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If N is the product of all multiples of 3 between 1 and 100, what is 17 Jul 2014, 06:35
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hamzakb
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rafi
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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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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If N is the product of all multiples of 3 between 1 and 100, what is 18 May 2015, 16:46
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reto
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
Show more
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, what is 23 May 2015, 02:20
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VeritasPrepKarishma
rafi
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.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.
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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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If N is the product of all multiples of 3 between 1 and 100, what is 23 May 2015, 03:24
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reto
VeritasPrepKarishma
rafi
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.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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!
Show more

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, what is 12 Nov 2016, 06:38
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rafi
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
Show more

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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If N is the product of all multiples of 3 between 1 and 100, what is 17 Mar 2024, 04:01
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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: https://gmatclub.com/forum/number-prope ... 84770.html
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­What is the meaning of a restrictive factor?
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If N is the product of all multiples of 3 between 1 and 100, what is 17 Mar 2024, 05:34
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Akriti_Khetawat

abmyers
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: https://gmatclub.com/forum/number-prope ... 84770.html
­What is the meaning of a restrictive factor?
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­
The restrictive factor among 2 and 5 would be the one with the lowest power in 3*6*9*12*15*...*99. Since we need the power of 10, which is equal to 2*5, the factor with the lowest power in 3*6*9*12*15*...*99 will determine the power of 10 in 3*6*9*12*15*...*99. As 5 appears with a lower power than 2 in 3*6*9*12*15*...*99, then 5 will be the restrictive factor, also known as the limiting factor, and its power, 7, would be the power of 10.
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If N is the product of all multiples of 3 between 1 and 100, what is 15 May 2024, 11:33
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rafi
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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­What i did

3∗6∗9∗...∗99=333∗(1∗2∗3∗4∗...∗32∗33)=3^33∗33!
Hence, you can just divide 33!/5= 6 33!/5^2= 33!/25=1   hence 6+1=7 and obviously there will be enough 2 which will make 5^7*2^7 as 10^7

hence C
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If N is the product of all multiples of 3 between 1 and 100, what is 21 May 2025, 23:21
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Given: N is the product of all multiples of 3 between 1 and 100.

To Find: The greatest integer m such that \(N/10^m\) is an integer.
  • Maximum value of m such that \(10^m\) divides N.
  • Since 10 = 2 × 5, we must find how many pairs of (2, 5) are in the prime factorization of N.

Solution:
Step 1: Express N
  • Multiples of 3: 3, 6, 9, ..., 99.
  • This is an arithmetic sequence going from 3(1) to 3(33). So, it has a total of 33 terms.
  • So, N =\( 3^{33}\)⋅(1⋅2⋅3⋅...⋅33) ----- [Taking 3 common from each multiple of 3.]
    • = \(3^{33}\)⋅33!

Step 2: Count how many times 2 and 5 appear in N
  • We already have N = \(3^{33}\) ⋅ 33!
    • We ignore the \(3^{33}\) because it doesn't contribute to powers of 2 or 5.
    • Now, observe that we only need to count the number of 5s in 33!. [Since any factorial has more 2s than 5s, so, number of pairs of 2 and 5 is determined by the limiting factor — the number of 5s.]
      • Power of 5 in 33!: The fastest way to find this is to repeatedly divide 33 by 5 and add all the quotients we get (till quotient > 0).
      • 6 + 1 = 7
  • So, number of times 10 = 2⋅5 divides N = 7

Correct Answer: C

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