Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Preform the addition.
2,442 5 + 404 5
![**Perform the addition.**
\[ 2,442_5 + 404_5 \]
In this problem, we are asked to perform the addition of two numbers in base-5 notation.
To solve this, follow these steps:
1. **Convert Both Numbers to Base-10:**
- \(2,442_5\):
\[2 \times 5^3 + 4 \times 5^2 + 4 \times 5^1 + 2 \times 5^0\]
\[= 2 \times 125 + 4 \times 25 + 4 \times 5 + 2 \times 1\]
\[= 250 + 100 + 20 + 2\]
\[= 372_{10}\]
- \(404_5\):
\[4 \times 5^2 + 0 \times 5^1 + 4 \times 5^0\]
\[= 4 \times 25 + 0 \times 5 + 4 \times 1\]
\[= 100 + 0 + 4\]
\[= 104_{10}\]
2. **Add the Base-10 Equivalents:**
\[372_{10} + 104_{10} = 476_{10}\]
3. **Convert the Sum Back to Base-5:**
- Divide 476 by 5 and keep track of the remainders:
\[
\begin{align*}
476 \div 5 &= 95 \text{ remainder } 1\\
95 \div 5 &= 19 \text{ remainder } 0\\
19 \div 5 &= 3 \text{ remainder } 4\\
3 \div 5 &= 0 \text{ remainder } 3\\
\end{align*}
\]
- Reading the remainders from bottom to top, we get \(3,401_5\).
Thus, the result of the addition in base-5 is:
\[ 2,442_5 + 404_5 = 3,401_5 \]](https://content.bartleby.com/qna-images/question/105fff1d-4db1-4cc6-9fff-4ea0e94f4214/f157d74d-d3f8-4167-9d80-537feb465941/x18tibf_processed.png)
Transcribed Image Text:**Perform the addition.**
\[ 2,442_5 + 404_5 \]
In this problem, we are asked to perform the addition of two numbers in base-5 notation.
To solve this, follow these steps:
1. **Convert Both Numbers to Base-10:**
- \(2,442_5\):
\[2 \times 5^3 + 4 \times 5^2 + 4 \times 5^1 + 2 \times 5^0\]
\[= 2 \times 125 + 4 \times 25 + 4 \times 5 + 2 \times 1\]
\[= 250 + 100 + 20 + 2\]
\[= 372_{10}\]
- \(404_5\):
\[4 \times 5^2 + 0 \times 5^1 + 4 \times 5^0\]
\[= 4 \times 25 + 0 \times 5 + 4 \times 1\]
\[= 100 + 0 + 4\]
\[= 104_{10}\]
2. **Add the Base-10 Equivalents:**
\[372_{10} + 104_{10} = 476_{10}\]
3. **Convert the Sum Back to Base-5:**
- Divide 476 by 5 and keep track of the remainders:
\[
\begin{align*}
476 \div 5 &= 95 \text{ remainder } 1\\
95 \div 5 &= 19 \text{ remainder } 0\\
19 \div 5 &= 3 \text{ remainder } 4\\
3 \div 5 &= 0 \text{ remainder } 3\\
\end{align*}
\]
- Reading the remainders from bottom to top, we get \(3,401_5\).
Thus, the result of the addition in base-5 is:
\[ 2,442_5 + 404_5 = 3,401_5 \]
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