Question

find the lowest number which when divided by 6, 8 and 12 leaves 3 as remainder in each case for class 4

Answer 1

To find the lowest number that, when divided by 6, 8, and 12, leaves a remainder of 3 in each case, we can follow these steps:

1. **Understand the Problem**:
We need a number \( N \) such that:
- \( N \equiv 3 \mod 6 \)
- \( N \equiv 3 \mod 8 \)
- \( N \equiv 3 \mod 12 \)

This means that if we subtract 3 from our number \( N \), the result should be divisible by 6, 8, and 12.

2. **Adjust the Value**:
Let's denote \( M = N - 3 \). Now we must find \( M \) such that:
- \( M \equiv 0 \mod 6 \)
- \( M \equiv 0 \mod 8 \)
- \( M \equiv 0 \mod 12 \)

This means \( M \) should be a common multiple of 6, 8, and 12.

3. **Calculate the Least Common Multiple (LCM)**:
The least common multiple (LCM) of 6, 8, and 12 can be determined as follows:
- Prime factorization:
- 6 = \( 2^1 \times 3^1 \)
- 8 = \( 2^3 \)
- 12 = \( 2^2 \times 3^1 \)
- Take the highest power of each prime:
- For 2: \( 2^3 \)
- For 3: \( 3^1 \)
- Therefore,
\[
\text{LCM}(6, 8, 12) = 2^3 \times 3^1 = 8 \times 3 = 24
\]

4. **Find \( M \)**:
We found that \( M = 24 \).

5. **Find \( N \)**:
Since \( N = M + 3 \):
\[
N = 24 + 3 = 27
\]

6. **Verification**:
Let's check if 27 meets the original conditions:
- \( 27 \div 6 = 4 \) remainder \( 3 \)
- \( 27 \div 8 = 3 \) remainder \( 3 \)
- \( 27 \div 12 = 2 \) remainder \( 3 \)

All conditions are satisfied.

Thus, the lowest number which when divided by 6, 8, and 12 leaves a remainder of 3 in each case is **27**.