prime numbers

Both of these statements are sufficient. There are no cases when N is composite and this is true.

I may be misunderstanding your question...

Jim@StratusPrep wrote:Both of these statements are sufficient. There are no cases when N is composite and this is true

Can you explain this ?

Thank you !

Given N is integer > 0.

St1: All prime factors of N greater than root of N

if n = 3, prime factor of 3 = 3; root3 = 1.7 PRIME FACTOR is GREATER THAN ROOT(N).
if n = 5, prime factor of 5 = 5; root5 = 2.1 PRIME FACTOR is GREATER THAN ROOT(N).
For PRIME FACTOR TO BE GREATER THAN ROOT(N), N must be prime.
hence SUFFICIENT

St2: All prime factors of N greater than n/2
if n = 3, prime factor of 3 = 3 and n/2 = 1.5 PF > N/2
if n = 5, prime factor of 5 = 5 and n/2 = 2.5 PF > N/2

if I check for n = (Npn-Prime)6, n/2 = 3 and prime factors of 6 = 2,3. One prime factor in this case (2) is LESS than n/2 and one prime factor (3) is EQUAL TO n/2. None of the prime factors is greater than n/2 PF <= N/2
Hence B also looks SUFFICIENT

IMO D

What about 1 ? Prime factors of 3 are 3 and 1.

burningman wrote:What about 1 ? Prime factors of 3 are 3 and 1

1 is not prime