Dear Uncle Colin,

I’m told that 19,683 is the cube of an integer. How would I figure out which one?

- Not A Problem I Expected, Really

Hi, NAPIER, and thanks for your message!

There are several approaches here, and I’m sure The Mathematical Ninja has more still.

First, the “obvious” way: modulo 10, all of the cubes are different, and in particular, $7^3 \pmod{10}=3$ – so our cube root ends in a 7.

Also, 19,683 is between $20^3=8,000$ and $30^3=27,000$, so the cube root must be 27.

### Digital root

The digits of 19,683 sum to 27, which means it’s a multiple of 3; knowing it’s a cube, it must be a multiple of 27 (and the cube root a multiple of 3).

So if $n^3 = 19,683$ and $n = 3a$, then $27a^3 = 19,683$, and $a^3 = 729$. That makes $a=9$, so $n=27$.

### Whoosh

Ah, sensei, I’ve been expecting you.

I really am losing my touch.

Never let it be said.

$\ln(20,000)=3\ln(10)+\ln(20)$ and $\ln(20)\approx 3$. The natural log of the cube root of our number is $1 + \ln(10)$, so $n \approx 10\mathrm{e} \approx 27$.

Time was, you’d have done error analysis on that.

Don’t push it, my friend. Don’t push it.

Hope that helps!

- Uncle Colin