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The digit sum of a natural number equals 100. How can I prove that the sum of $n^3$ can be equal to 1000000? I think it has something to do with binomial coefficients.

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    Have you tried a number with $11$ nines with a lot of irregular-length sequences of zeroes in-between (and a $1$ somewhere)? How about a hundred $1$'s? Might work. – 2017-01-23
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    Hm.. there are only 190,569,292 ways to partition 100, so we could take those which consist of digit partitions, permutate, generate their numbers, cube them and check for digit sum 1M. Or maybe not. – 2017-01-23
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    Hmm, I think I found one. It is 11010001... but backwards. Number 1 repeats every 1, 2, 4, 8, 16... th digit. But I need to prove it... – 2017-01-23
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    That one works for $n^2$, you need the gaps to increase slightly larger. – 2017-01-23

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We are going to work with with a number that only has $1$ and $0$ in its decimal representation.

We can write your number as $\sum\limits_{i=1}^{100}10^{a_i}$.

Notice that $(\sum\limits_{i=1}^{100} 10^{a_i})^3=6\sum\limits_{1\leq i< j < k\leq 100}10^{a_i+a_j+a_k}+3\sum\limits_{i\neq j}10^{a_i+2a_j}+\sum\limits_{i=1}^{100}10^{3a_i}$.

If all of the powers of $10$ inside each summand are different then the sum of digits is just going to be $6\binom{100}{3}+2\times3\binom{100}{2}+100=100,000$.

So all that we need to do is prove that there exists a set of $100$ positive integers $a_1

it is easy to prove by induction that such a set exists for any size. If the set $\{a_1,a_2,\dots,a_n\}$ works, then you can find an $a_{n+1}$ such that $\{a_1,a_2,\dots,a_n,a_{n+1}\}$ also works. We just need $a_{n+1}$ to be large enough. It suffices for $a_{n+1}$ to be larger than or equal to $3a_n$, because that way you can deduce how many copies of $a_{n+1}$ appear in $a+b+c$ just by seeing if the sum is larger than $a_{n+1}$, if it is larger than $2a_{n+1}$ and if it is equal to $3a_{n+1}$.

So for example the sequence $1,3,9,27,\dots $ does the trick (one can notice this also by looking at the base $3$ expansion of a number).

In conclusion, the number $\sum\limits_{i=1}^{100} 10^{3^i}$ does the trick.