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The question is asking for the smallest number that, when divided by 8, 12, 15, and 20, leaves a remainder of 5.

To solve this, we can break it down as follows:

1. **Find the Least Common Multiple (LCM)** of the divisors 8, 12, 15, and 20.
2. Add the remainder (5) to the LCM result to find the smallest number that will leave a remainder of 5.

### Step 1: Find the LCM of 8, 12, 15, and 20

Prime factorization:
- \(8 = 2^3\)
- \(12 = 2^2 \times 3\)
- \(15 = 3 \times 5\)
- \(20 = 2^2 \times 5\)

The LCM takes the highest power of each prime factor:
- Highest power of 2: \(2^3 = 8\)
- Highest power of 3: \(3^1 = 3\)
- Highest power of 5: \(5^1 = 5\)

So, \( \text{LCM} = 8 \times 3 \times 5 = 120 \).

### Step 2: Add the Remainder

The smallest number that, when divided by 8, 12, 15, and 20, leaves a remainder of 5 would be:
\[
120 + 5 = 125
\]

### Answer
The correct answer is **D. 125**.

Would you like further details or have any questions?

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Here are 5 related questions for practice:
1. What is the smallest number that leaves a remainder of 3 when divided by 5, 7, and 9?
2. How do you find the LCM of three numbers?
3. Why do we add the remainder after finding the LCM in questions like this?
4. Can a number have more than one remainder when divided by different numbers?
5. How does prime factorization help in finding the LCM?

**Tip:** When working with divisors, always start with their prime factorizations to simplify the process of finding the LCM.

Berapakah bilangan terkecil yang bila dibagi dengan 8, 12, 15, dan 20 akan bersisa 5?

Find the smallest number that leaves a remainder of 5 when divided by 8, 12, 15, and 20. This problem uses the Least Common Multiple (LCM) method and the concept of remainder to solve. Suitable for students in Grades 7-9.

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