gmatpapa wrote:

Baten80 wrote:

7!+3 to 7!+4

so 5043 to 5047 [5043, 5044, 5045, 5046 and 5047]

there is no prime numbers.

Ans. A

What others easy way?

The quick way is to realize that if a factor(or any of its multiples) is added to a multiple of that factor, the result will be divisible by that factor. For example: 3 is a factor of 9. If 3 is added to 9, the result will be divisible by 3. Is 5 is added to 25, the result will be divisible by 5.

Coming to the problem. You see, 7! will be divisible by all numbers from 1 through 7. In other words, all integers from 1 to 7 are factors of 7! So, if any number between 1 to 7 is added to 7!, the result will be divisible by the number that is added (if 3 is added to 7!, result will be divisible by 3. If four is added, the result will be divisible by 4 and so on..) Essentially, the numbers between 7!+3 and 7!+7, inclusive will be: 7!+3, 7!+4, 7!+5, 7!+6, 7!+7. All these numbers will be divisible by one or the other number between 3 to 7, hence making all of them non-prime.

Answer A.

Great Explanation!

let me add it up of my own explanation...

so the list is

(7!+3), (7!+ 4), (7! + 5), (7! + 6), (7! + 7)

Notice that 7! = 7 x 6 x 5 x 4 x 3 x 2 (there is 5 and 2 in there so the Unit Digit must be 0)

is one of (7!+ 4), (7! + 5), (7! + 6) a prime number ?

just focus on the unit digit ! a prime number is a number that is divisible only by 1 and its own number,

unit digit of (7!+ 4) is 4 (since unit digit of 7! is 0), so therefore it is divisible by 2 (because it is even) = NOT A PRIME NUMBER

unit digit of (7! + 5) is 5, so therefore it is divisible by 5 = NOT A PRIME NUMBER

unit digit of (7! + 6) is 6, so therefore it is divisible by 2 (because it is even) = NOT A PRIME NUMBER

There you have it !