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

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If N is the product of all multiples of 3 between 1 and 100 [#permalink]  17 Sep 2010, 02:14
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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

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Re: Product of sequence [#permalink]  17 Sep 2010, 02:29
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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.
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Re: Product of sequence [#permalink]  15 Feb 2011, 19:36
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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: number-properties-from-gmatprep-84770.html
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Re: Product of sequence [#permalink]  15 Feb 2011, 19:36
2
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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: number-properties-from-gmatprep-84770.html
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]  16 Nov 2012, 06:58
2
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I did it in a different way..... since it is multiplication of all 3 multiples....
3*6*9*..... *99=3(1*2*3*4*5*......33)=3*33!
3 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.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]  04 Jun 2013, 04:48
2
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Expert's post
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.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 25239 Followers: 3429 Kudos [?]: 25228 [1] , given: 2702 Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] 31 Dec 2012, 02:34 1 This post received 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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 25239 Followers: 3429 Kudos [?]: 25228 [1] , given: 2702 Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] 22 Aug 2013, 02:53 1 This post received KUDOS 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. How did you come up with this?Please help " 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 Similar questions to practice: if-n-is-the-greatest-positive-integer-for-which-2n-is-a-fact-144694.html what-is-the-largest-power-of-3-contained-in-103525.html if-n-is-the-product-of-all-positive-integers-less-than-103218.html if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html if-p-is-the-product-of-integers-from-1-to-30-inclusive-137721.html what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html if-6-y-is-a-factor-of-10-2-what-is-the-greatest-possible-129353.html if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html if-p-is-a-natural-number-and-p-ends-with-y-trailing-zeros-108251.html if-73-has-16-zeroes-at-the-end-how-many-zeroes-will-147353.html find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html how-many-zeros-are-the-end-of-142479.html how-many-zeros-does-100-end-with-100599.html find-the-number-of-trailing-zeros-in-the-product-of-108248.html if-60-is-written-out-as-an-integer-with-how-many-consecuti-97597.html if-n-is-a-positive-integer-and-10-n-is-a-factor-of-m-what-153375.html if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html Hope it helps. _________________ Math Expert Joined: 02 Sep 2009 Posts: 25239 Followers: 3429 Kudos [?]: 25228 [1] , given: 2702 Re: Number properties task, please, help! [#permalink] 25 Aug 2013, 10:29 1 This post received KUDOS 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. _________________ Moderator Affiliations: GMAT Club Joined: 21 Feb 2012 Posts: 1164 Location: India City: Pune GPA: 3.4 WE: Business Development (Manufacturing) Followers: 138 Kudos [?]: 872 [1] , given: 843 Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] 25 Sep 2013, 12:34 1 This post received KUDOS 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! _________________ Be the coolest guy in the MBA Forum - Be a threadmaster! Have a blog? Feature it on GMAT Club! All MBA Resources All 'Sticky' Topics at one place Please share your profiles for this application season: 2015 Profiles w/ Admit/Dings Results! Get the best GMAT Prep Resources with GMAT Club Premium Membership Next Generation GMATClub CATS with Brilliant Analytics. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4694 Location: Pune, India Followers: 1090 Kudos [?]: 4888 [1] , given: 163 Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] 25 Sep 2013, 20:23 1 This post received KUDOS 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Product of sequence [#permalink]  20 Oct 2010, 03:09
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.
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Re: Product of sequence [#permalink]  20 Oct 2010, 03:14
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.
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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?
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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!
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Re: Product of sequence [#permalink]  14 Feb 2011, 05:59
Thanks Brunel! I didnt consider that 75 has 2 factors of 5 thus adding 1 more to m.
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Re: Product of sequence [#permalink]  15 Nov 2012, 11:10
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?
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Re: Product of sequence [#permalink]  16 Nov 2012, 02:32
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.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]  16 Nov 2012, 07:47
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
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Re: If N is the product of all multiples of 3 between 1 and 100   [#permalink] 16 Nov 2012, 07:47
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