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!

Teresa Lisbon · Accepted Answer

$pmb{Hint}$ : 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 $n$.

xyzzyz · Answer

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

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

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