If n is a positive integer, which one of the following must have a remainder of 3 when divided by any of the number 4,5 and 6 ?

Options

A. `12n + 3`
B. `24 n + 3`
C. `80n + 3`
D. `120 n + 3`

Answer

To solve the problem, we need to find which expression among the given options must leave a remainder of 3 when divided by 4, 5, and 6.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find an expression of the form \( k + 3 \) where \( k \) is divisible by 4, 5, and 6. This means that \( k \) must be a common multiple of these numbers.

2. **Finding the Least Common Multiple (LCM)**:
- The LCM of 4, 5, and 6 needs to be calculated.
- The prime factorization of the numbers is:
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- Taking the highest power of each prime:
- \( 2^2 \) from 4
- \( 3^1 \) from 6
- \( 5^1 \) from 5
- Therefore, \( \text{LCM}(4, 5, 6) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \).

3. **Formulating the Expression**:
- Since \( k \) must be a multiple of 60, we can express it as \( k = 60m \) for some integer \( m \).
- Thus, the expression becomes \( k + 3 = 60m + 3 \).

4. **Evaluating the Options**:
- We will check each option to see if it can be expressed in the form \( 60m + 3 \).
- The options are:
- \( 2 \)
- \( 12n + 3 \)
- \( 24n + 3 \)
- \( 80n + 3 \)
- \( 120n + 3 \)

5. **Checking Each Option**:
- **Option 1: \( 2 \)**: Not of the form \( 60m + 3 \).
- **Option 2: \( 12n + 3 \)**: \( 12n \) is not a multiple of 60 for all \( n \).
- **Option 3: \( 24n + 3 \)**: \( 24n \) is not a multiple of 60 for all \( n \).
- **Option 4: \( 80n + 3 \)**: \( 80n \) is not a multiple of 60 for all \( n \).
- **Option 5: \( 120n + 3 \)**: \( 120n \) is a multiple of 60 for all \( n \).

6. **Conclusion**: The only option that satisfies the condition is \( 120n + 3 \).

### Final Answer:
The expression that must have a remainder of 3 when divided by 4, 5, and 6 is **\( 120n + 3 \)**.