The question asks to find the sum of digits in the product of:
$11 * 101 * 10,001 * 100,000,001 *10,000,000,000,000,001$.
Now I've somewhat worked out a potential (?) solution with factoring each term to a base 10 binomial:
$(10^1 + 1) * (10^2 + 1) * (10^4 + 1) * (10^8 + 1) * (10^{16} + 1)$.
...Noticing that all exponents were base 2.
i see from the multiplication of the $10^1$ term $ * 10^2$ term produces $10^3 + 10^2 + 10^1 + 1(10^0)$.
Seeing that the degree was determined by multiplying the two leading terms, would it be correct to assume that would happen for $(10^3 + 10^2...) * (10^4 + 1) = 10^7 + 10^6 + 10^5...$ ?
And would this pattern carry on all the way to the last binomial, where i would multiply all those base 10s in the five binomials to get $10^{31} + 10^{30}$ going all the way down?
And thus the product would be 32 ones and have a digitsum of 32?
Method seems too weird and thus sketchy for me. Any other potential concrete solutions?
Since each positive integer can be written as a sum of powers of $2$ in exactly one way, when expanded out you get the sum of $32$ different powers of $10$, from $10^0$ to $10^{31}$. So the result is $\sum_{j=0}^{31} 10^j = 11111111111111111111111111111111$.