composite numbers are non-prime numbers
(1) Each prime factor of q is also a factor of p.
let a and b prime factors for q and p.
if \(p = a^5b^3\) and \(q = a^3b^2\) then \(\frac{p}{q}\) is integer
if \(p = a^3b^2\) and \(q = a^5b^3\) then \(\frac{p}{q}\) is non-integer
insufficient ---> the trick here is that the statement does't say that "
each factor of q is a factor of p"
(2) The total number of factors of q is 12, and that of p is 24
let a,b and c prime factors
if \(p = a^5b^3\) [no. of factors = (5+1)(3+1)=24] and \(q = a^3b^2\) [no. of factors =(3+1)(2+1)=12] then \(\frac{p}{q}\) is integer
if \(p = a^2b^3c^1\) [no. of factors = (2+1)(3+1)(1+1)=24] and \(q = a^3b^2\) [no. of factors = (3+1)(2+1)=12] then \(\frac{p}{q}\) is non-integer
insufficientcombining 1 & 2:
the same assumptions stated to test statement (2) shows that combining is still
insufficient.
E
_________________