What is the largest $n$ such that $a = 2^{306} \cdot 3^{340}$ is a perfect $n$th power?

Hint: it will be \(\sqrt{2^{306}} \times \sqrt {3^{340}}\), that way \(n^2 = {2^{306}} \times {3^{340}}\)

No we are asked to find the $n$ for which is the square of that. We need the largest *n* th root. 

n=  Greatest Common Divisor of  306  and  340 

306=2×3^2×17 

340=2^2×5×17 

n=GCD(306,340)=2×17=34 

a=2^306×3^340=2^(9×34)×3^(10×34)=(2^9)^34×(3^10)^34 

a=(2^9×3^10)^34=30,233,088^34

n=34