ykaiim wrote:

If K is a positive integer, how many different prime numbers are factors of the expression K^2?

1) Three different prime numbers are factors of 4K^4.

2) Three different prime numbers are factors of 4K.

First of all

k^x (where

x is an integer

\geq{1}) will have as many different prime factors as integer

k. Exponentiation doesn't "produce" primes.

Next:

p^y*k^x (where

p is a prime and

y is an integer

\geq{1}) will have

as many different prime factors as integer k if k already has p as a factor OR one more factor than kif k doesn't have p as a factor .

So, the question basically is: how many different prime numbers are factors of

k?

(1) Three different prime numbers are factors of

4k^4 --> if

k itself has 2 as a factor (eg 30) than it's total # of primes is 3

but if k doesn't have 2 as a factor (eg 15) than it's total # of primes is 2. Not sufficient.

(2) Three different prime numbers are factors of

4k --> the same as above: if

k itself has 2 as a factor (eg 30) than it's total # of primes is 3

but if k doesn't have 2 as a factor (eg 15) than it's total # of primes is 2. Not sufficient.

(1)+(2) Nothing new, k can be 30 (or any other number with 3 different primes, out of which one factor is 2) than the answer is 3

or k can be 15 (or any other number with 2 different primes, out of which no factor is 2) than the answer is 2. Not sufficient.

Answer: E.

Another method that worked for me was to find integer K-if we find K we can find its factors.

Both statements have 4 as a multiplier which is not divisible by any of the prime factors on the LHS. Hence K cannot be found as an integer so E. Bunuel, does this logic make sense? Please enlighten!