not too sure in my reasoning but try
(1) 2n=2*3*4*5, cancel 2 and left with
n=3*4*5 here n has 3 different prime factors
but on the other hand
2n=2*2*3*5*7, cancel 2
n=2*3*5*7 here n has 4 different prime factors
so 1st insufficient
(2) if n^2 has 4 differrent prime facrors that n has the same prime factors
n=2*3*5*7-4 prime factors then
n^2=(2^2*3^2*5^2*7^2) again the same 4 prime factors
so B wins

hi clock,

can you please elucidate st 1 further?

"(1) 2n=2*3*4*5, cancel 2 and left with
n=3*4*5 here n has 3 different prime factors

in the case above, there are only 3 primes--2, 3, and 5 but st 1 writes that there are 4 different primes as factors of 2n?

i assume that if 4 different primes of 2n-- assume 2 is one of the 4 primes and n contributes the 3 remaining primes?[/quote]

hi gmatjedi
i am not sure that i got you question,but may be i am wrong (for sure)

if 2n=x, and x has exactly 4 different prime numbers, x may look like
x=2*3*5*7
and
2n=2*3*5*7
n=3*5*7 here n has 3 diff prime factors,(3,5,7) i did not see any flaw yet...

again 2x=y. and y has 4 diff prime numbers
y may looks like
y=2*2*2*2*3*5*7
with primes 2, 3, 5, 7-4in total
2n=2*2*2*2*3*5*7
n=2*2*2*3*5*7
here n has 4 diff primes (2,3,5,7)

i think it may be said
if n-even or it includes 2 as prime factors than it will have 4 diff primes, and if n odd than it will have 3 diff primes
what i am missing??

hey clock
thanks for your reply.
that makes sense-- the key is that the 2 can be a redundant prime in n and 2n

any time friend
but i think we need more replies from math champ , may be any formal approach exists to this problem,
plugging and guessing is a tedious procedure