Since they are natural numbers, if you knew the maximum number of digits any of them can have, say d, you could set a=1, b=10^d, c=10^2d, and you would be able to read the d-digit numbers directly. So, you use the first calculation to find the maximum number of digits, $(a,b,c)=(1,1,1)$. let $d =$ digits of this result $(x+y+z)$ Then, set $(a,b,c) = (1, 10^d, 10^{2d})$ Let the sum be $S$. Then $x =$ (first $d$ digits of $S$), $y = [d+1]$ to $2d$-digits of S, $z$ = $[2d+1$ to $3d]$ digits of $S$

Thus, we note that its posssible to solve for n natural numbers $x_1,x_2,...x_n$ with just $2$ questions.