Which least number should be subtracted from the number 6846 so that the number will be completely divisible by 6, 8 and 12?

Options

A. 4
B. 6
C. 8
D. 9

Answer

To find the least number that should be subtracted from 6846 so that the result is completely divisible by 6, 8, and 12, we can follow these steps:

### Step 1: Find the LCM of 6, 8, and 12
To determine the least number that should be subtracted, we first need to find the Least Common Multiple (LCM) of the numbers 6, 8, and 12.

- **Prime Factorization**:
- 6 = 2 × 3
- 8 = 2^3
- 12 = 2^2 × 3

- **Taking the highest power of each prime**:
- For 2: highest power is 2^3 (from 8)
- For 3: highest power is 3^1 (from 6 or 12)

- **Calculating the LCM**:
\[ \text{LCM} = 2^3 \times 3^1 = 8 \times 3 = 24 \]

### Step 2: Check the divisibility of 6846 by 24
Next, we need to check how 6846 behaves when divided by 24 to find the remainder.

- **Perform the division**:
\[ 6846 \div 24 = 285 \quad \text{(whole number part)} \]
\[ 285 \times 24 = 6840 \]

- **Finding the remainder**:
\[ \text{Remainder} = 6846 - 6840 = 6 \]

### Step 3: Determine the number to subtract
To make 6846 divisible by 24, we need to subtract the remainder we found:

\[ \text{Number to subtract} = \text{Remainder} = 6 \]

### Conclusion
Thus, the least number that should be subtracted from 6846 to make it completely divisible by 6, 8, and 12 is **6**.

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