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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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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]
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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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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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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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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n is a positive integer, and k is the product of all integers from 1 [#permalink]
Hope, this illustration will work.

Posted from my mobile device
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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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Factorise 1440 : = 10 X 144
2x5 x 12 x 12
2X5x 3 x 4 x 6 x 2
1 x 2x3x 5 x 6 x 8 ( to get consequtive number 1 to n = you need 4 and 7
Now K is the Multiple of 1440 : There fore K can be = 1440 x 28
and also 1 x 2 x 3x4x5x6x7x8 = 1440 X 28
So n should be latest : 8
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Re: n is a positive integer, and k is the product of all integers from 1 [#permalink]
Tajmir wrote:
Hope, this illustration will work.

Posted from my mobile device




that has to be the neatest penmanship I have ever seen... lol
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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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n is a positive integer, and k is the product of all integers from 1 to n inclusive which means k = n!

K is a multiple of 1440 => k = 1440 * p

1440 = 2*2*2*2*2*3*3*5

=> 1440 = \(2^5\) * \(3^2\) * 5

n! = \(2^5\) * \(3^2\) * 5 * p

We need at least five (2's), two(3's) and one(5)

=> 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2 = \(2^5\) * \(3^2\) * 5 * \(2^2\) * 7 = 1440 * \(2^2\) * 7

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

A. 8
B. 12
C. 16
D. 18
E. 24


The question tells k! is divided by \(1440\)

So k have to have all the factors that \(1440\) will have.

\(The \ factors \ of \ 1440=2^5, 3^2\ & \ 5\)

From the answer choices \(8!\) contains \(= 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2=2^7, 3^2, 5, \ & \ 7\)

The answer is \(A.\)
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Re: n is a positive integer, and k is the product of all integers from 1 [#permalink]
Hey are these type of questions still relevant in 2024?
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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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Rangerwarrior wrote:
Hey are these type of questions still relevant in 2024?

This particular question is very much in the syllabus of GMAT Focus.
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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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