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605-655 Level|   Functions and Custom Characters|   Number Properties|                  
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For any integer n greater than 1, [n denotes the product of 09 Mar 2014, 15:11
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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
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A
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Bunuel
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For any integer n greater than 1, [n denotes the product of 09 Mar 2014, 15:11
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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 integers from 1 to n, inclusive, we have [6 + 2 = 6! + 2 and [6 + 6 = 6! + 6.

Now, notice that we can factor out 2 from 6! + 2, so it cannot be a prime number. Similarly, we can factor out 3 from 6! + 3, 4 from 6! + 4, and so on. This applies to all numbers in the range from 6! + 2 to 6! + 6, inclusive.

Thus, there are no prime numbers in this range.

Answer: A.

Question to practice on the same concept:
https://gmatclub.com/forum/does-the-int ... 26735.html
https://gmatclub.com/forum/if-x-is-an-i ... 00670.html
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For any integer n greater than 1, [n denotes the product of 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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For any integer n greater than 1, [n denotes the product of 09 Mar 2014, 19:00
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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
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For any integer n greater than 1, [n denotes the product of 09 Mar 2014, 23:57
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Option A.
6!=720
6!+2=722
6!+6=726
There is no prime integer from 722 to 726.
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For any integer n greater than 1, [n denotes the product of 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)

Answer A
Time Taken 1:29
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For any integer n greater than 1, [n denotes the product of 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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For any integer n greater than 1, [n denotes the product of 16 Sep 2014, 13:47
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Bunuel
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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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 .
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For any integer n greater than 1, [n denotes the product of 16 Sep 2014, 13:55
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Bunuel
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
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 .
Show more

Check other function questions in our Special Questions Directory:

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions
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For any integer n greater than 1, [n denotes the product of 28 Aug 2015, 05:46
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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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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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For any integer n greater than 1, [n denotes the product of 21 Aug 2016, 22:56
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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 17 Aug 2020, 12:27
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Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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
Show more

[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 02 Sep 2022, 21:52
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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

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For any integer n greater than 1, [n denotes the product of 18 Dec 2022, 02:01
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BrentGMATPrepNow
Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 + 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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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
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For any integer n greater than 1, [n denotes the product of 18 Dec 2022, 08:20
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BrentGMATPrepNow
Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 + 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

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
Show more

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
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For any integer n greater than 1, [n denotes the product of 19 Dec 2022, 03:53
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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?
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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?
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That's correct.
If you can factor an integer (greater than one) out of an expression, then the expression is not prime
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For any integer n greater than 1, [n denotes the product of 11 Feb 2024, 01:26
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­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. 
 
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For any integer n greater than 1, [n denotes the product of 08 Oct 2025, 06:00
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Here, we're dealing with factorial notation and checking for primes in a specific range. But there's actually a really elegant pattern here that makes this much simpler than manually checking each number. Let me walk you through it.

Here's how to approach this:

Step 1: Calculate the factorial and identify your range

First, let's figure out what we're working with. The notation \(6!\) means \(6\) factorial, which is:
\(6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720\)

So you need to find prime numbers between:
  • \(6! + 2 = 720 + 2 = 722\)
  • \(6! + 6 = 720 + 6 = 726\)

That gives you five numbers to check: \(722, 723, 724, 725, 726\)

Step 2: Recognize the key pattern (this is the insight that changes everything)

Here's what you need to see: Since \(6!\) is the product \(1 \times 2 \times 3 \times 4 \times 5 \times 6\), it contains all these numbers as factors. This means:
  • \(6!\) is divisible by 2
  • \(6!\) is divisible by 3
  • \(6!\) is divisible by 4
  • \(6!\) is divisible by 5
  • \(6!\) is divisible by 6

Now here's the crucial part: When you add one of these factors back to \(6!\), you can factor it out!

Step 3: Apply the divisibility pattern

Let's see this in action:

\(6! + 2 = 722\)
Since \(6!\) is divisible by 2, we can write \(6! = 2k\) where \(k = 360\)
So \(6! + 2 = 2k + 2 = 2(k + 1) = 2(361)\)
Therefore, 722 is divisible by 2 → not prime

\(6! + 3 = 723\)
Since \(6!\) is divisible by 3: \(6! + 3 = 3(240 + 1) = 3(241)\)
Therefore, 723 is divisible by 3 → not prime

\(6! + 4 = 724\)
Since \(6!\) is divisible by 4: \(6! + 4 = 4(180 + 1) = 4(181)\)
Therefore, 724 is divisible by 4 → not prime

\(6! + 5 = 725\)
Since \(6!\) is divisible by 5: \(6! + 5 = 5(144 + 1) = 5(145)\)
Therefore, 725 is divisible by 5 → not prime

\(6! + 6 = 726\)
Since \(6!\) is divisible by 6: \(6! + 6 = 6(120 + 1) = 6(121)\)
Therefore, 726 is divisible by 6 → not prime

Step 4: Count the primes

Every single number in the range has an obvious factor greater than 1, which means none of them are prime.

Answer: (A) None

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The pattern I showed you here – that \(n! + k\) is divisible by \(k\) when \(k \leq n\) – is incredibly powerful for factorial problems. But there's more to learn about when this pattern applies, how to extend it to similar problems, and what variations you might encounter on the GMAT. You can check out the complete framework and systematic approach on Neuron by e-GMAT to understand how to recognize and apply these factorial divisibility patterns across different question types. You can also explore detailed solutions for other GMAT official questions here to build your pattern recognition skills systematically.

Hope this helps you see the elegant shortcut! Let me know if you have questions.
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