imho E is the answ

chieftang wrote:Let P = 200! + 1.

Then, the greatest prime factor of P must be:

(A) Less than 40
(B) Between 40 and 80
(C) Between 80 and 120
(D) Between 120 and 160
(E) Greater than 160


Source: GMAT Hints

IMO E

200! is the product of all integers from 1 to 200...
if we add it by 1.. the no will not be divisible by any integer b/w 1 and 200 inclusive.

There is no single approach to solving the problem, but since 200!+1 is 1 greater than 200! which is product of all numbers from 1 to 200 inclusive, I think, the greatest prime divisor should be greater than 160. (E) is the answer.

I think it will be: (E) Greater than 160

chieftang wrote:Let P = 200! + 1.

Then, the greatest prime factor of P must be:

(A) Less than 40
(B) Between 40 and 80
(C) Between 80 and 120
(D) Between 120 and 160
(E) Greater than 160


Source: GMAT Hints

Since the difference between them is 1, P and 200! are consecutive integers. Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common to both x and x+1. In other words, any two consecutive integers -- including P and 200! -- are COPRIMES.

200! = 1 * 2 * 3 * 4...198 * 199 * 200.
Every prime number less than 200 is a factor of 200!.
Since P and 200! are COPRIMES -- meaning they share no factors other than 1 -- no prime factor less than 200 can be a factor of P.

Thus, the smallest prime factor of P must be greater than 200.
Since the smallest prime factor of P is greater than 200, so must be the greatest prime factor of P.

The correct answer is E.

199 is the greatest Prime that divides 200!

so for 200!+1 it has to be greater than 199

(E) is the answer