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If N is the product of all positive integers less than 31,
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19 Oct 2010, 21:57
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If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? A. 3 B. 6 C. 7 D. 14 E. 26
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If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?A. 3 B. 6 C. 7 D. 14 E. 26 Check this: everythingaboutfactorialsonthegmat85592.htmlGiven: \(n=30!\). Question: if \(\frac{30!}{18^k}=integer\) then \(k_{max}=?\) We should determine the highest power of 18 in 30!. \(18=2*3^2\), so we should find the highest powers of 2 and 3 in 30!: Highest power of 2 in 30!: \(\frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26\), > \(2^{26}\); Highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14\) > \(3^{14}\); \(n=30!=2^{26}*3^{14}*p\), where \(p\) is the product of other multiples of 30! (other than 2 and 3) > \(n=30!=(2*3^{2})^7*2^{19}*p=18^7*2^{19}*p\) > so the highest power of 18 in 30! is 7 > \(\frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer\) > \(k=7\). Answer: C. Hope it's clear.
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Re: If N is the product of all positive integers less than 31,
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Updated on: 12 Feb 2014, 06:40
TomB wrote: If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?
A. 3 B. 6 C. 7 D. 14 E. 26 Hi all, this was my approach for solving this. Basically the question is asking us what is the highest power of 18 that can give us a number that is a factor of 31! So remember 18 after prime factorization is (3^2)(2). Now there are going to be less factors of 3 in 31!, than factors of 2. Therefore, lets find how many factors of 3 in 31! We can use this quick method 31/3^1 = 10 31/3^2= 3 31/3^3=1 Sum = 14 Just ignore the remainders. So we have that 3^14 must be the least. Now don't forget that 18 is 3^2k so k must be ONLY 7, because 2k will give us the 14. Hence answer is (C) Hope it helps Bunuel could you please validate this one? Thank you Cheers J
Originally posted by jlgdr on 08 Oct 2013, 12:05.
Last edited by jlgdr on 12 Feb 2014, 06:40, edited 1 time in total.



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Re: If N is the product of all positive integers less than 31,
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12 Oct 2013, 18:33
According to Bunuel \(18=2*3^2\)
I listed 14 3's 14 3's 3 3 3 3 3 3 3 3 3 3 3 3 3 3 26 2's 2 2 2 2 2 2 2 2 2 2 2 2 2 2 and so on
notice that 7 such combinations of 18 are possible. so answer= \(18^7\)



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Bunuel wrote: If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?A. 3 B. 6 C. 7 D. 14 E. 26 Check this: everythingaboutfactorialsonthegmat85592.htmlGiven: \(n=30!\). Question: if \(\frac{30!}{18^k}=integer\) then \(k_{max}=?\) We should determine the highest power of 18 in 30!. \(18=2*3^2\), so we should find the highest powers of 2 and 3 in 30!: Highest power of 2 in 30!: \(\frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26\), > \(2^{26}\); Highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14\) > \(3^{14}\); \(n=30!=2^{26}*3^{14}*p\), where \(p\) is the product of other multiples of 30! (other than 2 and 3) > \(n=30!=(2*3^{2})^7*2^{19}*p=18^7*2^{19}*p\) > so the highest power of 18 in 30! is 7 > \(\frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer\) > \(k=7\). Answer: C. Hope it's clear. Hi Bunuel, The logic here is the same as the logic in finding terminating "0" of a number right? But instead of checking for the level of "5" we do it for "3^2", right?



