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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]

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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

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[Reveal] Spoiler: OA

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Re: Product of sequence [#permalink]

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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.

Answer: C.

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

Hope it helps.
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Re: Product of sequence [#permalink]

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New post 20 Oct 2010, 04:09
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]

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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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Re: Number properties task, please, help! [#permalink]

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New post 14 Feb 2011, 06:11
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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Re: Number properties task, please, help! [#permalink]

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New post 14 Feb 2011, 06:17
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?

Answers:
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: Number properties task, please, help! [#permalink]

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Merging similar topics.


Also check:
number-property-108971.html
trailing-zeros-question-complicated-one-108249.html
trailing-zeros-question-108248.html
trailing-zeros-question-logical-approach-needed-108251.html
gmat-club-m12-100599.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
facorial-ps-105746.html
can-you-take-this-challenge-700-quant-103525.html
ps-103218.html
least-value-of-n-m09q33-76716.html
ds-product-of-first-30-positive-integers-50292.html
anything-wrong-in-this-problem-can-anyone-dare-to-solve-98777.html
hard-tricky-question-97597.html

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Re: Product of sequence [#permalink]

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New post 14 Feb 2011, 06:59
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]

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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]

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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]

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New post 15 Nov 2012, 12:10
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]

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

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I did it in a different way..... since it is multiplication of all 3 multiples....
3*6*9*..... *99=(3^33)(1*2*3*4*5*......33)=(3^33)*33! (3 power 33 because a 3 can be extracted from each number inside)
(3^33) 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.

Last edited by Amateur on 23 Jan 2015, 09:04, edited 2 times 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 16 Nov 2012, 08:47
i got answer as '6'
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]

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New post 30 Dec 2012, 19:21
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

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

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

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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 :!:


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

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I think the easiest way to do it is to count the number of 5's from 1 to 33.
3^ 33 ( 1 x 2x 3...... 33)

5 factors

5 - 5x1
10- 5x2
15- 5x3
20 - 5x4
25 - 5x5
30 - 5x6

Therefore 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 17 Aug 2013, 12:37
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

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stunning math.

my solution:
3.6.9.12..............................99 = 3^33 . (1.2.3.4.5...............................33)
factors that can bring zeros are = 2,5,10,12,15,20,22,25,30,32 (All from 2 , 0 and 5 )
But only 25 can produce two 5.

so, 2*5 , 10, 12*15, 20, 22*5, 30, 32*5 = 7 zeros.

so m=7 we can afford at most. 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 Aug 2013, 03:44
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;

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Re: If N is the product of all multiples of 3 between 1 and 100   [#permalink] 22 Aug 2013, 03:44

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