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n is a positive integer, and k is the product of all integers from 1

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n is a positive integer, and k is the product of all integers from 1  [#permalink]

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New post 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  [#permalink]

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New post 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.

Answer: A (8).
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Re: n is a positive integer, and k is the product of all integers from 1  [#permalink]

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New post 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 integers from 1  [#permalink]

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New post 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
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Re: n is a positive integer, and k is the product of all integers from 1  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 20 Oct 2012, 22:08
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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

Hence Answer A.
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Re: n is a positive integer, and k is the product of all integers from 1  [#permalink]

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New post 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  [#permalink]

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Re: n is a positive integer, and k is the product of all integers from 1   [#permalink] 22 Apr 2020, 11:17

n is a positive integer, and k is the product of all integers from 1

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