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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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21 Aug 2017, 16:18
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. Answer: C. Check this for more: http://gmatclub.com/forum/everythingab ... 85592.htmlHope it helps. hi maybe it is very obvious ....but can you please tell me why should we assume that there are at least as many 2s as are 5s in N! ...? thanks in advance ..
Last edited by gmatcracker2018 on 21 Aug 2017, 16:28, edited 1 time in total.



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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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21 Aug 2017, 19:03
VeritasPrepKarishma 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 :!: 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.
Last edited by gmatcracker2018 on 16 Nov 2017, 02:39, edited 1 time in total.



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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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14 Nov 2017, 23:13
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. Answer: C. Check this for more: http://gmatclub.com/forum/everythingab ... 85592.htmlHope it helps. Here is a somewhat similar way if you know how to find trailing zeros in factorial of a number. N =3*6*9*12*15...99 N=3(1*2*3*4*...*33) N=3*33! Now number of zeros in 33! is floor(33/5)+floor(33/25) =6+1=7. So number of trailing zeros in 3*33! will be 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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16 Nov 2017, 02:18
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PBad 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. Answer: C. Check this for more: http://gmatclub.com/forum/everythingab ... 85592.htmlHope it helps. Here is a somewhat similar way if you know how to find trailing zeros in factorial of a number. N =3*6*9*12*15...99 N=3(1*2*3*4*...*33) N=3*33! Now number of zeros in 33! is floor(33/5)+floor(33/25) =6+1=7. So number of trailing zeros in 3*33! will be 7. hi at first I also tried the problem this way, but later I realize this is not the correct approach to solving the problem say, 3 + 6 + 9 = 18 here you can take 3 as a common term so, 3 (1+ 2 + 3) = 18, that's okay but, 3 * 6 * 9 = 162 now, if you work it out as before, you will get 3 ( 1 * 2 * 3) = 18, as you can see, that's not okay thus, you can take any term common out of an expression, when the expression is in addition or subtraction form, not when the expression is in multiplication form So, please see the solution provided by Bunuel or Karishma thanks cheers through the kudos button if this 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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16 Nov 2017, 02:33
gmatcracker2017 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. Answer: C. Check this for more: http://gmatclub.com/forum/everythingab ... 85592.htmlHope it helps. hi maybe it is very obvious ....but can you please tell me why should we assume that there are at least as many 2s as are 5s in N! ...? thanks in advance .. hi that's okay, I have got it When we write, 1 *2 * 3 * 4 * 5, two comes well ahead of 5s. Thus, there could be more 2s than 5s, so 5 is a limiting factor here. So, finding the factors of 5 in any multiple is equivalent to finding the factors of 10 thanks, man



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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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16 Nov 2017, 02:44
gmatcracker2017 wrote: VeritasPrepKarishma 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 :!: 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. hi mam could you please speak few more words on below...? 3 * 6 * 9....99 = 3^ 33 * ( 1 * 2 * 3 * 4 * .....* 32 * 33) = 3^33 * 33! thanks in advance, mam I guess in the series, 3 * 6 * 9 * 12 * 15 * 18 * 21* 28, every number is a multiple of 3 So taking 3 out of each number, we get 3^8 now, 3^8 [1, {because 3 has already been extracted from (3 * 1)} * 2 {because 3 has already been extracted from (3 * 2)} * 3 {because 3 has already been extracted from (3 * 3)} * 4 {because 3 has already been extracted from (3 * 4)} * 5 {because 3 has already been extracted from (3 * 5)} * 6 {because 3 has already been extracted from (3 * 6)} * 7 {because 3 has already been extracted from (3 * 7)} * 8 {because 3 has already been extracted from (3 * 8)} ] So, 3 ^8 * 8! Is that okay, mam..?



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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03 Dec 2017, 15:39
I thought of this one, as I was going by dividing by 5's to get an idea. If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which N/10^m is an integer? A. 3 B. 6 C. 7 D. 8 E. 10 We know that we need to consider numbers which are multiples of 3. So dividing 100/3= approx 33 numbers are multiples of 3 (1/3 of numbers and a bit more, but we don't need to consider the bits left). 33/5 = 6 times (remainder we don't need to consider) > 6 zeroes. 6/5 = 1 (discard remainder) = 1 more zero 6 + 1 = 7 zeroes. Kudos, please if it made sense to you! Also, I request any correction of the concepts applied! Thank you



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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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28 Dec 2017, 06:38
To answer the question, we must count the number of 5's in the product of multiples of 3 between 1 and 100
5 will occur for mulitple of 15 LCM(3,5) between 1 and 100 => 6 multiples of 15 => prod has \(5^6\) 5 will also occur for multiple of 75 LCM(3,25) between 1 and 100 => 1 multiple of 75 => prod has one more 5
so totally, product has \(5^7\) = (\(5^6 * 5\)). Ofcourse, number of 2s will be definitely greater than number of 5s, but to form a 10, we need 2 and 5, so number of 10s is restricted by number of 5s
so maximum m, for which, prod / \(10^m\) is an integer is 7
Answer (C)
If the question had been, prod of multiples of 3 between 1 and 200
then, 5 will occur for mulitple of 15 LCM(3,5) between 1 and 200 => 13 multiples of 15 => product has \(5^{13}\) 5 will also occur for multiple of 75 LCM(3,25) between 1 and 200 => 2 multiples of 75 => product has another \(5^2\) 5 will also occur for multiple of 375 LCM(3,125) between 1 and 200 => has 0 multiples of 375 5 will also occur for multiple of 1875 LCM(3,625) between 1 and 200 => has 0 multiples of 1875 . . goes on




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