Find the least number, when divided by 12, 15, 20 and 54, leaves in each case a remainder of 8:

Question

Accepted Answer

To find the least number that, when divided by 12, 15, 20, and 54, leaves a remainder of 8, we need to follow these steps:

1. **Understand the Problem**: We are looking for a number $ N $ such that:
   $$
   N \equiv 8 \pmod{12}, \quad N \equiv 8 \pmod{15}, \quad N \equiv 8 \pmod{20}, \quad N \equiv 8 \pmod{54}
   $$
   This means that $ N - 8 $ is a common multiple of 12, 15, 20, and 54.

2. **Find the Least Common Multiple (LCM)**: Calculate the LCM of 12, 15, 20, and 54. The LCM is the smallest number that is divisible by each of these numbers.

- Prime factorization:
     - $ 12 = 2^2 \times 3 $
     - $ 15 = 3 \times 5 $
     - $ 20 = 2^2 \times 5 $
     - $ 54 = 2 \times 3^3 $

- LCM calculation:
     - Take the highest power of each prime that appears in the factorizations:
     - $ 2^2 $ (from 12 and 20)
     - $ 3^3 $ (from 54)
     - $ 5 $ (from 15 and 20)

- LCM = $ 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540 $

3. **Determine the Number**: Since $ N - 8 $ must be a multiple of the LCM (540), we have:
   $$
   N = 540k + 8
   $$
   For the smallest positive $ N $, set $ k = 1 $:
   $$
   N = 540 \times 1 + 8 = 548
   $$

4. **Verify the Solution**: Check that 548 leaves a remainder of 8 when divided by each of the numbers:
   - $ 548 \div 12 = 45 $ remainder 8
   - $ 548 \div 15 = 36 $ remainder 8
   - $ 548 \div 20 = 27 $ remainder 8
   - $ 548 \div 54 = 10 $ remainder 8

All conditions are satisfied, confirming that 548 is the correct answer.

5. **Explanation of Incorrect Options**:
   - **Option 1 (504)**: When divided by 12, 15, 20, or 54, 504 does not leave a remainder of 8.
   - **Option 2 (540)**: 540 is the LCM, but it leaves a remainder of 0, not 8.
   - **Option 3 (546)**: When divided by 12, 15, 20, or 54, 546 does not leave a remainder of 8.

Thus, Option 4 (548) is the correct answer, as it satisfies all the conditions of the problem.