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

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Question Stats: 66% (02:16) correct 34% (02:19) wrong based on 928 sessions

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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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Math Expert V
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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

$$k$$ 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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Re: the smallest possible value of n  [#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
##### General Discussion
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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

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.
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Got it
Thanks B !! Thanks Karisham
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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...
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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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i have problem in integer..please can anyone help me
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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

http://gmatclub.com/forum/gmat-math-book-87417.html

Best of luck
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Re: the smallest possible value of n  [#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

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

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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...  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

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

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1
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...  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

$$k$$ 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.

No, that's not correct. You got the correct answer by fluke.
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GRE 1: Q169 V154 Re: n is a positive integer, and k is the product of all integer  [#permalink]

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1
Great Question.

1440=> 2^5*5*3^2

Here we should use Brute Force.
8!=> It has 4+2+1=> 7 Two's
2=> Three's
1=> Five's

Hence A

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

N=8!=720*56
K multiple of 1440
Hence as per options a) is ans
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Re: n is a positive integer, and k is the product of all integer  [#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

k = 1 * 2 * 3 * 4..............n

Or, k = n!

Further we know k = m*1440

Or, k = m*( 2*6! )

Now, we have - n! = m*( 2*6! )

Or, n! = m *2*6*5*4*3*2*1

Thus, The nearest factorial of n! will be 8

Hence, correct answer will be 8!
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GRE 1: Q169 V154 Re: n is a positive integer, and k is the product of all integer  [#permalink]

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Another technique can be brute force=>
Checking from least integer first
n=8 -> it will have 2^7 ,3^2 and 5
Hence it will be a multiple of 1440.
Hence A.

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

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k is a multiple of 1440. That means K must have at least all the factors of 1440. For this reason, 6 or 7 can't be the value of n. n must comprise the extra factor '2'.
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Re: n is a positive integer, and k is the product of all integer  [#permalink]

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Hi

Thanks for the answers - very useful.

I had a question please: if n! is equal to k is equal to 1440x then doesn't that imply that 1440/n is equal to the product of all preceding numbers?

So if n could be 8 then 1440 divided by 8 should be the product of all number 1 through to 8.

But that isn't the case. Am I missing something? Any advice would be greatly appreciated!

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

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