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n is a positive integer, and k is the product of all integer [#permalink]
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28 Sep 2009, 19:49
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n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is A. 8 B. 12 C. 16 D. 18 E. 24
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Re: smallest possible value [#permalink]
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28 Sep 2009, 20:52
tejal777 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is Using mathematical brute force I got to 8 although I'm sure there is a much more elegant solution. I figured 1440 is not too big a number for that type of formula so trying a few numbers If n = 6: k = 1 x 2 x 3 x 4 x 5 x 6 = 720 K is a multiple of 1440 meaning k/1440 must not have remainder. Using the above if n = 7 then k wont be divisible by 1440 evenly, since it's odd. (since 720 is 1440/2 and hence 7x1440/2 = 7x1440/2) But if n = 8 then it will be divisible by 1440 since it's even. That took me under 2 mins. It's probably not the best solution but made sense to me



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Re: smallest possible value [#permalink]
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28 Sep 2009, 21:25
1440=2^5*3^2*5
k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s n=8 => n! is divided by 5, 3^2 remains 0
n=8



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Re: smallest possible value [#permalink]
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28 Sep 2009, 22:06
Mikko wrote: 1440=2^5*3^2*5
k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s n=8 => n! is divided by 5, 3^2 remains 0
n=8 Sorry I'm little confused. In this thread: whatistheleastvalueforn82903.htmlWe have the same question but for 990. Factoring 990 we get = \(2*3^2*5*11\) So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor. However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5. Is that correct?



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Re: smallest possible value [#permalink]
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28 Sep 2009, 23:50
yangsta8 wrote: Mikko wrote: 1440=2^5*3^2*5
k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s n=8 => n! is divided by 5, 3^2 remains 0
n=8 Sorry I'm little confused. In this thread: whatistheleastvalueforn82903.htmlWe have the same question but for 990. Factoring 990 we get = \(2*3^2*5*11\) So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor. However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5. ( I don't think that there is a correlation between Prime 5the factor and the 5 in 2^5) Is that correct? It's coincident that you got 11 is the largest prime and n! has enough factors of 2, 3 or 5 in \(2*3^2*5*11\) but in your topic, 5 is the largest Prime factor but not enough for the power 2^5 if stop at n=5 (only 2=2^1, 4=2^2=> max is 2^3). U can approach anyway as long as n! is divisible by all the powerfactors of the given number



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Re: smallest possible value [#permalink]
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Value of n [#permalink]
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24 Sep 2010, 08:45
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is A. 8 B. 12 C. 16 D. 18 E. 24 To me Ans is 2. LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help.
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Re: Value of n [#permalink]
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24 Sep 2010, 08:57
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Baten80 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
To me Ans is 2.
LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help. \(k\) is the product of all integers from 1 to \(n\) inclusive > \(k=n!\); \(k\) is a multiple of 1440 > \(n!=1440*p=2^5*3^2*5*p\), for some integer \(p\) > \(n!\) must have at least 2 in power 5, 3 in power 2 and 5 as its factors. Now, \(7!=2*3*(2^2)*5*(2*3)*7=2^4*3^2*5*7\) not enough power of 2 > next #: \(8!=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5*7=1440*(2^2*7)\) > so lowest value of \(n\) is 8. Answer: A (8).
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Re: Value of n [#permalink]
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24 Sep 2010, 09:01
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Baten80 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
To me Ans is 2.
LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help. \(1440=2^5*3^2*5\) => n should have at least one 5, two 3's and 5 2's. if you take n=6 => k = 6! = \(1*2*3*4*5*6 = 2^4 * 3^2 * 5\) => does not satisfy we need one more 2. so the next even number 8 is the answer.
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Re: Value of n [#permalink]
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24 Sep 2010, 11:55
Baten80 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
A. 8 B. 12 C. 16 D. 18 E. 24
To me Ans is 2.
LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help. k=n! And 1440 divides k 1440 = 2^5 * 3^2 * 5 Counting the powers of 2 in n!Upto 2 (1) Upto 4 (3) Upto 6 (4) Upto 8 (7) So n>=8 Counting powers of 3 in n!Upto 3 (1) Upto 6 (2) So n>=6 Counting powers of 5 in n!Upto 5 (1) So n>=5 Combining, minimum n=8
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Re: n is a positive integer, and k is the product of all integer [#permalink]
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18 Jun 2016, 07:49
1440 = 2^5 * 3^2 * 5^1 We need five 2s, two 3s and one 5 in the n!. 8! contains all of these and it is the least of all. A is the answer.
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Re: n is a positive integer, and k is the product of all integer [#permalink]
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23 Oct 2016, 11:21
tejal777 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
A. 8 B. 12 C. 16 D. 18 E. 24 1440 = 144 * 10 = 12 * 12 * 2 * 5 = 2^5 * 3^2 * 5 we need at least 5 factors of 2, two factors of three, and 1 factor of five. to have 2 factors of three, we need at least 6. 6! has 2^4, 3^2, and a 5. but we need one more factor of two..therefore...8 would work! A



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Re: n is a positive integer, and k is the product of all integer [#permalink]
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24 Jan 2017, 19:28
1) The prime factors of 1440 are (2)*(5)*(2*2)*(3)*2*(2*3) 2) If we try to create a sequence of consecutive integers from the above, we get 2*3*4*5*6, and one more 2 remains. 3) The next consecutive integer  7 does not have 2 as one of its prime factors. However, the following multiple 8  does require a 2. It means that n has to be at least 8.
The correct answer is A



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Re: n is a positive integer, and k is the product of all integer [#permalink]
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10 May 2017, 21:40
I used a very direct approach for this question. Scan answer choices picked the smallest one. Very simple. Right ??? My thought process for above action n is the multiple of 1440. Which means that 1440 is divisible by n. Scanning answer choices I can see all are multiple of 2. Question is asking for smallest so I picked option A Ans : Option A



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Re: n is a positive integer, and k is the product of all integer [#permalink]
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15 May 2017, 17:30
tejal777 wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
A. 8 B. 12 C. 16 D. 18 E. 24 Let’s break 1440 into prime factors: 1440 = 144 x 10 = 12 x 12 x 10 = 2^5 x 3^2 x 5^1 Thus, k/(2^5 x 3^2 x 5^1) = integer. We also know that k is the product of all integers from 1 to n inclusive, or in other words, k = n!. Let’s check our answer choices: A. If n = 8, then k = 8! and 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 = 7 x 5 x 3^2 x 2^7 does contain five 2s, two 3s and one 5. Answer: A
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