# Showing that sum of first $998$ cubes is divisible by $999$

Mathematics Asked by Rebronja on November 29, 2020

I’ve got this assignment which reads:

Show that $$displaystyle sum_{k=0}^{998} k^{3}$$ is divisible by $$999$$.

Now I know that a number is divisible by $$999$$ if the sum of its three digit numbers is divisible by $$999$$. My guess would be to try and calculate the sum and check for the number if it is divisible, but I am guessing there has to be more elegant way to go about showing this. I was wondering if anyone can give me a hint or tell me in which direction I should think. Thanks in advance!

$$pmb{Hint}$$ : from $$a^3+b^3 = (a+b)(a^2-ab+b^2)$$ we have

$$k^3 + (999-k)^3 = 999(...)$$

now write $$bbox[yellow,border:2px solid red] {sum_{k=0}^{998} k^3 = sum_{k=1}^{998} k^3 = sum_{k=1}^{499}big[k^3 + (999-k)^3big]}$$

You alternately have the sum of cubes formula : $$sum_{k=1}^{998} k^3 = frac{998^2999^2}{4}$$, in fact this shows that $$999^2$$ divides the answer.

Also note that since $$a^n+b^n$$ is a multiple of $$a+b$$ for all odd positive integers $$n$$, it follows that $$999$$ would divide $$sum_{k=0}^{998} k^n$$ for all odd positive integers $$n$$.

Correct answer by Teresa Lisbon on November 29, 2020

This easily follows from Faulhaber's formula: sum of first $$998$$ cubes is $$frac{(998cdot 999)^2}{4}$$, which is divisible by $$999$$.

Answered by xyzzyz on November 29, 2020