Find the smallest square number which is divisible by each of the numbers 6, 9 and 15

Question

Accepted Answer

To determine the smallest square number that is divisible by each of the numbers 6, 9, and 15, we need to follow a systematic approach:

1. **Find the Least Common Multiple (LCM):**
   - First, we find the prime factorization of each number:
     - 6 = 2 × 3
     - 9 = 3²
     - 15 = 3 × 5
   - The LCM is found by taking the highest power of each prime number that appears in the factorizations:
     - The highest power of 2 is 2¹.
     - The highest power of 3 is 3².
     - The highest power of 5 is 5¹.
   - Therefore, the LCM of 6, 9, and 15 is 2¹ × 3² × 5¹ = 90.

2. **Find the Smallest Square Number:**
   - A square number is a number that can be expressed as the square of an integer.
   - To find the smallest square number divisible by 90, we need to ensure that all prime factors in the LCM are raised to an even power.
   - The prime factorization of 90 is 2¹ × 3² × 5¹. To make each power even, we need:
     - 2² (since 2¹ needs one more 2 to become 2²)
     - 3² (already even)
     - 5² (since 5¹ needs one more 5 to become 5²)
   - Therefore, the smallest square number divisible by 90 is (2 × 3 × 5)² = 30² = 900.

3. **Explanation of Options:**
   - **Option 1: 225**
     - 225 = 15², but it is not divisible by 6 because it lacks the factor of 2.
   - **Option 2: 900**
     - 900 = 30², and it is divisible by 6, 9, and 15, as shown above.
   - **Option 3: 450**
     - 450 is not a perfect square (since it cannot be expressed as the square of an integer).
   - **Option 4: 36**
     - 36 = 6², but it is not divisible by 15 because it lacks the factor of 5.

Thus, the smallest square number divisible by 6, 9, and 15 is indeed 900, making Option 2 the correct answer.