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

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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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New post 17 Jul 2014, 06:17
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.


Hope it helps.


I found my answer by finding the number of multiples of 3 between 1 and 100 i.e 100/3 = 33.

Then I found the number of trailing zeroes in 33! = 7

so 10^7 can be the maximum for N/10^m to remain an integer.


Am I just lucky or can this also be a method of solving?
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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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New post 17 Jul 2014, 06:35
hamzakb 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.


Hope it helps.


I found my answer by finding the number of multiples of 3 between 1 and 100 i.e 100/3 = 33.

Then I found the number of trailing zeroes in 33! = 7

so 10^7 can be the maximum for N/10^m to remain an integer.


Am I just lucky or can this also be a method of solving?


If you solve this way it should be:
N = 3*6*9*12*15*...*99 = 3^33(1*2*3*...*33) = 3^33*33!.

The number of trailing zeros for 33! is 33/5 + 33/25 = 6 + 1 = 7.

Check Trailing Zeros and Power of a number in a factorial questions in our Special Questions Directory.

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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New post 17 Jul 2014, 06:37
abmyers wrote:
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


Red part above are not correct.

Should be: \(N = 3*6*9*12*15*...*99 = 3^{33}(1*2*3*...*33) = 3^{33}*33!\).
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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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New post 27 Jul 2014, 06:39
Please correct me if I am wrong.

After reading Bunuel's thread on trailing zeroes
I simply calculated m (= amount of trailing zeroes) this way:

\(\frac{100}{3*5}+\frac{100}{3* 25}= 6 + 1 = 7\)

dividing by 3*5 and 3*25 one ensures that only multiples of 3 are taken into consideration!
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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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New post 27 Jul 2014, 15:30
JK13 wrote:
Please correct me if I am wrong.

After reading Bunuel's thread on trailing zeroes
I simply calculated m (= amount of trailing zeroes) this way:

\(\frac{100}{3*5}+\frac{100}{3* 25}= 6 + 1 = 7\)

dividing by 3*5 and 3*25 one ensures that only multiples of 3 are taken into consideration!


No, that's not correct. N = 3*6*9*12*15*...*99 = 3^33*33!, not 100.

Check Trailing Zeros and Power of a number in a factorial questions in our Special Questions Directory.

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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New post 18 May 2015, 16:46
2
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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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#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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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#permalink]

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New post Updated on: 23 May 2015, 02:21
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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Originally posted by reto on 20 May 2015, 11:27.
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, what is  [#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, what is  [#permalink]

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New post 23 May 2015, 03:24
1
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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Re: If N is the product of all multiples of 3 between 1 and 100, what is  [#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, what is  [#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, what is  [#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, what is  [#permalink]

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New post 03 Dec 2017, 16:39
I thought of this one, as I was going by dividing by 5's to get an idea.

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which N/10^m is an integer?

A. 3
B. 6
C. 7
D. 8
E. 10

We know that we need to consider numbers which are multiples of 3. So dividing 100/3= approx 33 numbers are multiples of 3 (1/3 of numbers and a bit more, but we don't need to consider the bits left).

33/5 = 6 times (remainder we don't need to consider) --> 6 zeroes.

6/5 = 1 (discard remainder) = 1 more zero

6 + 1 = 7 zeroes.

Kudos, please if it made sense to you! Also, I request any correction of the concepts applied! Thank you :)
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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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New post 28 Dec 2017, 07:38
To answer the question, we must count the number of 5's in the product of multiples of 3 between 1 and 100

5 will occur for mulitple of 15 LCM(3,5) between 1 and 100 => 6 multiples of 15 => prod has \(5^6\)
5 will also occur for multiple of 75 LCM(3,25) between 1 and 100 => 1 multiple of 75 => prod has one more 5

so totally, product has \(5^7\) = (\(5^6 * 5\)).
Ofcourse, number of 2s will be definitely greater than number of 5s, but to form a 10, we need 2 and 5, so number of 10s is restricted by number of 5s

so maximum m, for which, prod / \(10^m\) is an integer is 7

Answer (C)

If the question had been, prod of multiples of 3 between 1 and 200

then,
5 will occur for mulitple of 15 LCM(3,5) between 1 and 200 => 13 multiples of 15 => product has \(5^{13}\)
5 will also occur for multiple of 75 LCM(3,25) between 1 and 200 => 2 multiples of 75 => product has another \(5^2\)
5 will also occur for multiple of 375 LCM(3,125) between 1 and 200 => has 0 multiples of 375
5 will also occur for multiple of 1875 LCM(3,625) between 1 and 200 => has 0 multiples of 1875
.
.
goes on
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Re: If N is the product of all multiples of 3 between 1 and 100, what is &nbs [#permalink] 28 Dec 2017, 07:38

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