The remainder when n is divided by 3 is 1 and the remainder when (n+1) is divided by 2 is 1. The remainder when (n-1) is divided by 6 is :

Options

A. 2
B. 3
C. 5
D. none of (a), (b), (c)

Answer

To solve the problem step by step, we need to analyze the conditions given in the question:

1. **Condition 1:** The remainder when \( n \) is divided by 3 is 1.
- This can be expressed mathematically as:
\[ n \equiv 1 \ (\text{mod} \ 3) \]
- This means \( n \) can be represented in the form:
\[ n = 3k + 1 \]
for some integer \( k \).

2. **Condition 2:** The remainder when \( n + 1 \) is divided by 2 is 1.
- This can be expressed as:
\[ n + 1 \equiv 1 \ (\text{mod} \ 2) \]
- Simplifying this, we find:
\[ n \equiv 0 \ (\text{mod} \ 2) \]
- This means \( n \) is an even number.

3. **Finding values of \( n \):**
- From the first condition, we know \( n = 3k + 1 \). Since \( n \) must also be even (from the second condition), we can substitute values for \( k \) to find suitable \( n \):
- For \( k = 0 \): \( n = 3(0) + 1 = 1 \) (odd)
- For \( k = 1 \): \( n = 3(1) + 1 = 4 \) (even)
- For \( k = 2 \): \( n = 3(2) + 1 = 7 \) (odd)
- For \( k = 3 \): \( n = 3(3) + 1 = 10 \) (even)
- The valid values of \( n \) that satisfy both conditions are \( n = 4, 10, \ldots \)

4. **Finding \( n - 1 \):**
- Now, we need to find the remainder when \( n - 1 \) is divided by 6:
- For \( n = 4 \):
\[ n - 1 = 4 - 1 = 3 \]
- The remainder when 3 is divided by 6 is 3.
- For \( n = 10 \):
\[ n - 1 = 10 - 1 = 9 \]
- The remainder when 9 is divided by 6 is 3.

5. **Conclusion:**
- In both cases, the remainder when \( n - 1 \) is divided by 6 is 3.

Thus, the final answer is:
\[ \text{The remainder when } (n - 1) \text{ is divided by 6 is } 3. \]