21. Determine what is wrong with the following addition problem. 23 six + 43six 66six
21. Determine what is wrong with the following addition problem. 23 six + 43six 66six
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Question:**
21. Determine what is wrong with the following addition problem.
\[
\begin{array}{c@{}c@{}c}
& 2 & 3_{\text{six}} \\
+ & 4 & 3_{\text{six}} \\
\hline
& 6 & 6_{\text{six}} \\
\end{array}
\]
**Explanation:**
This problem involves adding two numbers in base six and checking if the result is correct. The numbers being added are \(23_{\text{six}}\) and \(43_{\text{six}}\).
**Steps for Addition in Base Six:**
1. **Align the Numbers Vertically:**
- Align the numbers \(23_{\text{six}}\) and \(43_{\text{six}}\) with the ones place lined up.
2. **Add the Digits from Right to Left:**
- **Units/Ones Place:** Add \(3 + 3 = 6\).
- In base six, the digit 6 does not exist. It should be converted, resulting in \(10_{\text{six}}\). This means you write down 0 and carry over 1.
- **Tens Place:** Add \(2 + 4 = 6\), and don’t forget to include the carried over 1, totaling 7.
- Again, in base six, 7 does not exist, so this translates to \(11_{\text{six}}\). You write down 1 and carry over 1.
3. **Double-Check the Result:**
- After adjusting for carries, the solution should be \(110_{\text{six}}\), not \(66_{\text{six}}\) as stated.
**Error:**
The original calculation displayed an incorrect result. The proper addition process in base six reveals that the correct answer should be \(110_{\text{six}}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ea3e25c-1ac2-48ba-80de-6b7a587bd404%2Fd6d4d15e-4a23-4221-8d14-b72437485c92%2F916rj76_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
21. Determine what is wrong with the following addition problem.
\[
\begin{array}{c@{}c@{}c}
& 2 & 3_{\text{six}} \\
+ & 4 & 3_{\text{six}} \\
\hline
& 6 & 6_{\text{six}} \\
\end{array}
\]
**Explanation:**
This problem involves adding two numbers in base six and checking if the result is correct. The numbers being added are \(23_{\text{six}}\) and \(43_{\text{six}}\).
**Steps for Addition in Base Six:**
1. **Align the Numbers Vertically:**
- Align the numbers \(23_{\text{six}}\) and \(43_{\text{six}}\) with the ones place lined up.
2. **Add the Digits from Right to Left:**
- **Units/Ones Place:** Add \(3 + 3 = 6\).
- In base six, the digit 6 does not exist. It should be converted, resulting in \(10_{\text{six}}\). This means you write down 0 and carry over 1.
- **Tens Place:** Add \(2 + 4 = 6\), and don’t forget to include the carried over 1, totaling 7.
- Again, in base six, 7 does not exist, so this translates to \(11_{\text{six}}\). You write down 1 and carry over 1.
3. **Double-Check the Result:**
- After adjusting for carries, the solution should be \(110_{\text{six}}\), not \(66_{\text{six}}\) as stated.
**Error:**
The original calculation displayed an incorrect result. The proper addition process in base six reveals that the correct answer should be \(110_{\text{six}}\).
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