About the cube's digit sum 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.
 A: 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 $a_1<a_2<\dots <a_{100}$ such that if you add three of these terms in different combinations then you get different results.
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.
