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.
Regards
Samir
