The least number , which when divided by 4, 5 and 6 leaves remainder 1, 2 and 3 respectively is

Options

A. 57
B. 59
C. 61
D. 63

Answer

To solve the problem of finding the least number that, when divided by 4, 5, and 6, leaves remainders of 1, 2, and 3 respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find a number \( N \) such that:
- \( N \mod 4 = 1 \)
- \( N \mod 5 = 2 \)
- \( N \mod 6 = 3 \)

2. **Rewrite the Conditions**: We can rewrite these conditions in terms of \( N \):
- \( N = 4k + 1 \) for some integer \( k \)
- \( N = 5m + 2 \) for some integer \( m \)
- \( N = 6n + 3 \) for some integer \( n \)

3. **Adjust the Remainders**: To simplify the problem, we can adjust the remainders:
- For \( N \mod 4 = 1 \), we can say \( N - 1 \) is divisible by 4.
- For \( N \mod 5 = 2 \), we can say \( N - 2 \) is divisible by 5.
- For \( N \mod 6 = 3 \), we can say \( N - 3 \) is divisible by 6.

This leads us to the following equivalent conditions:
- \( N - 1 \equiv 0 \mod 4 \)
- \( N - 2 \equiv 0 \mod 5 \)
- \( N - 3 \equiv 0 \mod 6 \)

4. **Find a Common Value**: We can observe that:
- \( N - 1 \) is a multiple of 4,
- \( N - 2 \) is a multiple of 5,
- \( N - 3 \) is a multiple of 6.

This means:
- \( N - 1 = 4k \)
- \( N - 2 = 5m \)
- \( N - 3 = 6n \)

If we let \( N - 3 = x \), then:
- \( N = x + 3 \)
- \( x \equiv 1 \mod 4 \) (since \( x + 3 - 1 \equiv 0 \mod 4 \))
- \( x \equiv 2 \mod 5 \) (since \( x + 3 - 2 \equiv 0 \mod 5 \))
- \( x \equiv 0 \mod 6 \) (since \( x + 3 - 3 \equiv 0 \mod 6 \))

5. **Finding the LCM**: The least common multiple of 4, 5, and 6 is:
\[ \text{LCM}(4, 5, 6) = 60 \]

6. **Finding the Number**: Since \( N - 3 \) must be a multiple of 60, we can express \( N \) as:
\[ N = 60k + 3 \]
To satisfy the conditions, we need to find the smallest \( k \) such that \( N \) also satisfies the remainder conditions.

The smallest value occurs when \( k = 1 \):
\[ N = 60 \cdot 1 + 3 = 63 \]

7. **Final Adjustment**: However, we need to check if \( N - 3 \) is indeed a multiple of 60. The correct adjustment is:
\[ N = 60 - 3 = 57 \]

### Conclusion:
The least number that satisfies all conditions is \( \boxed{57} \).