For an integer n greater than 1, n* denotes the product

Hi Bunuel,

Based on the concept you mentioned that we can take a factor out from each number which results in no prime numbers between 7!+2 and 7!+7, can we say that if we add to 7! a number which is not a factor of 7* then the resultant number will be prime no.?

e.g. can we say 7!+11 or 7!+13 are prime numbers? Thanks in advance.

Hi gmatkiller88,

Not necessarily so. Here is a counter example, 3! + 19 = 25, not prime.

Ok. So I guess in such cases we need to check each number individually to decide if the number is prime or not. Thanks.

I did it this way
7!+2=2(7*6*5*4*3*1+1)=this will not be prime
7!+3=3(7*6*5*4*2*1+1)=this will not be prime
.......
7!+7=7(7*6*5*4*2*1+1)=same,not prime as it is multiple of 7
answer=0/A

+1 for kudosssss

7* = 7! = (7)(6)(5)(4)(3)(2)(1)

So, 7* + 2 = (7)(6)(5)(4)(3)(2)(1) + 2
= 2[(7)(6)(5)(4)(3)(1) + 1]
= some multiple of 2
So, 7* + 2 is NOT a prime number

7* + 3 = (7)(6)(5)(4)(3)(2)(1) + 3
= 3[(7)(6)(5)(4)(2)(1) + 1]
= some multiple of 3
So, 7* + 3 is NOT a prime number

7* + 4 = (7)(6)(5)(4)(3)(2)(1) + 4
= 4[(7)(6)(5)(3)(2)(1) + 1]
= some multiple of 4
So, 7* + 4 is NOT a prime number
.
.
.
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7* + 7 = (7)(6)(5)(4)(3)(2)(1) + 7
= 7[(6)(5)(4)(3)(2)(1) + 1]
= some multiple of 7
So, 7* + 7 is NOT a prime number

ASIDE: You can assume that I was able to perform the same steps with 7* + 5 and 7* + 6

Answer: A

Cheers,
Brent

Notice that n* (which is the same thing as n!) is divisible by all positive integers from 1 to n.

7* + 2 is divisible by 2 (since both 7* and 2 are divisible by 2)

7* + 3 is divisible by 3 (since both 7* and 3 are divisible by 3)

…

In general, 7* + k is divisible by k when k is between 2 and 7, inclusive (since 7* is divisible by k and k is divisible by k).

Therefore, none of the numbers between 7* + 2 and 7* + 7, inclusive, is a prime.

Answer: A

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 \(7∗\) + 2 and \(7∗\) + 7, inclusive

\(7∗\) = Product of all the integers from 1 to 7, inclusive = 1*2*3*4*5*6*7 = 7!

\(7∗\) + 2 = 7! + 2 = 1*2*3*4*5*6*7 + 2 = 2*(1*3*4*5*6*7 + 1) = a Multiple of 2 => NOT a Prime Number
Similarly, 7! is also a multiple of all numbers from 3 to 7
=> all of the numbers from 7! + 2 to 7! + 7 will be Non-Prime numbers.

So, Answer will be A.
Hope it helps!

(Working below)

\(7∗\) + 3 = 7! + 3 = 1*2*3*4*5*6*7 + 3 = a Multiple of 3 => NOT a Prime Number
\(7∗\) + 4 = 7! + 4 = 1*2*3*4*5*6*7 + 4 = a Multiple of 4 => NOT a Prime Number
\(7∗\) + 5 = 7! + 5 = 1*2*3*4*5*6*7 + 5 = a Multiple of 5 => NOT a Prime Number
\(7∗\) + 6 = 7! + 6 = 1*2*3*4*5*6*7 + 6 = a Multiple of 6 => NOT a Prime Number
\(7∗\) + 7 = 7! + 7 = 1*2*3*4*5*6*7 + 7 = a Multiple of 7 => NOT a Prime Number

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