What will be the least number which when divided by 12, 21 and 35 leaves 6 as remainder in each case?

Options

A. 426
B. 326
C. 536
D. 436

Answer

To find the least number which, when divided by 12, 21, and 35, leaves a remainder of 6 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 12 = 6 \)
- \( N \mod 21 = 6 \)
- \( N \mod 35 = 6 \)

This means that \( N - 6 \) should be divisible by 12, 21, and 35.

### Step 2: Set Up the Equation
Let \( M = N - 6 \). Therefore, we need to find \( M \) such that:
- \( M \mod 12 = 0 \)
- \( M \mod 21 = 0 \)
- \( M \mod 35 = 0 \)

This implies that \( M \) is a common multiple of 12, 21, and 35.

### Step 3: Find the Least Common Multiple (LCM)
To find \( M \), we need to calculate the LCM of the numbers 12, 21, and 35.

1. **Prime Factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 21 = 3^1 \times 7^1 \)
- \( 35 = 5^1 \times 7^1 \)

2. **Identify the Highest Powers of Each Prime**:
- The highest power of \( 2 \) is \( 2^2 \) from 12.
- The highest power of \( 3 \) is \( 3^1 \) (common in 12 and 21).
- The highest power of \( 5 \) is \( 5^1 \) from 35.
- The highest power of \( 7 \) is \( 7^1 \) from 21 and 35.

3. **Calculate the LCM**:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]
- Calculate step by step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 5 = 60 \)
- \( 60 \times 7 = 420 \)

Thus, \( \text{LCM}(12, 21, 35) = 420 \).

### Step 4: Find \( N \)
Now that we have \( M = 420 \), we can find \( N \):
\[ N = M + 6 = 420 + 6 = 426 \]

### Conclusion
The least number which, when divided by 12, 21, and 35, leaves a remainder of 6 is **426**.

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