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For any integer n greater than 1, n* denotes the product of [#permalink]
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01 May 2012, 08:58
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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
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Re: For any integer n greater than 1......... [#permalink]
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01 May 2012, 10:42
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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, 11:05
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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, N* denotes the product of [#permalink]
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01 May 2012, 13:27
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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. Answer: A. Question to practice on the same concept: doestheintegerkhaveafactorpsuchthat1pk126735.htmlifxisanintegerdoesxhaveafactornsuchthat100670.htmlHope 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, 22:45
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. Answer: A None



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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, 07:48
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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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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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 Answer None  Kindly press"+1 Kudos" to appreciate



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Re: For any integer n greater than 1, n* denotes the product of [#permalink]
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21 Feb 2017, 05:24
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 Alternate approach is to multiply all the prime numbers between 1 and 6 and add 2 or 6. *6 + 2 = 2 * 3 * 5 + 2 = 32 *6 + 10 = 2 * 3 * 5 + 6 = 36 33, 34, 35 are not prime. (A)




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