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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
 17 Sep 2010, 03:14
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63% (01:17) wrong based on 23 sessions
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 
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Re: Product of sequence [#permalink]
17 Sep 2010, 03:29
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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.htmlHope it helps.
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Re: Product of sequence [#permalink]
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]
20 Oct 2010, 04:14
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Number properties task, please, help! [#permalink]
14 Feb 2011, 05:37
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
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Re: Number properties task, please, help! [#permalink]
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]
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]
14 Feb 2011, 06:34
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Re: Product of sequence [#permalink]
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]
15 Feb 2011, 20:36
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, 20:36
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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 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]
16 Nov 2012, 03:32
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, 07:58
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]
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]
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]
31 Dec 2012, 03:34
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.
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Re: If N is the product of all multiples of 3 between 1 and 100
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31 Dec 2012, 03:34
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