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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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22 Aug 2013, 03:53



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Re: Number properties task, please, help! [#permalink]
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25 Aug 2013, 08:14
I have spent most of the day on this topic and questions. I am still feeling little apprehensive about the time these consume and the difficulty of these questions. Any suggestions how to gain confidence in this area? Bunuel wrote:



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Re: Number properties task, please, help! [#permalink]
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25 Aug 2013, 10:50



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Re: Number properties task, please, help! [#permalink]
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25 Aug 2013, 11:23
Bunuel wrote: Can you please tell what do you find most challenging in them? Thank you. Check other similar questions here: ifnistheproductofallmultiplesof3between1and10118720.html#p1259389I 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.



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Re: Number properties task, please, help! [#permalink]
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25 Aug 2013, 11:29



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Re: Number properties task, please, help! [#permalink]
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25 Aug 2013, 11:34
Thank you!!! You have always been helpful. Yes, I hope that will help. I will try to practise every available question on this topic.he Thanks for your outstanding work on GC math forums. Bunuel wrote: pavan2185 wrote: Bunuel wrote: Can you please tell what do you find most challenging in them? Thank you. Check other similar questions here: ifnistheproductofallmultiplesof3between1and10118720.html#p1259389I 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.



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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25 Sep 2013, 12:18
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?
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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25 Sep 2013, 13:34
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 nBack 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!
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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25 Sep 2013, 21:23
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.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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25 Sep 2013, 23:35
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We know that for a number to be divisible by 10 must have at least one zero. Let's break the 10 into its prime factors, ie. 5 and 2. Now, we need to find pairs of 2 and 5 in the numerator. Here, 5 is our limiting factor, as it appears less than 2 does. therefore two cont the number of 5s, we must count the 5s in all multiples of 3 between 1 and 100.
15= One 5 30= One 5 45= One 5 60= One 5 75 = Two 5s (5 x 5 x3=75) 90= One 5.\
Answer is 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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30 Sep 2013, 12:07
N= 3*6*9*............99 N= 3 { 1*2*3*4*5*.........*10..........*15....*20...*25...*30.....*33} The numbers of times 5 can appear in above product is 5=1 10=1 15=1 20=1 25=2 30=1 Total 7 So N/10^m => N/(5^m *2^m) Thus m=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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19 Jan 2014, 08:40
Hello all, I am just wondering if the question was a bit different like this :" If N is the product of all multiples of 5 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?",
then we will have": \(N = {5^{33}} * {33!}\). If so, then my answer would be 31. Please allow me to share my take on this.
Basically, I need to find how many 2s and how many 5s are there in N. I use Bunuel's formula and I got the followings:
Number of 2s in 33!: 33/2 + 33/4 + 33/8 + 33/16 + 33/32 = 31. (Since \({5^{33}}\) does not have any 2) Number of 5s in 33!: 33/5 + 33/25 = 7. However, I have \({5^{33}}\), that leaves me with \(5^{40}\) in N.
Since I need \({10^m}\), I will need as many 5 AND 2 as possible in N. I have 31 pairs of 2 and 5. So, m = 31.
If someone please confirm my thought, it would be greatly appreciated! Thanks.



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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17 Jul 2014, 06:17
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. Hope it helps. I found my answer by finding the number of multiples of 3 between 1 and 100 i.e 100/3 = 33. Then I found the number of trailing zeroes in 33! = 7 so 10^7 can be the maximum for N/10^m to remain an integer. Am I just lucky or can this also be a method of solving?



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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17 Jul 2014, 06:35
hamzakb 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. Hope it helps. I found my answer by finding the number of multiples of 3 between 1 and 100 i.e 100/3 = 33. Then I found the number of trailing zeroes in 33! = 7 so 10^7 can be the maximum for N/10^m to remain an integer. Am I just lucky or can this also be a method of solving? If you solve this way it should be: N = 3*6*9*12*15*...*99 = 3^33(1*2*3*...*33) = 3^33*33!. The number of trailing zeros for 33! is 33/5 + 33/25 = 6 + 1 = 7. Check Trailing Zeros and Power of a number in a factorial questions in our Special Questions Directory. Hope it helps.
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If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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17 Jul 2014, 06:37



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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27 Jul 2014, 06:39
Please correct me if I am wrong.
After reading Bunuel's thread on trailing zeroes I simply calculated m (= amount of trailing zeroes) this way:
\(\frac{100}{3*5}+\frac{100}{3* 25}= 6 + 1 = 7\)
dividing by 3*5 and 3*25 one ensures that only multiples of 3 are taken into consideration!



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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27 Jul 2014, 15:30



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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28 Jul 2014, 02:06
Quote: JK13 wrote: Please correct me if I am wrong.
After reading Bunuel's thread on trailing zeroes I simply calculated m (= amount of trailing zeroes) this way:
\(\frac{100}{3*5}+\frac{100}{3* 25}= 6 + 1 = 7\)
dividing by 3*5 and 3*25 one ensures that only multiples of 3 are taken into consideration!
No, that's not correct. N = 3*6*9*12*15*...*99 = 3^33*33!, not 100.
Check Trailing Zeros and Power of a number in a factorial questions in our Special Questions Directory.
Hope it helps. Dear Bunuel, thanks for the your fast answer. I understand that N is not 100 in this question, when looking for ALL trailing zeros in a number (which here would ofc be 3^33*33!). My approach was more based on finding the number of those trailing zeros in 100! that are a result of multiples of 3. Isn't that then the same amount of trailing zeros, that we want to find in this question? Another example could be: "Look for the trailing zeros in the product of all multiples of 7 between 1 and 100" \(\frac{100}{7*5}+\frac{100}{7*25}= 2 + 0 = 2\)



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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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13 Aug 2014, 05:39
Question seems to be very confusing at first. I solved it like this: 3*6*9*12*15.........*96*99 = 3(1*2*3*4.........32*33) = 3*33! Now number of training 0's will be = Number of trailing 0's in 33! i.e. 33/5+33/25 = 7 Hence 7 is the answer.
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Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
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18 Aug 2014, 11:11
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 Hi Bunuel, I used following approach to solve this question. Please advise if the approach and my assumption are correct : Product of all multiples of 3 between 1 and 100 = 3*6*9*12...*99 This can be reduced to 3^33 * 33! As 3^33 will never contribute a 0 to the result (I think. Please confirm), We can just go ahead by calculating the number of trailing zeroes in 33! # of trailing 0s in 33! = 7 and hence 7 is the answer. Please confirm if this is the right approach.




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