A = 28^12, B = 18^8 and C = 21^6. How many natural numbers are there by which at least one among A, B and C is divisible?

Options

A. 499
B. 501
C. 504
D. 505

Answer

To find how many natural numbers are there by which at least one among \( A \), \( B \), and \( C \) is divisible, we will follow these steps:

### Step 1: Factorize \( A \), \( B \), and \( C \)

1. **Factorize \( A = 28^{12} \)**:
- \( 28 = 2^2 \times 7^1 \)
- Therefore, \( A = (2^2 \times 7^1)^{12} = 2^{24} \times 7^{12} \)

2. **Factorize \( B = 18^{8} \)**:
- \( 18 = 2^1 \times 3^2 \)
- Therefore, \( B = (2^1 \times 3^2)^{8} = 2^{8} \times 3^{16} \)

3. **Factorize \( C = 21^{6} \)**:
- \( 21 = 3^1 \times 7^1 \)
- Therefore, \( C = (3^1 \times 7^1)^{6} = 3^{6} \times 7^{6} \)

### Step 2: Count the total number of factors for \( A \), \( B \), and \( C \)

1. **Total factors of \( A \)**:
- The formula for the number of factors is \( (e_1 + 1)(e_2 + 1) \) where \( e_i \) are the powers of the prime factors.
- For \( A = 2^{24} \times 7^{12} \):
\[ \text{Total factors of } A = (24 + 1)(12 + 1) = 25 \times 13 = 325 \]

2. **Total factors of \( B \)**:
- For \( B = 2^{8} \times 3^{16} \):
\[ \text{Total factors of } B = (8 + 1)(16 + 1) = 9 \times 17 = 153 \]

3. **Total factors of \( C \)**:
- For \( C = 3^{6} \times 7^{6} \):
\[ \text{Total factors of } C = (6 + 1)(6 + 1) = 7 \times 7 = 49 \]

### Step 3: Use the principle of inclusion-exclusion

We need to find the total number of natural numbers that divide at least one of \( A \), \( B \), or \( C \).

Using the principle of inclusion-exclusion:
\[ \text{Total} = \text{Factors of } A + \text{Factors of } B + \text{Factors of } C - \text{Common factors of } A \text{ and } B - \text{Common factors of } B \text{ and } C - \text{Common factors of } A \text{ and } C + \text{Common factors of } A, B, C \]

### Step 4: Find common factors

1. **Common factors of \( A \) and \( B \)**:
- \( \text{HCF}(A, B) = 2^{\min(24, 8)} \times 3^{\min(0, 16)} \times 7^{\min(12, 0)} = 2^{8} \)
- Total factors = \( 8 + 1 = 9 \)

2. **Common factors of \( B \) and \( C \)**:
- \( \text{HCF}(B, C) = 3^{\min(0, 6)} \times 7^{\min(0, 6)} = 1 \)
- Total factors = \( 1 \)

3. **Common factors of \( A \) and \( C \)**:
- \( \text{HCF}(A, C) = 2^{\min(24, 0)} \times 3^{\min(0, 6)} \times 7^{\min(12, 6)} = 7^{6} \)
- Total factors = \( 6 + 1 = 7 \)

4. **Common factors of \( A, B, C \)**:
- \( \text{HCF}(A, B, C) = 1 \)
- Total factors = \( 1 \)

### Step 5: Substitute into the inclusion-exclusion formula

\[ \text{Total} = 325 + 153 + 49 - 9 - 1 - 7 + 1 \]
\[ \text{Total} = 325 + 153 + 49 - 9 - 1 - 7 + 1 = 511 \]

### Final Answer

The total number of natural numbers by which at least one among \( A \), \( B \), and \( C \) is divisible is **511**.

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