Answer :
To find the smallest square number which is divisible by 6, 9, and 15, we can follow a methodical approach:
1. Find the Least Common Multiple (LCM) of 6, 9, and 15:
First, we need to compute the LCM of the given numbers. The LCM is the smallest number that is a multiple of each of the given numbers.
- Prime factorization:
- 6 = \(2 \times 3\)
- 9 = \(3^2\)
- 15 = \(3 \times 5\)
- Determine the LCM:
- Take the highest power of each prime that appears in the factorization:
- For 2, the highest power is \(2^1\).
- For 3, the highest power is \(3^2\).
- For 5, the highest power is \(5^1\).
- Therefore, the LCM is \(2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90\).
2. Consider the smallest square number divisible by the LCM (90):
A square number must have all primes in its factorization with even exponents. The LCM we obtained is 90, which is \(2^1 \times 3^2 \times 5^1\).
- To make it a perfect square, each prime factor must have an even exponent. For 90:
- The prime factor 2 has an exponent of 1, which we need to make 2 (even), so we need one more 2.
- The prime factor 3 already has an exponent of 2 (even), so no change is needed.
- The prime factor 5 has an exponent of 1, which we need to make 2 (even), so we need one more 5.
Therefore, the smallest square number that includes these factors is:
[tex]\[ (2^1 \times 3^2 \times 5^1)^2 = (2 \times 9 \times 5)^2 = (90)^2 = 8100 \][/tex]
Hence, the smallest square number which is divisible by 6, 9, and 15 is:
[tex]\[ \boxed{8100} \][/tex]
1. Find the Least Common Multiple (LCM) of 6, 9, and 15:
First, we need to compute the LCM of the given numbers. The LCM is the smallest number that is a multiple of each of the given numbers.
- Prime factorization:
- 6 = \(2 \times 3\)
- 9 = \(3^2\)
- 15 = \(3 \times 5\)
- Determine the LCM:
- Take the highest power of each prime that appears in the factorization:
- For 2, the highest power is \(2^1\).
- For 3, the highest power is \(3^2\).
- For 5, the highest power is \(5^1\).
- Therefore, the LCM is \(2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90\).
2. Consider the smallest square number divisible by the LCM (90):
A square number must have all primes in its factorization with even exponents. The LCM we obtained is 90, which is \(2^1 \times 3^2 \times 5^1\).
- To make it a perfect square, each prime factor must have an even exponent. For 90:
- The prime factor 2 has an exponent of 1, which we need to make 2 (even), so we need one more 2.
- The prime factor 3 already has an exponent of 2 (even), so no change is needed.
- The prime factor 5 has an exponent of 1, which we need to make 2 (even), so we need one more 5.
Therefore, the smallest square number that includes these factors is:
[tex]\[ (2^1 \times 3^2 \times 5^1)^2 = (2 \times 9 \times 5)^2 = (90)^2 = 8100 \][/tex]
Hence, the smallest square number which is divisible by 6, 9, and 15 is:
[tex]\[ \boxed{8100} \][/tex]
