Tmoni26 wrote:

I am probably wrong but I will my solution out there and someone will tell me where I fell down

So 27^3 - 9^3 - 3^6, get them into bases of 3s

27^3 = (3^3)^3

9^3 = (3^2)^3 = (3^3)^2

3^6 = (3^3)^2

So we have (3^3)^3 -(3^3)^2 - (3^3)^2

Factor out 3^3, we have 3^3 (3^3 - 3^2 - 3^2) ----> 3^3 (27 -9-9) ----> (27) (9) -->

Highest prime factor is 3

Hi Tmoni,

Your approach was right, but where you "fell down" is when you factored out the \(3^3\).

Factoring out \(3^3\) from \((3^3)^3-(3^3)^2-(3^3)^2\) will give you \((3^3)[(3^3)^2-(3^3)-(3^3)]\)

Consider replacing \(3^3\) with \(x\), then the expression would look like \(x^3-x^2-x^2\)

Now if you factor out an \(x\) what happens? You get \(x(x^2-x-x)\)

Remember, \(x^2 = x*x\), so \((3^3)^2 = (3^3)*(3^3)\). When you factor out one \(3^3\), you still have one \(3^3\) left over, not \(3^2\)

So in fact you could have factored out \((3^3)^2\) and gotten \((3^3)^2[3^3-1-1] = (3^3)^2*[25] = 3^6*5^2\)

Now the expression is broken down into its prime factors and we can see that the greatest prime factor is 5.

Answer: C

Does that help?

Cheers,

_________________

Dave de Koos

GMAT aficionado