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

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

Jun 11, 2022

#1
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Hint: it will be $$\sqrt{2^{306}} \times \sqrt {3^{340}}$$, that way $$n^2 = {2^{306}} \times {3^{340}}$$

Jun 11, 2022
#2
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No we are asked to find the $n$ for which is the square of that. We need the largest *n* th root.

Jun 12, 2022
#3
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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

Jun 12, 2022