Each question has four options(A), (B), (C) and (D) for answers. Select the right answer and write in English letters in the box against each question in the enclosed answer sheet.
A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is

Options

A. 50
B. 53
C. 19
D. 214

Answer

To solve the problem step by step, we need to find a number \( x \) that, when divided by 4, 5, and 6, leaves remainders of 2, 3, and 4 respectively.

### Step 1: Set up the equations based on the remainders
We can express the conditions given in the problem as:
- When \( x \) is divided by 4, the remainder is 2:
\[ x \equiv 2 \mod 4 \]
- When \( x \) is divided by 5, the remainder is 3:
\[ x \equiv 3 \mod 5 \]
- When \( x \) is divided by 6, the remainder is 4:
\[ x \equiv 4 \mod 6 \]

### Step 2: Rewrite the equations
We can rewrite these congruences in terms of \( x \):
1. From \( x \equiv 2 \mod 4 \), we can express \( x \) as:
\[ x = 4k + 2 \quad \text{(for some integer } k\text{)} \]
2. From \( x \equiv 3 \mod 5 \), we can express \( x \) as:
\[ x = 5m + 3 \quad \text{(for some integer } m\text{)} \]
3. From \( x \equiv 4 \mod 6 \), we can express \( x \) as:
\[ x = 6n + 4 \quad \text{(for some integer } n\text{)} \]

### Step 3: Solve the equations
We will substitute \( x = 4k + 2 \) into the second equation \( x \equiv 3 \mod 5 \):
\[ 4k + 2 \equiv 3 \mod 5 \]
Subtracting 2 from both sides gives:
\[ 4k \equiv 1 \mod 5 \]
To solve for \( k \), we can find the multiplicative inverse of 4 modulo 5, which is 4 (since \( 4 \times 4 \equiv 1 \mod 5 \)):
\[ k \equiv 4 \mod 5 \]
This means:
\[ k = 5j + 4 \quad \text{(for some integer } j\text{)} \]
Substituting \( k \) back into \( x \):
\[ x = 4(5j + 4) + 2 = 20j + 16 + 2 = 20j + 18 \]

### Step 4: Substitute into the third equation
Now we substitute \( x = 20j + 18 \) into the third equation \( x \equiv 4 \mod 6 \):
\[ 20j + 18 \equiv 4 \mod 6 \]
Calculating \( 20 \mod 6 \) gives \( 2 \), and \( 18 \mod 6 \) gives \( 0 \):
\[ 2j + 0 \equiv 4 \mod 6 \]
This simplifies to:
\[ 2j \equiv 4 \mod 6 \]
Dividing by 2 gives:
\[ j \equiv 2 \mod 3 \]
Thus:
\[ j = 3m + 2 \quad \text{(for some integer } m\text{)} \]

### Step 5: Substitute back to find \( x \)
Substituting \( j \) back into \( x \):
\[ x = 20(3m + 2) + 18 = 60m + 40 + 18 = 60m + 58 \]
To find the least such number, we set \( m = 0 \):
\[ x = 58 \]

### Conclusion
The least number \( x \) that satisfies all the conditions is:
\[ \boxed{58} \]