Let N be a positive integer not equal to 1. Then note that none of the numbers 2, 3, ... , N is a divisor of (N! -1). From this we can conclude that:
(A) (N! - 1) is a prime number;
(B) at least one of the numbers N+1 , N+2 , ...., N! - 2 is a divisor of (N! -1);
(C) the smallest number between N and N! which is a divisor of (N!-1), is a prime number;
(D) none of the foregoing statements is necessarily correct;
Discussion:
N! - 1 could be a prime (example N = 4 gives N! - 1 = 23). It might not be a prime as well (example N = 5 gives N! - 1 = 119 which is divisible by 7).
Hence none of option A or B is necessarily true.
However option (C) is true as N! - 1 is not a prime, it's first prime divisor occurs between N and N! (since none of the primes from 1 to N divides it. Even if it is a prime, the smallest number between N and N! that divides it is itself which is a prime.
Answer: C
Let N be a positive integer not equal to 1. Then note that none of the numbers 2, 3, ... , N is a divisor of (N! -1). From this we can conclude that:
(A) (N! - 1) is a prime number;
(B) at least one of the numbers N+1 , N+2 , ...., N! - 2 is a divisor of (N! -1);
(C) the smallest number between N and N! which is a divisor of (N!-1), is a prime number;
(D) none of the foregoing statements is necessarily correct;
Discussion:
N! - 1 could be a prime (example N = 4 gives N! - 1 = 23). It might not be a prime as well (example N = 5 gives N! - 1 = 119 which is divisible by 7).
Hence none of option A or B is necessarily true.
However option (C) is true as N! - 1 is not a prime, it's first prime divisor occurs between N and N! (since none of the primes from 1 to N divides it. Even if it is a prime, the smallest number between N and N! that divides it is itself which is a prime.
Answer: C
