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Convert and solve the following addition problem in base 6:

[tex]\( 1,005_6 + 3,142_6 \)[/tex]


Sagot :

To solve the sum of the numbers [tex]\(1,005_6 + 3,142_6\)[/tex] in base 6, let's follow these steps:

1. Convert each base 6 number to base 10:
- For the number [tex]\(1,005_6\)[/tex]:
- The rightmost digit is the [tex]\(0\)[/tex]th position and so forth to the left.
- [tex]\(1 \cdot 6^3 + 0 \cdot 6^2 + 0 \cdot 6^1 + 5 \cdot 6^0\)[/tex]
- [tex]\(1 \cdot 216 + 0 \cdot 36 + 0 \cdot 6 + 5 \cdot 1\)[/tex]
- [tex]\(216 + 0 + 0 + 5 = 221_{10}\)[/tex]
- For the number [tex]\(3,142_6\)[/tex]:
- Applying the same positional process:
- [tex]\(3 \cdot 6^3 + 1 \cdot 6^2 + 4 \cdot 6^1 + 2 \cdot 6^0\)[/tex]
- [tex]\(3 \cdot 216 + 1 \cdot 36 + 4 \cdot 6 + 2 \cdot 1\)[/tex]
- [tex]\(648 + 36 + 24 + 2 = 710_{10}\)[/tex]

2. Add the numbers in base 10:
- [tex]\(221_{10} + 710_{10} = 931_{10}\)[/tex]

3. Convert the sum back to base 6:
- To convert 931 from base 10 to base 6, we repeatedly divide the number by 6 and keep track of the remainders.
- [tex]\(931 \div 6 = 155\)[/tex] remainder [tex]\(1\)[/tex]
- [tex]\(155 \div 6 = 25\)[/tex] remainder [tex]\(5\)[/tex]
- [tex]\(25 \div 6 = 4\)[/tex] remainder [tex]\(1\)[/tex]
- [tex]\(4 \div 6 = 0\)[/tex] remainder [tex]\(4\)[/tex]
- Reading the remainders in reverse gives us [tex]\(4151_6\)[/tex]

Therefore, the sum of [tex]\(1,005_6 + 3,142_6\)[/tex] is [tex]\(4,151_6\)[/tex].
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