To solve the problem of adding the numbers [tex]\(14_{\text{six}}\)[/tex] and [tex]\(15_{\text{six}}\)[/tex] in base-6, follow these steps:
1. Convert Each Number from Base-6 to Base-10:
- The number [tex]\(14_{\text{six}}\)[/tex] can be converted to base-10 as follows:
[tex]\[
1 \times 6^1 + 4 \times 6^0 = 6 + 4 = 10_{\text{ten}}
\][/tex]
- The number [tex]\(15_{\text{six}}\)[/tex] can be converted to base-10 as follows:
[tex]\[
1 \times 6^1 + 5 \times 6^0 = 6 + 5 = 11_{\text{ten}}
\][/tex]
2. Perform the Addition of the Base-10 Numbers:
- Now we add [tex]\(10_{\text{ten}}\)[/tex] and [tex]\(11_{\text{ten}}\)[/tex]:
[tex]\[
10 + 11 = 21_{\text{ten}}
\][/tex]
3. Convert the Sum from Base-10 back to Base-6:
- To convert [tex]\(21_{\text{ten}}\)[/tex] into base-6, we find the remainders when dividing by 6:
- [tex]\(21 \div 6 = 3\)[/tex] with a remainder of [tex]\(3\)[/tex], so the least significant digit is [tex]\(3\)[/tex].
- Next, take the quotient [tex]\(3\)[/tex] and divide by [tex]\(6\)[/tex]:
[tex]\[
3 \div 6 = 0 \text{ (quotient) with a remainder of 3}
\][/tex]
- Thus, [tex]\(21_{\text{ten}}\)[/tex] converts to [tex]\(33_{\text{six}}\)[/tex] (read the remainders from top to bottom).
Therefore, the final result of adding [tex]\(14_{\text{six}}\)[/tex] and [tex]\(15_{\text{six}}\)[/tex] is:
[tex]\[
33_{\text{six}}
\][/tex]
