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

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

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28 Sep 2009, 19:49
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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
[Reveal] Spoiler: OA

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28 Sep 2009, 20:52
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

Using mathematical brute force I got to 8 although I'm sure there is a much more elegant solution.
I figured 1440 is not too big a number for that type of formula so trying a few numbers

If n = 6: k = 1 x 2 x 3 x 4 x 5 x 6 = 720

K is a multiple of 1440 meaning k/1440 must not have remainder.

Using the above if n = 7 then k wont be divisible by 1440 evenly, since it's odd. (since 720 is 1440/2 and hence 7x1440/2 = 7x1440/2)
But if n = 8 then it will be divisible by 1440 since it's even.

That took me under 2 mins. It's probably not the best solution but made sense to me
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28 Sep 2009, 21:25
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8
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28 Sep 2009, 22:06
Mikko wrote:
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread:
what-is-the-least-value-for-n-82903.html
We have the same question but for 990. Factoring 990 we get = $$2*3^2*5*11$$
So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5.

Is that correct?
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28 Sep 2009, 23:50
yangsta8 wrote:
Mikko wrote:
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread:
what-is-the-least-value-for-n-82903.html
We have the same question but for 990. Factoring 990 we get = $$2*3^2*5*11$$
So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5. (I don't think that there is a correlation between Prime 5-the factor and the 5 in 2^5)

Is that correct?

It's coincident that you got 11 is the largest prime and n! has enough factors of 2, 3 or 5 in $$2*3^2*5*11$$ but in your topic, 5 is the largest Prime factor but not enough for the power 2^5 if stop at n=5 (only 2=2^1, 4=2^2=> max is 2^3).

U can approach anyway as long as n! is divisible by all the power-factors of the given number
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24 Sep 2010, 05:06
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Is not the question asked the smallest possible value of n? Is not 2?
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24 Sep 2010, 08:45
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

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2
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24 Sep 2010, 08:57
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Baten80 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

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

$$k$$ 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.

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24 Sep 2010, 09:01
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Baten80 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

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

$$1440=2^5*3^2*5$$ => n should have at least one 5, two 3's and 5 2's.

if you take n=6 => k = 6! = $$1*2*3*4*5*6 = 2^4 * 3^2 * 5$$ => does not satisfy we need one more 2.

so the next even number 8 is the answer.
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24 Sep 2010, 11:55
Baten80 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

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

k=n!
And 1440 divides k
1440 = 2^5 * 3^2 * 5

Counting the powers of 2 in n!
Upto 2 (1)
Upto 4 (3)
Upto 6 (4)
Upto 8 (7)
So n>=8

Counting powers of 3 in n!
Upto 3 (1)
Upto 6 (2)
So n>=6

Counting powers of 5 in n!
Upto 5 (1)
So n>=5

Combining, minimum n=8
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Re: n is a positive integer, and k is the product of all integer [#permalink]

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26 May 2016, 23:30
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Re: n is a positive integer, and k is the product of all integer [#permalink]

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18 Jun 2016, 07:49
1440 = 2^5 * 3^2 * 5^1

We need five 2s, two 3s and one 5 in the n!. 8! contains all of these and it is the least of all. A is the answer.
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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 2016, 11:21
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

1440 = 144 * 10 = 12 * 12 * 2 * 5 = 2^5 * 3^2 * 5
we need at least 5 factors of 2, two factors of three, and 1 factor of five.
to have 2 factors of three, we need at least 6.
6! has 2^4, 3^2, and a 5.
but we need one more factor of two..therefore...8 would work!

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

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24 Jan 2017, 19:28
1) The prime factors of 1440 are (2)*(5)*(2*2)*(3)*2*(2*3)
2) If we try to create a sequence of consecutive integers from the above, we get 2*3*4*5*6, and one more 2 remains.
3) The next consecutive integer - 7 does not have 2 as one of its prime factors. However, the following multiple 8 - does require a 2. It means that n has to be at least 8.

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

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10 May 2017, 21:40
I used a very direct approach for this question.

Scan answer choices picked the smallest one. Very simple. Right ???

My thought process for above action

n is the multiple of 1440. Which means that 1440 is divisible by n.

Scanning answer choices I can see all are multiple of 2.

Question is asking for smallest so I picked option A

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

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15 May 2017, 17:30
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!.

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.

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Re: n is a positive integer, and k is the product of all integer   [#permalink] 15 May 2017, 17:30
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