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.

Answer: A.

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
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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.

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
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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
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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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Kindly press"+1 Kudos" to appreciate

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 !

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.

Don't fall for 721 and 723.
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Another way to look at this is
6!+2= 2[(6*5*4*3*1)+1] primes cant have another factor apart from 1 and itself..2 clearly is a factor .Hence not a prime

Similarly for +3, +4, +5 and +6 , each of the numbers (3,4,5,6) will be a factor of the resulting number

Hence, No prime exist in this range and the correct answer is A) None

[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
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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

[6 = 6! = 720

720 + 2 = 722
720 + 6 = 726

722,723,724,725,726 <-- All composites

A.
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Hey, saukrit, interesting approach
Let us consider the number 11, which is a prime and not a factor of 6!.
So, following your logic 6! +11= 731, which comprises 17x43, ought to be a prime, is not it?

Hi, I could not understand the question, n denotes to product from 1 to n is fine but how to figure out in 6+2 and 6+6 that 6 = n and hence 6!

Posted from my mobile device

Dear JenniferMassey
pursuant to the question stem:
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?


It signifies that [n = n!
then [6 + 2 = n! + 2

Hope it helps