IMO D.

If a number has only 2 prime numbers as factors and has more than 9 factors - few of other factors will definitely be multiple of these 2 prime numbers. (LET ME KNOW IF I AM WRONG)

Stmt 2 - m is a multiple of p^3 and a mulitple of t so it would be multiple of p^2*t

Ya I agree with kajcha,the ans should be D

stmt 1: m has 9 +ve factors
now all the factors of m would be all the possible products of its prime factors i.e p,t

now we will have 9 possible factors

the factors are p,t

p^2, pt , t^2

p^2*t, t^2*p ....

so m is a multiple of p^2*t

i.e. SUFF

stmt 2: p^3 is a factor of m so will be p^2 hence p^2*t

can you guys please explain why 1st stat is true?say prime factors are 5 and 7 and other factors are all series of 7 expect 35 like 14,21,28,42,49 then why is it still necessary for that number to be a multiple of 5*5=25..
Thanks and sorry for ignorance !!

Good point gmatguy... In that case ans should be B

What's the OA?

Hi gmatguy/Kajcha,
You are correct but if you see my post I have mentioned these, to find all the factors of a nos we need its prime factors, so all the other factors will be all posible multiples of the prime factors

in ur example u mentioned factors as 5 & 7, now stmt 1 says that the nos has 9 +ve factors, now s you said these other factors would be constructed using higher powers of the prime factors 5 & 7

i.e here we need to construct total 7 nos by using 5 & 7 as building blocks i.e the prime factors.

so the factors are 5,7 taken 1 at a time i.e 2 factors

5*7, 5*5,7*7 taken two at a time so we have a total of 5

5*5*5, 5*5*7,7*7*5, 7*7*7 which gives us total 9 factors

Here consider any one as p , we will have p^2*t present in the factors.
so A is SUFF

now you also mentioned that factors of 7 like 14,21 etc will also be factors

now as the nos has prime factors as p & t only so we have

the nos can be something like p^a * t ^b

where a,b can be some powers starting from 1

now in ur e.g we can say that the nos is 5^a*7^b

so u quoted that 14,21 ..will be its factors

so we have 5^a*7^b / 7*2 (14 =7*2)

here 7 will cancel out (divide) from the denominator, but will how you remove 2 here, it cant divde 7 or 5, so the ans will not be integer,so 14 is not a factor

I hope this helps.

Geeez.. I need to think through the question first...

The question says only 2 prime numbers are factors... In the example discussed by gmatguy.. if 14 and 21 are factors then 2 and 3 are factors too and that is against the question..

thanks Samir for clarification....

Anytime Kajcha

Sorry for the delayed response.

The OA answer is B.

A is not sufficient because. the nine factors could be

p, t, pt, pt^2, pt^3, pt^4.....


p^2 t need not be one of them.

Hi karthick451,
Thanks for the OA, should have thought that, but anyways it was a good question, taught a useful lesson.

I really liked this question. So I am bringing it up again.

Lets say p & t are the only prime factors of a number. So the guaranteed minimal set of factors is::

{1, p, t, pt} (thats 4 factors)
which is the product of the sets {1, p} * {1, t}

Going up one notch the sets grows to::
{1, p, t, t^2, pt, pt^2} (thats 6 factors)
here t is repeated twice, but not p.

This is the product of the sets {1, p} and {1, t, t^2}
And so forth.

Let me know if you agree.