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

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Math Expert
Joined: 02 Sep 2009
Posts: 44412
Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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22 Aug 2013, 03:53
1
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Expert's post
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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Hope it helps.
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25 Aug 2013, 08:14
I have spent most of the day on this topic and questions.

I am still feeling little apprehensive about the time these consume and the difficulty of these questions.

Any suggestions how to gain confidence in this area?

Bunuel wrote:
Math Expert
Joined: 02 Sep 2009
Posts: 44412

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25 Aug 2013, 10:50
pavan2185 wrote:
I have spent most of the day on this topic and questions.

I am still feeling little apprehensive about the time these consume and the difficulty of these questions.

Any suggestions how to gain confidence in this area?

Bunuel wrote:

Can you please tell what do you find most challenging in them? Thank you.

Check other similar questions here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1259389
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25 Aug 2013, 11:23
Bunuel wrote:
Can you please tell what do you find most challenging in them? Thank you.

Check other similar questions here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1259389

I understand the basic concept you explained in the mathbook by Gmatclub and various explanations you have given,but I am finding it difficult to apply on hard questions that involve multiple factorilas and questions that do not specifically give any fcatorial but give a complex product of numbers.
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Joined: 02 Sep 2009
Posts: 44412

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25 Aug 2013, 11:29
1
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Expert's post
pavan2185 wrote:
Bunuel wrote:
Can you please tell what do you find most challenging in them? Thank you.

Check other similar questions here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1259389

I understand the basic concept you explained in the mathbook by Gmatclub and various explanations you have given,but I am finding it difficult to apply on hard questions that involve multiple factorilas and questions that do not specifically give any fcatorial but give a complex product of numbers.

In that case, I must say that practice should help.
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25 Aug 2013, 11:34
Thank you!!! You have always been helpful.

Yes, I hope that will help. I will try to practise every available question on this topic.he

Thanks for your outstanding work on GC math forums.

Bunuel wrote:
pavan2185 wrote:
Bunuel wrote:
Can you please tell what do you find most challenging in them? Thank you.

Check other similar questions here: if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187-20.html#p1259389

I understand the basic concept you explained in the mathbook by Gmatclub and various explanations you have given,but I am finding it difficult to apply on hard questions that involve multiple factorilas and questions that do not specifically give any fcatorial but give a complex product of numbers.

In that case, I must say that practice should help.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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25 Sep 2013, 12:18
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 [#permalink]

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25 Sep 2013, 13:34
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Expert's post
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 [#permalink]

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25 Sep 2013, 21:23
1
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Expert's post
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 [#permalink]

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25 Sep 2013, 23:35
1
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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.\

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

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30 Sep 2013, 12:07
N= 3*6*9*............99

N= 3 { 1*2*3*4*5*.........*10..........*15....*20...*25...*30.....*33}

The numbers of times 5 can appear in above product is

5=1
10=1
15=1
20=1
25=2
30=1

Total 7

So N/10^m => N/(5^m *2^m)

Thus m=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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19 Jan 2014, 08:40
Hello all, I am just wondering if the question was a bit different like this :" If N is the product of all multiples of 5 between 1 and 100, what is the greatest integer m for which $$\frac{N}{10^m}$$ is an integer?",

then we will have": $$N = {5^{33}} * {33!}$$. If so, then my answer would be 31. Please allow me to share my take on this.

Basically, I need to find how many 2s and how many 5s are there in N. I use Bunuel's formula and I got the followings:

Number of 2s in 33!: 33/2 + 33/4 + 33/8 + 33/16 + 33/32 = 31. (Since $${5^{33}}$$ does not have any 2)
Number of 5s in 33!: 33/5 + 33/25 = 7. However, I have $${5^{33}}$$, that leaves me with $$5^{40}$$ in N.

Since I need $${10^m}$$, I will need as many 5 AND 2 as possible in N. I have 31 pairs of 2 and 5. So, m = 31.

If someone please confirm my thought, it would be greatly appreciated! Thanks.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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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.

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 [#permalink]

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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.

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

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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 [#permalink]

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27 Jul 2014, 06:39
Please correct me if I am wrong.

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 [#permalink]

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27 Jul 2014, 15:30
JK13 wrote:
Please correct me if I am wrong.

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 [#permalink]

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28 Jul 2014, 02:06
Quote:
JK13 wrote:
Please correct me if I am wrong.

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.

I understand that N is not 100 in this question, when looking for ALL trailing zeros in a number (which here would ofc be 3^33*33!).
My approach was more based on finding the number of those trailing zeros in 100! that are a result of multiples of 3.
Isn't that then the same amount of trailing zeros, that we want to find in this question?

Another example could be: "Look for the trailing zeros in the product of all multiples of 7 between 1 and 100"
$$\frac{100}{7*5}+\frac{100}{7*25}= 2 + 0 = 2$$
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]

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13 Aug 2014, 05:39
Question seems to be very confusing at first.

I solved it like this:
3*6*9*12*15.........*96*99 = 3(1*2*3*4.........32*33) = 3*33!

Now number of training 0's will be = Number of trailing 0's in 33!
i.e. 33/5+33/25 = 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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18 Aug 2014, 11:11
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.
Re: If N is the product of all multiples of 3 between 1 and 100   [#permalink] 18 Aug 2014, 11:11

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