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n is a positive integer, and k is the product of all integers from 1
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04 Nov 2010, 17:15
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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: n is a positive integer, and k is the product of all integers from 1
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04 Nov 2010, 17:28
shrive555 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
8 12 16 18 24
please share your method. Thanks \(k\) 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: n is a positive integer, and k is the product of all integers from 1
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20 Oct 2012, 23:29
kapsycumm 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 Factorizing 1440, we get \(2^5 * 3^2 * 5^1\) So, k should have at least five 2s, two 3s and one 5. a) 8! = 2*3*4*5*6*7*8. Above condition is satisfied. Hence answer is A




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Re: n is a positive integer, and k is the product of all integers from 1
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04 Nov 2010, 17:35
shrive555 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
8 12 16 18 24
please share your method. Thanks When we say k is the product of all integers from 1 to n, we mean k = n! Now 1440 = 1.2.3.4.5.6.2 Now if k is a multiple of 1440, it has to be at least 8!. This is so because it cannot be 6! due to the extra 2 factor at the end; it cannot be 7! either, again because the extra 2 in the end will be unaccounted for. If k = 8! = 1.2.3.4.5.6.7.8, then k = 1440*7*4 i.e. k will be a multiple of 1440. Answer A
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Re: n is a positive integer, and k is the product of all integers from 1
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06 Nov 2010, 09:53
Took me 0:37 to do 1440x2 and divide that by the smallest choice to see if it fit. 2880/8=360, so A.
(I know my methods are unorthodox, but what can you do.)



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Re: n is a positive integer, and k is the product of all integers from 1
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07 Nov 2010, 07:37
rockzom wrote: Took me 0:37 to do 1440x2 and divide that by the smallest choice to see if it fit. 2880/8=360, so A.
(I know my methods are unorthodox, but what can you do.) 'Unorthodox' is one way of putting it. I'm not sure I understand why you've done what you did, but we're not looking here for the smallest answer choice which divides some multiple of 1440; it's coincidence that you get the right answer if you try that. If '4' or '5' had instead been the first answer choice, you'd find that 2880/4 or 2880/5 is an integer, but those aren't correct answers to the question. There's a reason Bunuel and Karishma carried out the steps they did.
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Re: n is a positive integer, and k is the product of all integers from 1
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20 Oct 2012, 22:08
kapsycumm 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 K= multiple of 1440 ie product of some integer P X 1440. & K= n! = 1.2.3.4......n Now lets factorize 1440 = 12 * 12 * 10 Now my objective is to find this multiple 12*12*10 with min possible numbers in n!. Lets go on now...... n= 1* 2*( 3*4)* 5* 6*7*8(4 X 2) Hence 8! is sufficient to get a multiple of 12*12*10 Hence Answer A.



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Re: n is a positive integer, and k is the product of all integers from 1
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15 May 2017, 16:30
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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Re: n is a positive integer, and k is the product of all integers from 1
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