Finding the Smallest Square Number Divisible by 8, 15, and 20

The question asks for the smallest square number among the given options that is divisible by each of 8, 15, and 20. A number that is divisible by 8, 15, and 20 must be a common multiple of these numbers. The smallest such number is the Least Common Multiple (LCM) of 8, 15, and 20. Any number divisible by 8, 15, and 20 must be a multiple of their LCM.

Calculating the LCM of 8, 15, and 20

To find the LCM, we first find the prime factorization of each number:

• The prime factorization of 8 is $2 \times 2 \times 2 = 2^3$.
• The prime factorization of 15 is $3 \times 5 = 3^1 \times 5^1$.
• The prime factorization of 20 is $2 \times 2 \times 5 = 2^2 \times 5^1$.

The LCM is found by taking the highest power of each prime factor that appears in any of the factorizations.

• Highest power of 2: $2^3$ (from 8)
• Highest power of 3: $3^1$ (from 15)
• Highest power of 5: $5^1$ (from 15 and 20)

So, the LCM of 8, 15, and 20 is $2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

Any number divisible by 8, 15, and 20 must be a multiple of 120.

Finding the Smallest Square Multiple of 120

We are looking for the smallest square number that is a multiple of 120. A square number is a number that can be expressed as the product of two identical integers (e.g., $36 = 6 \times 6$). In terms of prime factors, a number is a perfect square if and only if the exponents of all prime factors in its prime factorization are even.

The prime factorization of 120 is $2^3 \times 3^1 \times 5^1$. To make this a perfect square, we need to multiply it by factors that will make all the exponents even. The current exponents are 3 (for 2), 1 (for 3), and 1 (for 5).

• To make the exponent of 2 even, we need to multiply by $2^{4-3} = 2^1$. The new exponent will be $3+1=4$, which is even.
• To make the exponent of 3 even, we need to multiply by $3^{2-1} = 3^1$. The new exponent will be $1+1=2$, which is even.
• To make the exponent of 5 even, we need to multiply by $5^{2-1} = 5^1$. The new exponent will be $1+1=2$, which is even.

So, the smallest number we need to multiply 120 by to make it a perfect square is $2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30$.

The smallest square number divisible by 120 (and thus by 8, 15, and 20) is $120 \times 30 = 3600$.

The prime factorization of 3600 is $2^3 \times 3^1 \times 5^1 \times 2^1 \times 3^1 \times 5^1 = 2^{3+1} \times 3^{1+1} \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$. Since all exponents (4, 2, 2) are even, 3600 is a perfect square. Specifically, $3600 = (2^2 \times 3^1 \times 5^1)^2 = (4 \times 3 \times 5)^2 = 60^2$.

Checking the Given Options

Now let's check the given options to see which one is the smallest square number that is a multiple of 120.

From the table, we see that both 3600 and 14400 are square numbers divisible by 8, 15, and 20. The question asks for the smallest square number from the given options that meets this condition. Comparing 3600 and 14400, 3600 is the smaller number.

Conclusion

The smallest square number among the given options that is divisible by each of 8, 15, and 20 is 3600. This matches our calculation for the smallest square multiple of the LCM of 8, 15, and 20.

Revision Table: Understanding Square Numbers and Divisibility

Additional Information: General Method for Smallest Square Multiple

To find the smallest square number divisible by a set of numbers (say $n_1, n_2, \dots, n_k$):

1. Find the LCM of the numbers: $L = \text{LCM}(n_1, n_2, \dots, n_k)$.
2. Find the prime factorization of the LCM: $L = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_m^{a_m}$.
3. To get the smallest square multiple of $L$, you need to multiply $L$ by factors such that all exponents in the resulting prime factorization are even. For each prime factor $p_i$ with an odd exponent $a_i$, you need to multiply by $p_i^{1}$ (or generally, $p_i^{k_i}$ where $a_i + k_i$ is the smallest even number $\ge a_i$). If $a_i$ is already even, you don't need to multiply by any power of $p_i$.
4. The required smallest square number is $L \times \prod p_i^{k_i}$ for all $p_i$ where $a_i$ was odd.

In our case, $L = 120 = 2^3 \times 3^1 \times 5^1$. Exponents 3, 1, 1 are all odd. We need to multiply by $2^1 \times 3^1 \times 5^1$. The smallest square multiple is $120 \times (2^1 \times 3^1 \times 5^1) = 120 \times 30 = 3600$. The prime factorization is $2^{3+1} \times 3^{1+1} \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$, which is a perfect square.