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