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n is a positive integer, and k is the product of all integer [#permalink]

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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

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.

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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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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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...

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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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

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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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]

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23 Oct 2013, 11:49

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Re: n is a positive integer, and k is the product of all integer [#permalink]

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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.

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 [#permalink]

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04 Aug 2015, 22:33

Hello from the GMAT Club BumpBot!

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Re: n is a positive integer, and k is the product of all integer [#permalink]

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13 Aug 2016, 00:08

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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