
I am reviewing this question and even though I understand now 99% of it, I would like complete my knowledge to 100%.

I found the explanation by Economist:

***********************************************************************************************************************************************
First step: understand what the question is asking: Break down N: N=3⋅6⋅9⋅...30⋅33..60⋅63..90⋅93....⋅96⋅99 - all the multiples of 3 within the range.

What is the greatest integer m? m is the powers of 10 in the denominator: the greater the m, the more powers of 10 we "stick" in the denominator. So why is m even limited? why can't m be 1,000, for example?

The limitation is that N/10m needs to be an integer. So that all the powers of 10 in the denominator must be reduced by "10s" in the numerator N. The question is effectively asking "N can be divided by how many 10s?"

Don't make the mistake of focusing only on the multiples of 10 in N (30, 60, 90). Remember that a 10 can also be 'constructed' from the building blocks of 2·5 - for example, 15 and 6 (both included in N) multiplied together equal 90, which is another power of 10 than can eliminate a 10 on the denominator. You could theoretically count all of the different ways of making a 10 in N, but that will take too long, and you might miss something.

Try a more systematic approach: Break down 10 to 2⋅5. So the question now becomes: "how many pairs of 2s and 5s are there in N?"

Further simplify the question. Which do you have more, the 5s or the 2s? Note that there are many more 2-factors than 5-factors (every even number in N will have a 2), so there is no need to count all the 2s - the 5s (that are also multiples of 3) are scarce, so count only them. You can now count how many multiples of 5 are there in N, and know that for every 5 you will have a 2 to pair up and make a 10. The question now becomes: "how many 5s are there in N?"

Since N is composed of multiples of 3, you are effectively looking for multiples of both 5 and 3: The multiples of 5 that are also multiples of 3 are basically multiples of 5·3=15. There are 6 multiples of 15 between 1 and a 100: 15, 30, 45, 60, 75, 90.

Focus on those, and count how many powers of 5 are in them. Count the number of 5s: 15, 30, 45, 60, and 90 each have one '5'. However, 75=3⋅25 = 3⋅52 has two building blocks of 5. This gives you a total of seven '5' building blocks, to which you will be able to find a '2' from some other factor on N, so N can be divided by 10 seven times at most.

***********************************************************************************************************************************************

In my view, this approach as described above is time consumming. I like the approach which VeritasPrepKarishma and abmyers have pointed out. They seem to have almost the same approach with a slight difference:

abmyers approach (quote), my comment in blue color:

"N = The product of the sequence of 3*6*9*12....*99" (so far so good, everything is understood)

"N therefore is also equal to 3* (1*2*3*.....*33)" (ok we just simplify by taking out the common factor of 3, understood!)

"Therefore N = 3* 33!" (still makes sense)

No lets shift to VeritasPrepKarishma's approach:

"Once you are done, note that this question can be easily broken down into the factorial form." (Yes, I like that approach, go on!)

"\(3∗6∗9∗...∗99=3^{33}∗(1∗2∗3∗4∗...∗32∗33)=3^{33}∗33!\)"

Wait a second, what about this difference here? 3^33*33! > Could you please explain? From my logical point of view, the approach from abmyers calculation was right by just factoring out 3, but why we have 3^33 here in your solution? I am sure there is a very logical reasoning behind and I would love to see that explanation. Thank you very much :)

While I was writing this, I think it just suddently dawned on me :-) We need to have \(3^{33}\) because if we factor out 3 from the product of the sequence (3*6*9*12....*99), we do that for 3, for 6, for 9 .... therefore in total we do this 99-3/2 = 32 + 1 times!

Maybe you can just confirm this?

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

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New post 22 Aug 2015, 04:02
Bunuel wrote:
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.

Answer: C.

Check this for more:
everything-about-factorials-on-the-gmat-85592.html

Hope it helps.



N = 3^33 (33!) then we must find number of 5 in 33!

33/5 + 33/25 + ... = 6 + 1 + 0... = 7

answer is C

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

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New post 22 Aug 2015, 04:48
Expert's post
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This post was
BOOKMARKED
jimwild wrote:
Bunuel wrote:
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.

Answer: C.

Check this for more:
everything-about-factorials-on-the-gmat-85592.html

Hope it helps.



N = 3^33 (33!) then we must find number of 5 in 33!

33/5 + 33/25 + ... = 6 + 1 + 0... = 7

answer is C


That's correct. Check here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1383974
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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New post 22 Aug 2015, 05:15
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 :!:

There is a short cut
N=3^33 *33!
Now you have to find how many 10s are there in 33!
10=2*5; since 5>2, divide 33 successively by 5 and add all the quotients
33/5 = 6
6/5 = 1
6+1 = 7 is the answer.

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

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New post 23 Aug 2015, 23:49
reto wrote:
"\(3∗6∗9∗...∗99=3^{33}∗(1∗2∗3∗4∗...∗32∗33)=3^{33}∗33!\)"

Wait a second, what about this difference here? 3^33*33! > Could you please explain? From my logical point of view, the approach from abmyers calculation was right by just factoring out 3, but why we have 3^33 here in your solution? I am sure there is a very logical reasoning behind and I would love to see that explanation. Thank you very much :)

While I was writing this, I think it just suddently dawned on me :-) We need to have \(3^{33}\) because if we factor out 3 from the product of the sequence (3*6*9*12....*99), we do that for 3, for 6, for 9 .... therefore in total we do this 99-3/2 = 32 + 1 times!

Maybe you can just confirm this?

