# Narcissistic cube asks who are we?

I am a five digit narcissistic cube who likes to display my root right upfront. I also have hidden it in the back of my four digit cube brother. Who are we?

$$32768 = 32^3$$ and $$5832 = 18^3$$

Process of finding them:

The cube root of a 5 digit number would have to be 2 digits. So say, $$AB^3 = ABCDE$$. Then, divide by $$AB$$ to get an approximation $$AB^2 \approx 1000$$. $$\sqrt{1000} \approx 31.6$$, so it makes sense to try slightly larger (since our approximation is a clear underestimate). We can try $$32^3 = 2^{15} = 32768$$ and sure enough it satisfies our condition. ($$33^3 = 1089 \cdot 33 > 1100 \cdot 32 = 35200$$ can be tested to not work, so this is our five digit cube).
To get the four-digit number, we know that $$32$$ appears at the end of the cube. The cubes have residues $$0, 1, 8, 7, 4, 5, 6, 3, 2, 9 \bmod 10$$, so we know that the cube root must be $$8 \bmod 10$$. Since our four-digit number must have a two-digit cube root, $$18^3$$ makes the most sense to try ($$28^3 > 784 \cdot 20 > 15000$$ would be way too large). So therefore we have our answer.

Afternote

Narcissistic numbers have an alternative meaning, so this was a little misleading for awhile.
Perhaps I should justify $$1089 \cdot 33 > 1100 \cdot 32$$. Well, $$1100$$ is roughly 1% larger than 1089 (it's 11 more), whereas $$32$$ is roughly 3% smaller than $$33$$.

• You just beat me to the writeup. :) – Rubio May 8 at 20:04
• Alternative proof: 1089*33=(1100-11)*(32+1)=1100*32+1100*1-11*32-11*1 – Acccumulation May 9 at 17:52