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Re: If N is the product of all positive integers less than 31,
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18 Aug 2016, 01:20
TomB wrote: If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?
A. 3 B. 6 C. 7 D. 14 E. 26 Responding to a pm: Quote: If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?
The maximum powers of a prime number 3, in 10!: 103+1032=3+1=4103+1032=3+1=4 (take only quotients into account). So, we have that the maximum power of 3 in 10! is 4, thus maximum power of 3 in (10!)^2 will be 8: (34)2=38(34)2=38. As discussed 8 is the maximum power of 6 as well.
But the below question we are taking both the factors 2 and 3 .Please can you explain the differnece
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.
Given: n=30! Question: if 30!/18k=integer kmax=?
We should determine the highest power of 18 in 30!.
18=2∗3^2 so we should find the highest powers of 2 and 3 in 30!:
Highest power of 2 in 30!: 30/2+30/4+30/8+30/16=15+7+3+1=26, > 2^26
Highest power of 3 in 30!: 30/3+30/9+30/27=10+3+1=14 > 3^14
n=30!=2^26∗3^14∗ where p is the product of other multiples of 30! (other than 2 and 3) > n=30!=(2∗3^2)^7∗2^19∗p> so the highest power of 18 in 30! is 7
Both the methods are the same. In some cases, you need to work on more than one factor, in others you don't. Here is how we decide:  If each factor appears only once in the number, you just need to find the number of times the largest factor appears. The largest factor will have the lowest exponent. e.g. 6 (2*3), 10 (2*5), 30 (2*3*5) etc  If the larger factor appears multiple times in the number, again you just need to find the number of times the largest factor appears. e.g. 18 (2 * 3 *3), 50 (2 * 5 * 5) etc  If the smaller factor appears more than the greater factor in the number, then you need to work on both factors. e.g. 12 (2 * 2 * 3), 175 (5 * 5 * 7) etc In this case, the smaller factor appears more often that the greater factor in the factorial but you actually need more of each smaller factor to make each number. So you don't know what will dominate. For more on this, check out: http://www.veritasprep.com/blog/2011/06 ... actorials/The post as well as the comments.
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Re: If N is the product of all positive integers less than 31,
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18 Aug 2016, 05:55
30! must be divisible by (3*3*2)^k. As I will find a lot of 2's among 30!'s factors I will focus on the 3's (that also occurs twice instead of only once as a factor in 18). Every multiple of 3 will provide me with atleast one factor of 3. There aren't that many so I list them quickly. 3 6 9* 12 15 18* 21 24 27** 30 The stars are marked as they are multiples of 9* or 27** and thus provides two or three 3's respectively. 3=3 (occurs 10 times) +10 3*3=9 (occurs 3 times) +3(adding just one additional three each as they were not counted for in the multiples of 3) 3*3*3=27 (occurs 1 time) +1 (again, adding just one as they were previously not counted) Total: 14. For every k we add two 3's. So we must divide 14 by 2 to get the highest integer K before we receive a noninteger result. Answer: 7 > Answer choice: C.
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Re: If N is the product of all positive integers less than 31,
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28 Feb 2017, 11:17
here n=30! now we have to find highest power of 18 in 30!. this is done in this way 18 = 2*3^2 so, we find the highest power of 2 and 3^2 in 30! individually for 2
30/2 + 30/4 + 30/8 + 30/16 15+7+3+1 26
similarly for 3 30/3+ 30/9+30/27 10+3+1 14 so, there are 14/2= 7, power of 3^2 in 30! so required answer is 7 as we need seven (3^2) and seven (2) to make their products as power of 18.



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Re: If N is the product of all positive integers less than 31,
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21 Aug 2017, 20:02
VeritasPrepKarishma wrote: TomB wrote: If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?
A. 3 B. 6 C. 7 D. 14 E. 26 Responding to a pm: Quote: If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?
The maximum powers of a prime number 3, in 10!: 103+1032=3+1=4103+1032=3+1=4 (take only quotients into account). So, we have that the maximum power of 3 in 10! is 4, thus maximum power of 3 in (10!)^2 will be 8: (34)2=38(34)2=38. As discussed 8 is the maximum power of 6 as well.
But the below question we are taking both the factors 2 and 3 .Please can you explain the differnece
If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.
Given: n=30! Question: if 30!/18k=integer kmax=?
We should determine the highest power of 18 in 30!.
18=2∗3^2 so we should find the highest powers of 2 and 3 in 30!:
Highest power of 2 in 30!: 30/2+30/4+30/8+30/16=15+7+3+1=26, > 2^26
Highest power of 3 in 30!: 30/3+30/9+30/27=10+3+1=14 > 3^14
n=30!=2^26∗3^14∗ where p is the product of other multiples of 30! (other than 2 and 3) > n=30!=(2∗3^2)^7∗2^19∗p> so the highest power of 18 in 30! is 7
Both the methods are the same. In some cases, you need to work on more than one factor, in others you don't. Here is how we decide:  If each factor appears only once in the number, you just need to find the number of times the largest factor appears. The largest factor will have the lowest exponent. e.g. 6 (2*3), 10 (2*5), 30 (2*3*5) etc  If the larger factor appears multiple times in the number, again you just need to find the number of times the largest factor appears. e.g. 18 (2 * 3 *3), 50 (2 * 5 * 5) etc  If the smaller factor appears more than the greater factor in the number, then you need to work on both factors. e.g. 12 (2 * 2 * 3), 175 (5 * 5 * 7) etc In this case, the smaller factor appears more often that the greater factor in the factorial but you actually need more of each smaller factor to make each number. So you don't know what will dominate. For more on this, check out: http://www.veritasprep.com/blog/2011/06 ... actorials/The post as well as the comments. hi mam okay with first and second scenario .... For the third scenario, however, say, 10!/12... here, we have 2 2s and 1 3...so the smaller factor appears more often than does the bigger factor... 10/2 + 10/4 + 10/8 = 5+2+1 = 8 2^8 and 10/3 + 10/9 = 3+1 = 4 3^4 So the answer is 4 the highest power 12 can have is 4..... please correct if I am missing something, mam ... thanks in advance ...



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Re: If N is the product of all positive integers less than 31,
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