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

Question

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!

Answer 1

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

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

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

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 $n$.

Answer 2

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

Answer 3

Starting with the pair $ \ 1^3 \ \ \text{and} \ \ 998^3 \ \ , $ we have

$$ 1^3 \ + \ (999 - 1)^3 \ \ = \ \ 1^3 \ + \ (999^3 \ - \ 3·999^2 · 1 \ + \ 3·999·1^2 \ - \ 1^3) $$ $$ = \ \ 999 \ · \ (999^2 \ - \ 3·999 \ + \ 3) \ \ . $$

Similarly. $$ n^3 \ + \ (999 - n)^3 \ \ = \ \ n^3 \ + \ (999^3 \ - \ 3·999^2 · n \ + \ 3·999·n^2 \ - \ n^3) $$ $$ = \ \ 999 \ · \ (999^2 \ - \ 3·999·n \ + \ 3n^2) \ \ . $$

The "middle" pair is $ \ n = 499 \ , \ (999-n) = 500 \ \ , $ so all of the integer-cubes are covered in this way.

Teresa Lisbon's hint is related, but this applies a binomial expansion, rather than the difference-of-two-cubes.

$ \ \ $

ADDENDUM: This can be generalized. If $ \ M \ $ and $ \ p \ $ are odd integers, then $$ \sum_{k \ = \ 1}^{M-1} \ k^p \ \ \text{is divisible by M} \ \ , $$ since $$ n^p \ + \ (M-n)^p \ \ = \ \ n^p \ + \ [ \ M^p \ - \ \binom{p}{1}·M^{p-1}·n \ + \ \binom{p}{2}·M^{p-2}·n^2 \ + \ \ldots $$ $$ - \ \binom{p}{p-2}·M^2·n^{p-2} \ + \ \binom{p}{p-1}·M·n^{p-1} \ - \ n^p \ ] \ \ , $$ with all integers from $ \ 1 \ \ \text{to} \ \ (M-1) \ \ $ covered since the middle-pair is $ \ \frac{M-1}{2} \ \ , \ \ \frac{M+1}{2} \ \ . $

In fact, it is not necessary to use a "run" of consecutive pairings: the sum of any subset of such pairs $ \ n^p \ , \ (M - n)^p \ $ is divisible by $ \ M \ \ . $