Thank you and happy weekend


Yes, that's correct. We do this (99 - 3)/3 + 1 = 33 times. That is how you get \(3^{33}\).
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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New post 07 Nov 2016, 11:31
We need to figure out how many times 3 goes into 101 (i.e. 100-1+1=101) --> Answer: 33 times

So we know we're going to see 33 multiples of 3, thus 3(1x2x3x...x33) = 3x33! =n

What do we know about factorials? Bunuel gave us a nice formula a while back for seeing the max exponential value of a number we can get out of a factorial. Let's use it! --> Equation will look like this now [3x33!]/10^m --> applying formula for factorials --> 33/5 + 33/25 = 6+1 = 7. We used 5 because the number of 5's that are at our disposal in this factorial will significantly limit the number of 10's we can create (because 5 is a prime number). Thus 7 is our answer.

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

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New post 10 Nov 2016, 12:26
This is such an amazing Question.
Here is what i did=>
A ten is formed via a 2 and a 5
Clearly number of 2's are sufficient.
so we need to just count the number of 5's in the product => 3*6*9*......99
Hmm
Lets count
15=> one
30=> one
45=> one
60=> one
75=> two
90=> one
Total => Seven
Hence C.

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

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New post 10 Nov 2016, 12:42
So to have multiples of 10 you need a 2 and a 5. Lets start by listing out some multiples of 3

3,6,9,12,15
there are way more two factors than 5 so we should find the factors of 5.

To find the factors of 3 we have 3-99
99-3=96/3=32+1=33

33/5=6.something

So we have 6 factors of 5 that are also multiples of 3.
15,30,45,60,75,90.

All of these have one 5 except for 75 which has two. There are 7 5s and thus the answer is 7

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

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New post 10 Nov 2016, 12:43
So to have multiples of 10 you need a 2 and a 5. Lets start by listing out some multiples of 3

3,6,9,12,15
there are way more two factors than 5 so we should find the factors of 5.

To find the factors of 3 we have 3-99
99-3=96/3=32+1=33

33/5=6.something

So we have 6 factors of 5 that are also multiples of 3.
15,30,45,60,75,90.

All of these have one 5 except for 75 which has two. There are 7 5s and thus the answer is 7

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

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New post 12 Nov 2016, 06:38
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


We can rewrite our question as:

What is the greatest integer m for which N/(2^m x 5^m) is an integer? So, in order for 10^m to divide into N, we need m factors of 2 and m factors of 5. Since we know there are fewer factors of 5s than 2s within the multiples of 3 between 1 and 100, let’s determine the number of factors of 5s within the multiples of 3 between 1 and 100.

3 x 5 = 15 (1 factor of 5)

3 x 10 = 30 (1 factor of 5)

3 x 15 = 45 (1 factor of 5)

3 x 20 = 60 (1 factor of 5)

3 x 25 = 75 (2 factors of 5)

3 x 30 = 90 (1 factor of 5)

Thus, there are 7 factors of 5 within the multiples of 3 between 1 and 100, and thus the maximum value of m is 7.

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

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New post 22 Jan 2017, 08:58
1)To find the number of multiples of 10 in the product of all multiples of 3 between 1 and 100 we need to find the number of pairs of prime factors 2 and 5.
2)The number of multiples of 5&3 is (90-15)/15+1=75/15+1=6. The number of multiples of 5*5*3 is 1 (it is 75). 6+1=7
3)There are more multiples of 2 than multiples of 5, but to check it, we can use the same formula as in the previous step. For 2*3=6: (96-6)/6+1=16 (is already more than 7) + 2*2*3=12: (96-12)/12+1=8 + 2*2*2*3=24: (72-24)/24+1=3 +2*2*2*2*3=48: (96-48)/48+1=2 +2*2*2*2*2*3=96 (96-96)/96+1=1. The total number of factors of 2 within all the multiples that lead up to product N is 16+12+8+3+2+1=42
4)Consequently the number of multiples of 10 is equal to the number of multiples of 5 which is 7

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

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New post 26 Jan 2017, 16:10
1) To paraphrase the question - we need to find out how many sets of prime factors 2*5 there are in the product of all the integers between 1 and 100.
2) Since there are more prime factors 2 than prime factors 5 in this set of consecutive integers. we only need to find the number of prime factors 5.
3) Since only multiples of 3 make us the numerator, we need to find all the numbers that are both multiples of 5 and multiples of 3. These are multiples of 15.
4) 100/15=6
5) We also need to find the multiples that contain two prime factors 5: 3*5*5=75. 100/75=1
6) 6+1=7

The correct answer is C.

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

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New post 28 Feb 2017, 11:35
N = 3*6*9*12*.....99.
N= 3(1*2*3*4*....33).
N= 3* 33!
so,
no. of trailing zeros in 33! is 33/5 +33/25.
this gives 7. so C is answer.
As value of m must be 7 to have N/10^m to integer.

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

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New post 12 Jul 2017, 17:51
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 :!:


In order to solve this question we first need to know how many multiples of 3 are in between 1 and 100

(largest multiple - smallest multiple)/ (multiple) +1
(99-3)/3 + 1 =33

N = 3 x 6 x 9 ... 99- but we can express this differently

If we look at even the product of all multiples of 3 from 3 to 15

3 x 3(2) x 3(3) x 3(4) x 3(5) x = 3^5 x 5!

In this case we would have 3^33 x 33!

Just find the number of trailing zeros in 33!-

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

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Re: If N is the product of all multiples of 3 between 1 and 100   [#permalink] 12 Jul 2017, 17:51

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