Sure, let's break down the addition of these two numbers in base 5 step by step.
1. Convert each base-5 number to base-10:
- For [tex]\(242_5\)[/tex]:
- The rightmost digit (2) is multiplied by [tex]\(5^0\)[/tex], which equals 2.
- The middle digit (4) is multiplied by [tex]\(5^1\)[/tex], which equals 20.
- The leftmost digit (2) is multiplied by [tex]\(5^2\)[/tex], which equals 50.
- Adding these values together, [tex]\(242_5 = 2 \cdot 5^0 + 4 \cdot 5^1 + 2 \cdot 5^2 = 2 + 20 + 50 = 72_{10}\)[/tex].
- For [tex]\(343_5\)[/tex]:
- The rightmost digit (3) is multiplied by [tex]\(5^0\)[/tex], which equals 3.
- The middle digit (4) is multiplied by [tex]\(5^1\)[/tex], which equals 20.
- The leftmost digit (3) is multiplied by [tex]\(5^2\)[/tex], which equals 75.
- Adding these values together, [tex]\(343_5 = 3 \cdot 5^0 + 4 \cdot 5^1 + 3 \cdot 5^2 = 3 + 20 + 75 = 98_{10}\)[/tex].
2. Add the base-10 equivalents:
[tex]\[
72_{10} + 98_{10} = 170_{10}
\][/tex]
3. Convert the sum from base-10 back to base-5:
- Divide 170 by 5:
- [tex]\(170 \div 5 = 34\)[/tex] with a remainder of 0, so the rightmost digit is 0.
- Divide 34 by 5:
- [tex]\(34 \div 5 = 6\)[/tex] with a remainder of 4, so the next digit is 4.
- Divide 6 by 5:
- [tex]\(6 \div 5 = 1\)[/tex] with a remainder of 1, so the next digit is 1.
- Lastly, 1 divided by 5 is less than 1, so the leftmost digit remains 1.
Therefore, [tex]\(170_{10}\)[/tex] converts to [tex]\(1140_5\)[/tex].
4. Conclusion:
The sum of [tex]\(242_5\)[/tex] and [tex]\(343_5\)[/tex] is [tex]\(1140_5\)[/tex].
[tex]\[
\begin{array}{r}
242_{\text {five }} \\
+343_{\text {five }} \\
\hline
1140_{\text {five }} \\
\end{array}
\][/tex]
The sum is [tex]\(1140_5\)[/tex].
