this question is from Thomas Koshy's book called "Discrete Mathematics With Applications":
Any idea how to do this question? I can tell that the base of the system is at least 3 (since we are adding two 2-digit numbers and receiving a 3-digit number as a result). But I'm stuck on what to do next.
Actually, even in binary, it’s possible to add two digit numbers to get a three digit number, e.g. $11+10=101$. What says more convincingly that the base is at least $3$ is that there are $3$ different symbols (and presumably the Venusians didn’t do anything so cruel as to use two different symbols for the same digit).
As is the case in the base $b$ number system, when adding two digits from $0, \ldots, b-1$, the most you can get is $2b-2$, which carries a $1$ to the next digit (to carry a $2$ or more, the sum would have to be at least $2b$). Even if a $1$ is carried into this sum, the sum is still at most $2b-1$, so only at most $1$ can be carried to the next digit.
So, because the third ($b^2$) digit appears from the sum of two two digit numbers, it must be the result of a $1$ carried from summing the second ($b$) digits. Thus, the square symbol is $1$.
This means the swirls are greater than $1$. Adding two of them yields a $1$ in the first digit ($b^0$), so they must carry a $1$ to the second digit ($b$). This also implies that the base is odd, as it equals two swirls plus one.
So, in the second ($b$) digit, we are summing $1+1+1$ (one from the carry). We know this initiates a carry of its own (to create the three digit number), so the base is at most $3$. Therefore, the base is exactly $3$, and the star-like shape is $0$, which makes the swirls $2$. That is, the full sum, in base $3$, is $12+12=101$.