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

Answer = (A) None

[6 = 6 * 5 * 4 * 3 * 2 * 1 = 720

[6 + 2 = 722

[6 + 6 = 726

722 , 723, 724, 725, 726

Not a single from above is a prime number, Answer = A

Option A.
6!=720
6!+2=722
6!+6=726
There is no prime integer from 722 to 726.

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)

Answer A
Time Taken 1:29
Difficulty Level 550

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

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 .

Check other function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

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

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 !

[6 + 2 = (6)(5)(4)(3)(2)(1) + 2 = 2[(6)(5)(4)(3)(1) + 1], which is a multiple of 2. So, [6 + 2 is NOT prime
[6 + 3 = (6)(5)(4)(3)(2)(1) + 3 = 3[(6)(5)(4)(2)(1) + 1], which is a multiple of 3. So, [6 + 3 is NOT prime
[6 + 4 = (6)(5)(4)(3)(2)(1) + 4 = 4[(6)(5)(3)(2)(1) + 1], which is a multiple of 4. So, [6 + 4 is NOT prime
[6 + 5 = (6)(5)(4)(3)(2)(1) + 5 = 5[(6)(4)(3)(2)(1) + 1], which is a multiple of 5. So, [6 + 5 is NOT prime
[6 + 6 = (6)(5)(4)(3)(2)(1) + 6 = 6[(5)(4)(3)(2)(1) + 1], which is a multiple of 6. So, [6 + 6 is NOT prime

Answer: A

Cheers,
Brent

Given that [n denotes the product of all the integers from 1 to n, inclusive and we need to find how many prime numbers are there between [6 + 2 and [6 + 6, inclusive

[6 = Product of all the integers from 1 to 6, inclusive = 1*2*3*4*5*6 = 6!

[6 + 2 = 6! + 2 = 1*2*3*4*5*6 + 2 = 2*(1*3*4*5*6 + 1) = a Multiple of 2 => NOT a Prime Number
Similarly, 6! is also a multiple of all numbers from 3 to 6
=> all of the numbers from 6! + 2 to 6! + 6 will be Non-Prime numbers.

So, Answer will be A.
Hope it helps!

(Working below)

[6 + 3 = 6! + 3 = 1*2*3*4*5*6 + 3 = a Multiple of 3 => NOT a Prime Number
[6 + 4 = 6! + 4 = 1*2*3*4*5*6 + 4 = a Multiple of 4 => NOT a Prime Number
[6 + 5 = 6! + 5 = 1*2*3*4*5*6 + 5 = a Multiple of 5 => NOT a Prime Number
[6 + 6 = 6! + 6 = 1*2*3*4*5*6 + 6 = a Multiple of 6 => NOT a Prime Number

Watch the following video to learn the Basics of Functions and Custom Characters

Hi BrentGMATPrepNow, not quite understand the way factorization work here. Why is the need for factorization and not sure why is not 2[(3)(5/2)(2)(3/2)(1/2) + 1] like usual factorization?
Thanks Brent

You are missing using the distributive property.
The distributive property applies when we are multiplying an expression consisting of ADDING and SUBTRACTING.
In these cases, we multiply each term inside the brackets by the term outside the brackets.
For example 2(3x + 5) = 6x + 10 [ here we multiplied 3x by 2, and we multiplied 5 by 2]
Important: Notice that I treated 3x as a single expression. That is, I did not multiply 3 by 2 AND x by 2. Instead, I multiplied the single term, 3x, by 2.

Similarly, the expression (6)(5)(4)(3)(2)(1) + 2 has two terms: (6)(5)(4)(3)(2)(1) and 2.
When we factor out a 2 we get: 2[(6)(5)(4)(3)(1) + 1]

In your factorization, 2[(3)(5/2)(2)(3/2)(1/2) + 1], you are assuming that we multiply (3) by 2, and multiply (5/2) by (2), which is incorrect.

To see the problem with your calculations first recognize that (1)(1)(1)(1)(1) = 1
So, we would expect that (2)(1)(1)(1)(1)(1) = 2
However, the way you are multiplying suggest that we multiply each 1 by 2 to get: (2)(1)(1)(1)(1)(1) = (2)(2)(2)(2)(2) = 32

Thanks Brent BrentGMATPrepNow for clarification and noted that the distributive property applies only to ADDING and SUBTRACTING from an expression.
Regarding factorization here, to clarify so if any number can be factored out from an expression, then it's not prime?

That's correct.
If you can factor an integer (greater than one) out of an expression, then the expression is not prime

­It is really helpful to remember- 
" If a multiple of a number is added to another multiple of the same number, the result is a multiple of the number " 

Here since 6!+ 3 is a multiple of 3, all the numbers between 6!+2 and 6!+6 will be multiples of the numbers they are being added to. 
Therefore, none of the above numbers will be a prime number.