Of course! Let's solve the problem step-by-step:
We need to find the sum of two numbers represented in base-6 (also known as senary or hexary). The numbers given are [tex]\(45_{\text{six}}\)[/tex] and [tex]\(14_{\text{six}}\)[/tex].
### Step 1: Convert each number from base-6 to decimal (base-10)
First, let's convert [tex]\(45_{\text{six}}\)[/tex] to its decimal (base-10) equivalent:
- In base-6, the digit 4 is in the "6^1" place and the digit 5 is in the "6^0" place.
So, [tex]\(45_{\text{six}} = 4 \cdot 6^1 + 5 \cdot 6^0 = 4 \cdot 6 + 5 \cdot 1 = 24 + 5 = 29_{\text{ten}}\)[/tex].
Next, let's convert [tex]\(14_{\text{six}}\)[/tex] to its decimal (base-10) equivalent:
- In base-6, the digit 1 is in the "6^1" place and the digit 4 is in the "6^0" place.
So, [tex]\(14_{\text{six}} = 1 \cdot 6^1 + 4 \cdot 6^0 = 1 \cdot 6 + 4 \cdot 1 = 6 + 4 = 10_{\text{ten}}\)[/tex].
### Step 2: Perform the addition in decimal (base-10)
Now we add the two decimal numbers obtained:
[tex]\[ 29_{\text{ten}} + 10_{\text{ten}} = 39_{\text{ten}} \][/tex]
### Step 3: Convert the sum from decimal (base-10) back to base-6
Next, we convert [tex]\(39_{\text{ten}}\)[/tex] to its base-6 equivalent:
- To convert from decimal to base-6, we divide the number by 6 and keep track of the remainders.
[tex]\( 39 \div 6 = 6 \)[/tex] with a remainder of 3 (3 is in the "6^0" place).
[tex]\( 6 \div 6 = 1 \)[/tex] with a remainder of 0 (0 is in the "6^1" place).
[tex]\( 1 \div 6 = 0 \)[/tex] with a remainder of 1 (1 is in the "6^2" place).
Reading the remainders from top to bottom, we get [tex]\(103_{\text{six}}\)[/tex].
### Conclusion
Therefore, the sum of [tex]\(45_{\text{six}}\)[/tex] and [tex]\(14_{\text{six}}\)[/tex] is:
[tex]\[ 45_{\text{six}} + 14_{\text{six}} = 103_{\text{six}} \][/tex]
Thus, the final answer is [tex]\(\boxed{103}\)[/tex].
