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!

If you are allowed to use the formula $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$, then the answer is straightforward. — angryavian, Sep 02 '20 at 04:19
See https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Sep 02 '20 at 04:25

Sarvesh Ravichandran Iyer · Accepted Answer · 2020-09-02T05:01:35.137

15

$\pmb{Hint}$ : 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 $n$.

edited Sep 02 '20 at 05:01

answered Sep 02 '20 at 04:21

Sarvesh Ravichandran Iyer

76,105

Thank you very much for comprehensive answer! – Rebronja Sep 02 '20 at 05:15
1

@Rebronja You are welcome, little did I expect the reception this answer got. – Sarvesh Ravichandran Iyer Sep 02 '20 at 05:16
@TeresaLisbon rep and The Algorithm™ work in mysterious ways. – qwr Sep 02 '20 at 05:30
Maybe because it's more than a "hint" ;-) – Déjà vu Sep 02 '20 at 05:35
1

@e2-e4 Cannot deny that, if I look back at it now! – Sarvesh Ravichandran Iyer Sep 02 '20 at 05:37

score 4 · Answer 2 · answered Sep 02 '20 at 04:18

4

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

answered Sep 02 '20 at 04:18

xyzzyz

7,612

score 0 · Answer 3 · 2021-08-18T09:15:44.167

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 \ \ . $

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

3 Answers3