M=P*T IS M MULTIPLE OF P^2T

1.M HAS MORE THAN 9 +VE FACTORS

NO INFORMATION SO INSUFF

2.M IS MULTIPLE OF P^3

m IS MULTIPLE OF P^3
p^2T SUPOSSE T=2,3,5 THEN P^4,p^6......
p^4=P^3.P
P^6=P^3.P^3
SO M IS A MULTIPLE OF P^2T

SUFF. HENCE B

If the prime nos P & T are the only prime factors of the integer M, is M a multiple of P^2T?

So by p^2T , I meant (p^2) * T. Is this how you arrived at your answer ? If yes, I did not get your explanation...can you elaborate a bit.

thanks

Skywalker wrote:If the prime nos P & T are the only prime factors of the integer M, is M a multiple of P^2T?

1) M has more than 9 +ve factors.
2) M is a multiple of P^3

OA B
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Does the OA says that it gives a clear NO as the answer ?

Main question: M/P^2T = int?

(1) Is clearly insufficient!

(2) This statement is saying that M/p^3 = int therefore m/p^2 has to be true.

Think of it like this: 8/2^3 ---> 8/2^2

Now, if m/p^2 = int. then t/P^2T = also int because remember that t is a factor of M

remember this rule: M is a multiple of its factors AND its factors factors. So 8 is a multiple of 4 AND 2 because 4/2=int.

Therefore (B) is sufficient!