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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
Problem Solving Question: 144 Category:Arithmetic Properties of numbers Page: 81 Difficulty: 600
Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
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Re: For any integer n greater than 1, [n denotes the product of
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09 Mar 2014, 15:11
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18
SOLUTION
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
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.
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09 Mar 2014, 20:04
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[n = n!
6! + 2 --> still a multiple of 2 6! + 3 --> still a multiple of 3 6! + 4 --> still a multiple of 4 6! + 5 --> still a multiple of 5 6! + 6 --> still a multiple of 6
None of the expressions are prime. The answer is A.
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11 Mar 2014, 00:13
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6! = 720 6! +2 = 722 6! +6 = 726
722, 723, 724, 725, 726 are the numbers Even numbers can't be prime (except for no. 2) as they are multiple of 2. We have 2 odd number 723 and 725, out of which 725 can be directly eliminated. 723 is the only no. to be tested and by divisibility test, we can find it is divisible by 3 (7+2+3 =12)
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11 Mar 2014, 05:00
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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
Sol: [6 is nothing but 6!
So we need to find no. of prime numbers between 6!+2 and 6!+6 The numbers in between will be 6!+3,6!+4 and 6!+5
Consider any number. Let's take 6!+3. It can be written as 3*( 1*2*4*5*6+ 1)-----> This number is multiple of 3. Hence not prime
Similarly 6!+5 will be a multiple of 5 and hence not prime. (5*(1*2*3*4*6+1))
Basically there is no prime no. Ans is A
Also 6!=720 so the numbers are 722,723,724,725,726
722,724,726 are divisible by 2 and hence not prime 725 is divisble by 5 723 is divisible by 3
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Re: For any integer n greater than 1, [n denotes the product of
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16 Sep 2014, 13:47
Bunuel wrote:
SOLUTION
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
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.
Nice question. Looking at the explanation , it is not so difficult; however, I was not able to understand the question hence marked the answer wrong. Should have read the question more carefully. I beseech you to post such questions more frequently. These are immensely informative .
Re: For any integer n greater than 1, [n denotes the product of
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16 Sep 2014, 13:55
dransa wrote:
Bunuel wrote:
SOLUTION
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
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.
Nice question. Looking at the explanation , it is not so difficult; however, I was not able to understand the question hence marked the answer wrong. Should have read the question more carefully. I beseech you to post such questions more frequently. These are immensely informative .
Re: For any integer n greater than 1, n* denotes the product of
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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
Answer None
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Re: For any integer n greater than 1, [n denotes the product of
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21 Aug 2016, 22:56
Nice question...Gmat played a trick here with wording and question formation... Most people thought [6 + 2 and [6 + 6 = [8 and [12 and there is only 1 prime number between 8 and 12 that is 11... Marking B and move on...
Other way it can be done wrong is if people start finding prime number between 8! and 12 !
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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
Problem Solving Question: 144 Category:Arithmetic Properties of numbers Page: 81 Difficulty: 600
Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.
Thank you!
There are two ways to solve this problem:
First, directly calculate the two values. The numbers are relatively small, so it's not a huge challenge.
[6 + 2 = 6*5*4*3*2 + 2 = 722
[6 + 6 = [6 + 2 + 4 = 726
722, 723, 724, 725, and 726 are all possible values. None are prime.
The second method is to realize that you can factor out 2 and 6 from each expression: 6! + 2 = 2*(6*5*4*3*1 + 1) and any number with a factor besides 1 and itself is not prime. Ditto for 6! + 6.
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