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n is a positive integer, and k is the product of all integer [#permalink]
04 Nov 2010, 17:15
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Question Stats:
66% (02:43) correct
34% (03:54) wrong based on 472 sessions
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
Re: property of Integers [#permalink]
04 Nov 2010, 17:28
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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.
Re: property of Integers [#permalink]
04 Nov 2010, 17:35
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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 _________________
Re: property of Integers [#permalink]
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: property of Integers [#permalink]
06 Nov 2010, 10:36
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...
Re: property of Integers [#permalink]
07 Nov 2010, 07:37
Expert's post
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. _________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Re: property of Integers [#permalink]
08 Nov 2010, 07:21
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
Re: the smallest possible value of n [#permalink]
20 Oct 2012, 22:08
1
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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
Re: the smallest possible value of n [#permalink]
20 Oct 2012, 23:29
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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 integer [#permalink]
23 Oct 2013, 11:49
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: n is a positive integer, and k is the product of all integer [#permalink]
22 Jul 2014, 20:47
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.
Re: n is a positive integer, and k is the product of all integer [#permalink]
23 Jul 2014, 01:20
1
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Expert's post
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. _________________
Re: n is a positive integer, and k is the product of all integer [#permalink]
04 Aug 2015, 22:33
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
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