9th root

Algebra Level 3

$\sqrt[9]{ 756680642578125}=x$

Solve for $x$ .

The answer is 45.

1 solution

Chew-Seong Cheong
Mar 29, 2018

Let $N=756680642578125$ . We note that $N$ which ends with 5 is divisible by 5. Since is digital sum is 72 is divisible by 9, $N$ is divisible by 9. If $x$ is a positive integer, let us assume that $N$ is divisible by $5^9$ and $9^9$ or $N$ is divisible by $45^9$ . We note that $\dfrac N{45^9} = 1$ $\implies N = 45^9$ $\implies x = \sqrt[9] N = \sqrt[9]{45^9}= \boxed{45}$ .

The statement "If $x$ is a positive integer we can assume that $N$ is divisible by $5^{9}$ or $9^{9}$ or $N$ is divisible by $45^{9}$ " is false. All we can say is " $N$ is divisible by $3^{9}$ and $5^{9}$ or $N$ is divisible by $15^{9}$ .

Krutarth Patel - 5 months, 3 weeks ago

Thanks. You are right. I will change the solution,

Chew-Seong Cheong - 5 months, 3 weeks ago

9th root

The answer is 45.

1 solution

0 pending reports