Find the smallest positive integer n

[anemone]

Find the smallest positive integer $n $ for which $n^{16}$ exceeds $16^{18}$.

 

[kaliprasad]

Find the smallest positive integer $n $ for which $n^{16}$ exceeds $16^{18}$.

23 as below

Spoiler

n^16 > 16^18
or n^8 > 16^9 or 4^18
or n^4 > 4^9 or 2^18
or n^2 > 2^9 or 512
n = 22 => n^2 = 484 and n = 23 => n^2 = 529

 

[soroban]

Hello, anemone!

$$\text{Find the smallest positive integer }n$$
. . $$\text{ for which }n^{16}\text{ exceeds }16^{18}.$$

kaliprasad is correct.
I used a different approach.

Spoiler

We want: .$$n^{16} \;> \; 16^{18}$$

. . . . . . . .$$n^{16} \;>\; (2^4)^{18}$$

. . . . . . . .$$n^{16} \;>\;2^{72}$$

. . . . . . . . . $$n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}$$

. . . . . . . . . $$n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}$$

. . . . . . . . . $$n \;>\; 16\sqrt{2} \;=\; 22.627417$$

Therefore: . $$n \;=\;23$$

 

[kaliprasad]

Hello, anemone!


kaliprasad is correct.
I used a different approach.

Spoiler

We want: .$$n^{16} \;> \; 16^{18}$$

. . . . . . . .$$n^{16} \;>\; (2^4)^{18}$$

. . . . . . . .$$n^{16} \;>\;2^{72}$$

. . . . . . . . . $$n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}$$

. . . . . . . . . $$n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}$$

. . . . . . . . . $$n \;>\; 16\sqrt{2} \;=\; 22.627417$$

Therefore: . $$n \;=\;23$$

above approach is more straight forward

 

[CellCoree]

The smallest positive integer $n$ for which $n^{16}$ exceeds $16^{18}$ is $17$. This can be determined by setting up the inequality $n^{16} > 16^{18}$ and taking the 16th root of both sides, giving $n > 16^{18/16} = 16^{9/8}$. Since $16^{9/8} \approx 17.34$, the smallest integer that satisfies this inequality is $n = 17$. This result can also be verified by computing $17^{16} \approx 2.16 \times 10^{22}$ and $16^{18} \approx 4.61 \times 10^{22}$, showing that $n^{16}$ indeed exceeds $16^{18}$ for $n = 17$.