Answer :
To solve the problem of adding the two numbers [tex]\(221_{\text{six}}\)[/tex] and [tex]\(45_{\text{six}}\)[/tex] and obtaining the result in base-6, we will break it down into several steps.
### Step 1: Convert [tex]\(221_{\text{six}}\)[/tex] to Decimal (Base-10)
The number [tex]\(221_{\text{six}}\)[/tex] is in base-6. To convert it to decimal:
[tex]\[ 221_{\text{six}} = 2 \times 6^2 + 2 \times 6^1 + 1 \times 6^0 \][/tex]
[tex]\[ = 2 \times 36 + 2 \times 6 + 1 \times 1 \][/tex]
[tex]\[ = 72 + 12 + 1 \][/tex]
[tex]\[ = 85_{\text{ten}} \][/tex]
### Step 2: Convert [tex]\(45_{\text{six}}\)[/tex] to Decimal (Base-10)
The number [tex]\(45_{\text{six}}\)[/tex] is also in base-6. To convert it to decimal:
[tex]\[ 45_{\text{six}} = 4 \times 6^1 + 5 \times 6^0 \][/tex]
[tex]\[ = 4 \times 6 + 5 \times 1 \][/tex]
[tex]\[ = 24 + 5 \][/tex]
[tex]\[ = 29_{\text{ten}} \][/tex]
### Step 3: Add the Decimal Numbers
Now, we add the two numbers we have converted to decimal:
[tex]\[ 85_{\text{ten}} + 29_{\text{ten}} = 114_{\text{ten}} \][/tex]
### Step 4: Convert Decimal Sum Back to Base-6
To convert [tex]\(114_{\text{ten}}\)[/tex] back to base-6, we repeatedly divide by 6 and record the remainders:
[tex]\[ 114 \div 6 = 19 \text{ remainder } 0 \][/tex]
[tex]\[ 19 \div 6 = 3 \text{ remainder } 1 \][/tex]
[tex]\[ 3 \div 6 = 0 \text{ remainder } 3 \][/tex]
Reading the remainders from bottom to top, [tex]\(114_{\text{ten}}\)[/tex] converts to:
[tex]\[ 310_{\text{six}} \][/tex]
### Conclusion
Thus, the sum of [tex]\(221_{\text{six}}\)[/tex] and [tex]\(45_{\text{six}}\)[/tex] is:
[tex]\[ \boxed{310_{\text{six}}} \][/tex]
### Step 1: Convert [tex]\(221_{\text{six}}\)[/tex] to Decimal (Base-10)
The number [tex]\(221_{\text{six}}\)[/tex] is in base-6. To convert it to decimal:
[tex]\[ 221_{\text{six}} = 2 \times 6^2 + 2 \times 6^1 + 1 \times 6^0 \][/tex]
[tex]\[ = 2 \times 36 + 2 \times 6 + 1 \times 1 \][/tex]
[tex]\[ = 72 + 12 + 1 \][/tex]
[tex]\[ = 85_{\text{ten}} \][/tex]
### Step 2: Convert [tex]\(45_{\text{six}}\)[/tex] to Decimal (Base-10)
The number [tex]\(45_{\text{six}}\)[/tex] is also in base-6. To convert it to decimal:
[tex]\[ 45_{\text{six}} = 4 \times 6^1 + 5 \times 6^0 \][/tex]
[tex]\[ = 4 \times 6 + 5 \times 1 \][/tex]
[tex]\[ = 24 + 5 \][/tex]
[tex]\[ = 29_{\text{ten}} \][/tex]
### Step 3: Add the Decimal Numbers
Now, we add the two numbers we have converted to decimal:
[tex]\[ 85_{\text{ten}} + 29_{\text{ten}} = 114_{\text{ten}} \][/tex]
### Step 4: Convert Decimal Sum Back to Base-6
To convert [tex]\(114_{\text{ten}}\)[/tex] back to base-6, we repeatedly divide by 6 and record the remainders:
[tex]\[ 114 \div 6 = 19 \text{ remainder } 0 \][/tex]
[tex]\[ 19 \div 6 = 3 \text{ remainder } 1 \][/tex]
[tex]\[ 3 \div 6 = 0 \text{ remainder } 3 \][/tex]
Reading the remainders from bottom to top, [tex]\(114_{\text{ten}}\)[/tex] converts to:
[tex]\[ 310_{\text{six}} \][/tex]
### Conclusion
Thus, the sum of [tex]\(221_{\text{six}}\)[/tex] and [tex]\(45_{\text{six}}\)[/tex] is:
[tex]\[ \boxed{310_{\text{six}}} \][/tex]
