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n is a positive integer, and k is the product of all integer
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04 Nov 2010, 18:15
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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: property of Integers
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04 Nov 2010, 18:28
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: the smallest possible value of n
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21 Oct 2012, 00:29
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: property of Integers
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04 Nov 2010, 18:35
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: property of Integers
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04 Nov 2010, 19:52
Got it Thanks B !! Thanks Karisham
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Re: property of Integers
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06 Nov 2010, 10:53
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: property of Integers
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06 Nov 2010, 11:36
rockzom, but if the first answer is not the correct one, you'll have to divide with all the 5 answer choices. Then it will take you more than 37 seconds...



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Re: property of Integers
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07 Nov 2010, 08:37
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: property of Integers
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07 Nov 2010, 20:28
i have problem in integer..please can anyone help me



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Re: property of Integers
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08 Nov 2010, 08:21
annie89 wrote: i have problem in integer..please can anyone help me Read Number Theory topic in Math Book of Gmat Club. That's a wonderful book. you can find the link in ' Bunel's ' signature. well let me make it more easy for you. here is link http://gmatclub.com/forum/gmatmathbook87417.htmlBest of luck
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Re: the smallest possible value of n
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20 Oct 2012, 23:08
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 integer
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22 Jul 2014, 21:47
Hi Bunuel, I have a question here. I use the same method as you > finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck... Thanks for your help in advance. Bunuel wrote: 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 integer
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23 Jul 2014, 02:20
nueyada wrote: Hi Bunuel, I have a question here. I use the same method as you > finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck... Thanks for your help in advance. Bunuel wrote: 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). No, that's not correct. You got the correct answer by fluke.
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Re: n is a positive integer, and k is the product of all integer
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28 Dec 2016, 07:38
Great Question.
1440=> 2^5*5*3^2
Here we should use Brute Force. 8!=> It has 4+2+1=> 7 Two's 2=> Three's 1=> Five's
Hence A
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Re: n is a positive integer, and k is the product of all integer
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28 Dec 2016, 08:40
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
A. 8 B. 12 C. 16 D. 18 E. 24 N=8!=720*56 K multiple of 1440 Hence as per options a) is ans
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Re: n is a positive integer, and k is the product of all integer
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28 Dec 2016, 11:49
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
A. 8 B. 12 C. 16 D. 18 E. 24 k = 1 * 2 * 3 * 4..............n Or, k = n! Further we know k = m*1440 Or, k = m*( 2*6! ) Now, we have  n! = m*( 2*6! ) Or, n! = m *2*6*5*4*3*2*1 Thus, The nearest factorial of n! will be 8 Hence, correct answer will be 8!
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Re: n is a positive integer, and k is the product of all integer
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09 Jan 2017, 22:12
Another technique can be brute force=> Checking from least integer first n=8 > it will have 2^7 ,3^2 and 5 Hence it will be a multiple of 1440. Hence A.
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n is a positive integer, and k is the product of all integer
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16 Feb 2017, 10:10
k is a multiple of 1440. That means K must have at least all the factors of 1440. For this reason, 6 or 7 can't be the value of n. n must comprise the extra factor '2'.
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Re: n is a positive integer, and k is the product of all integer
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21 Feb 2017, 01:27
Hi
Thanks for the answers  very useful.
I had a question please: if n! is equal to k is equal to 1440x then doesn't that imply that 1440/n is equal to the product of all preceding numbers?
So if n could be 8 then 1440 divided by 8 should be the product of all number 1 through to 8.
But that isn't the case. Am I missing something? Any advice would be greatly appreciated!
Thank you, Rukshan



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