Express the sum of 112012 three and 2112 three in base six.

Question

ScanSolve AI Tutor · Accepted Answer

To express the sum in base six, we first convert each number to base 10, then add them, and finally convert the sum to base six.

Convert $112012_{\text{three}}$ to base 10.
$112012_{\text{three}} = 1 \times 3^5 + 1 \times 3^4 + 2 \times 3^3 + 0 \times 3^2 + 1 \times 3^1 + 2 \times 3^0$ $= 1 \times 243 + 1 \times 81 + 2 \times 27 + 0 \times 9 + 1 \times 3 + 2 \times 1$ $= 243 + 81 + 54 + 0 + 3 + 2$ $= 383_{\text{ten}}$
Convert $21(21)_{\text{four}}$ to base 10.
Assuming $21(21)_{\text{four}}$ means $2121_{\text{four}}$ (as is common for number base notation where digits are listed sequentially).
$2121_{\text{four}} = 2 \times 4^3 + 1 \times 4^2 + 2 \times 4^1 + 1 \times 4^0$ $= 2 \times 64 + 1 \times 16 + 2 \times 4 + 1 \times 1$ $= 128 + 16 + 8 + 1$ $= 153_{\text{ten}}$
Find the sum of the two numbers in base 10.
$\text{Sum} = 383_{\text{ten}} + 153_{\text{ten}}$ $\text{Sum} = 536_{\text{ten}}$
Convert $536_{\text{ten}}$ to base six.
We use repeated division by 6 and record the remainders.
$536 \div 6 = 89 \text{ remainder } 2$ $89 \div 6 = 14 \text{ remainder } 5$ $14 \div 6 = 2 \text{ remainder } 2$ $2 \div 6 = 0 \text{ remainder } 2$
Reading the remainders from bottom to top, we get $2252_{\text{six}}$ .

The sum of $112012_{\text{three}}$ and $21(21)_{\text{four}}$ in base six is:

Answer