If a positive odd integer n has 8 positive factors, how many distinct

The number of positive factors of an integer will be the product of the exponents of the prime factors when one is added to each one. [ie. \(x^3 y^2\) will have \((3+1)*(2+1) = 12\) factors].

As \(n\) has 8 positive factors, \(n\) must be in one of 3 forms when prime factorised: \(abc\), \(a^3 b\) or \(a^7\). In other words, \(n\) either has 1, 2 or 3 prime factors.

(1) n is less than 150

We know that \(n\) is odd, has 8 factors and is <150. \(n\) cannot be in the form \(a^7\). The smallest possible odd value for \(a^7\) would be \(3^7\) which is far greater than 150.

As \(n\) is odd, all of its factors will be odd. Therefore the smallest value \(n\) can be will be the product of the first four odd numbers: \(1*3*5*7 = 105\), and the next smallest would be \(1*3*5*9 = 135\). These both meet the characteristics of \(n\): <150, odd and 8 factors. However, they have a different number of distinct prime factors -> \(105: 3,5,7 (abc)\) & \(135: 3^3, 5 (a^3 b)\)

INSUFFICIENT

(2) n is greater than 100

Here the possibilities of \(n\) are vast and could take any of the three forms \(abc\), \(a^3 b\) or \(a^7\)

INSUFFICIENT

(1) + (2)

Using the statements together still does not give enough information to identify how many distinct prime factors \(n\) has. It is either 2 or 3.

INSUFFICIENT

Answer E