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# For any integer n greater than 1, n* denotes the product of

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For any integer n greater than 1, n* denotes the product of [#permalink]

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01 May 2012, 09:58
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For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between 6* + 2 and 6* + 6, inclusive?

A. None
B. One
C. Two
D. Three
E. Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks
[Reveal] Spoiler: OA

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Re: For any integer n greater than 1......... [#permalink]

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01 May 2012, 11:42
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carcass wrote:
For any integer n greater than 1, |_ N denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between |_ 6 + 2 and |_ 6 + 6, inclusive?

(A) None
(8) One
(C) Two
(D) Three
(E) Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks

There doesn't exist any prime numbers between any |_N +2 and |_N +N ,inclusive.This is bcoz |_N +x is always is divisible by
x(for x < N).

Hope that helps.
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Re: For any integer n greater than 1......... [#permalink]

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01 May 2012, 12:05
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NightFury wrote:
carcass wrote:
For any integer n greater than 1, |_ N denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between |_ 6 + 2 and |_ 6 + 6, inclusive?

(A) None
(8) One
(C) Two
(D) Three
(E) Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks

There doesn't exist any prime numbers between any |_N +2 and |_N +N ,inclusive.This is bcoz |_N +x is always is divisible by
x(for x < N).

Hope that helps.

Knowing this principle obviously allows you to solve to problem pretty much immediately. However, it is possible to multiply this out in under 2 minutes. To solve for |_6, I went from high to low. 6x5=30x4=120x3=360x2=720. I wrote that out when practicing this question, but I feel like I would have saved additional time by not doing so. Once you get 720, you know you are looking for a prime number from 722, 723, 724, 725, and 726. Eliminating the evens and 725 leaves you with 723, which turns out to be divisible by 3. Answer=A.

As I said to begin though, remembering NightFury's rule speeds things up immensely.
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Re: For any integer n greater than 1......... [#permalink]

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01 May 2012, 13:34
Thanks to both.

Was a tricky question
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Re: For any integer n greater than 1, N* denotes the product of [#permalink]

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01 May 2012, 14:27
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carcass wrote:
For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between 6* + 2 and 6* + 6, inclusive?

A. None
B. One
C. Two
D. Three
E. Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks

Given that n* denotes the product of all the integers from 1 to n, inclusive so, 6*+2=6!+2 and 6*+6=6!+6.

Now, notice that we can factor out 2 our of 6!+2 so it cannot be a prime number, we can factor out 3 our of 6!+3 so it cannot be a prime number, we can factor out 4 our of 6!+4 so it cannot be a prime number, ... The same way for all numbers between 6*+2=6!+2 and 6*+6=6!+6, inclusive. Which means that there are no primes in this range.

Question to practice on the same concept:
does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

Hope it helps.
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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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04 Jan 2013, 23:45
carcass wrote:
For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between 6* + 2 and 6* + 6, inclusive?

A. None
B. One
C. Two
D. Three
E. Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks

Since 6! Will contain a 2 and a 5 the last digit will be a “0” so 6! = XXXX0. Now we have to check numbers XXXX2…XXXX6 => XXXX2 –> div by 2, XXXX3 ->Sum of digits div by 3 so divisible by 3,XXXX4 - >div by 2, XXXX5 -> div by 5, and XXXX6 -> div by 6.
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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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05 Jan 2013, 08:48
My approach is:

6! is dividable through 1,2,3,4,5 and 6

x is dividable through y if x=z+y where z is dividable through y

for example:

12 is dividable through 4 since 12=8+4. One can also visualize this on a number line.

therefore no number between 6! and 6!+6 is a prime, which includes the intervall 6!+2 to 6!+6
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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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08 Jan 2013, 03:17
Although all the approaches are correct. Here is one more, but less cluttered approach.

Since 6!= 720. So in nut shell, we have to find out prime # between 722 & 726. There are no prime # between 722 & 726.

-Shalabh Jain
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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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09 Mar 2014, 18:53
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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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29 May 2015, 09:27
Hello from the GMAT Club BumpBot!

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Re: For any integer n greater than 1, n* denotes the product of [#permalink]

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28 Aug 2015, 05:46
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carcass wrote:
For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many prime numbers are there between 6* + 2 and 6* + 6, inclusive?

A. None
B. One
C. Two
D. Three
E. Four

I'm not sure how to attack this problem. 6! come to play ......but I do not really understand how to figure out.

thanks

Although as Bunuel and others method is good for generic numbers, it is very easy to calculate 6!=720. I always try to expand small factorials this way...
6!= 6x5x4x3x2x1= 6x3x2x4x5=36x20=720....

so the question is asking for primes between 6!+2 and 6!+6. which is 722 and 726
722/724/726 divisible by 2 so not prime
723 divisible by 3 so not prime
725 divixible by 5 so not prime

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Re: For any integer n greater than 1, n* denotes the product of   [#permalink] 28 Aug 2015, 05:46
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