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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] New post 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 :?:
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[Reveal] Spoiler: OA

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Re: Product of sequence [#permalink] New post 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.

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] New post 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] New post 20 Oct 2010, 03:14
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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] New post 14 Feb 2011, 05:11
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] New post 14 Feb 2011, 05:17
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] New post 14 Feb 2011, 05:34
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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] New post 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] New post 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] New post 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] New post 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] New post 16 Nov 2012, 02:32
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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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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] New post 16 Nov 2012, 06:58
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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] New post 16 Nov 2012, 07:47
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] New post 30 Dec 2012, 18:21
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] New post 31 Dec 2012, 02:34
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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] New post 04 Jun 2013, 04:48
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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] New post 16 Jul 2013, 23:35
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink] New post 16 Aug 2013, 23:58
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] New post 17 Aug 2013, 11:37
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] 17 Aug 2013, 11:37
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