+130070 2510 + 3610 = 6110
61 / 3 = 20(3) R 1
20 / 3 = 6(3) R 2
6 / 3 = 2(3) R 0
2 / 3 = 0(3) R 2
Read the red numbers from bottom to top :
20213 = 2(3)^3 + 0(2)^2 + 2(3)^1 + 1(3)^0 = 54 + 6 + 1 = 61
+1252 \(25_{10}+36_{10}=61_{10}\)
Now, we see that \(3^3\) is the greatest power that fits in \(61\), and it fits twice.
\(61-3^3\times 2=7\)
Next, \(3^2\) doesn't fit, but \(3^1\) fits twice.
\(7-3^1 \times 2=1\)
Finally, \(3^0\) fits one time.
Therefore, the answer is \(2021_3\) because they fit twice, then never, then twice, then once.
