Find the greatest number that will divide 390, 495 and 300 with out leaving a remainder

Options

A. 5
B. 15
C. 25
D. 35

Answer

To find the greatest number that will divide 390, 495, and 300 without leaving a remainder, we need to calculate the Highest Common Factor (HCF) of these three numbers. Here’s a step-by-step solution:

### Step 1: Prime Factorization of Each Number

**1. Factor 390:**
- Divide by 2: \( 390 \div 2 = 195 \)
- Divide by 3: \( 195 \div 3 = 65 \)
- Divide by 5: \( 65 \div 5 = 13 \)
- 13 is a prime number.

So, the prime factorization of 390 is:
\[ 390 = 2^1 \times 3^1 \times 5^1 \times 13^1 \]

**2. Factor 495:**
- Divide by 3: \( 495 \div 3 = 165 \)
- Divide by 3 again: \( 165 \div 3 = 55 \)
- Divide by 5: \( 55 \div 5 = 11 \)
- 11 is a prime number.

So, the prime factorization of 495 is:
\[ 495 = 3^2 \times 5^1 \times 11^1 \]

**3. Factor 300:**
- Divide by 2: \( 300 \div 2 = 150 \)
- Divide by 2 again: \( 150 \div 2 = 75 \)
- Divide by 3: \( 75 \div 3 = 25 \)
- Divide by 5: \( 25 \div 5 = 5 \)
- Divide by 5 again: \( 5 \div 5 = 1 \)

So, the prime factorization of 300 is:
\[ 300 = 2^2 \times 3^1 \times 5^2 \]

### Step 2: Identify Common Factors

Now that we have the prime factorizations:
- \( 390 = 2^1 \times 3^1 \times 5^1 \times 13^1 \)
- \( 495 = 3^2 \times 5^1 \times 11^1 \)
- \( 300 = 2^2 \times 3^1 \times 5^2 \)

We will identify the common prime factors and take the lowest power of each:

- For \( 2 \): The minimum power is \( 0 \) (since 495 does not have 2).
- For \( 3 \): The minimum power is \( 1 \) (common in all three).
- For \( 5 \): The minimum power is \( 1 \) (common in all three).
- For \( 13 \) and \( 11 \): Not common in all three.

### Step 3: Calculate the HCF

Now, we multiply the common prime factors raised to their lowest powers:
\[ \text{HCF} = 3^1 \times 5^1 = 3 \times 5 = 15 \]

### Conclusion

Thus, the greatest number that will divide 390, 495, and 300 without leaving a remainder is **15**.

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