Finding the Smallest Perfect Square Divisible by Multiple Numbers

The question asks for the smallest number that satisfies two conditions:

1. It is a perfect square.
2. It is divisible by each of the numbers 6, 8, and 15.

For a number to be divisible by 6, 8, and 15, it must be a multiple of the Least Common Multiple (LCM) of these numbers. Let's first find the LCM of 6, 8, and 15.

Calculating the LCM of 6, 8, and 15

We find the prime factorization of each number:

• Prime factorization of 6: $\displaystyle 6 = 2 \times 3$
• Prime factorization of 8: $\displaystyle 8 = 2 \times 2 \times 2 = 2^3$
• Prime factorization of 15: $\displaystyle 15 = 3 \times 5$

The LCM is found by taking the highest power of all prime factors that appear in any of the factorizations.

• Highest power of 2: $2^3$ (from 8)
• Highest power of 3: $3^1$ (from 6 and 15)
• Highest power of 5: $5^1$ (from 15)

So, the LCM is $\displaystyle 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

Any number divisible by 6, 8, and 15 must be a multiple of 120. We are looking for the smallest multiple of 120 that is also a perfect square.

Finding the Smallest Perfect Square Multiple of 120

A perfect square is a number whose prime factorization only contains prime factors raised to an even power.

The prime factorization of 120 is $\displaystyle 2^3 \times 3^1 \times 5^1$.

To make this a perfect square, we need to multiply 120 by the smallest possible number such that all the exponents in the resulting prime factorization become even.

• The exponent of 2 is 3 (odd). To make it even, we need to multiply by at least one more factor of 2 ($2^1$), so the exponent becomes $3+1=4$.
• The exponent of 3 is 1 (odd). To make it even, we need to multiply by at least one more factor of 3 ($3^1$), so the exponent becomes $1+1=2$.
• The exponent of 5 is 1 (odd). To make it even, we need to multiply by at least one more factor of 5 ($5^1$), so the exponent becomes $1+1=2$.

The smallest number we need to multiply by is $\displaystyle 2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30$.

The smallest perfect square that is a multiple of 120 is $\displaystyle 120 \times 30 = (2^3 \times 3^1 \times 5^1) \times (2^1 \times 3^1 \times 5^1) = 2^{3+1} \times 3^{1+1} \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$.

Now, we calculate the value:

$\displaystyle 2^4 = 16$

$\displaystyle 3^2 = 9$

$\displaystyle 5^2 = 25$

The smallest perfect square is $\displaystyle 16 \times 9 \times 25 = 144 \times 25 = 3600$.

Let's check if 3600 is a perfect square and divisible by 6, 8, and 15:

• Is 3600 a perfect square? $\displaystyle 3600 = 60^2$, so yes. Its prime factorization is $\displaystyle 2^4 \times 3^2 \times 5^2$, with all exponents being even.
• Is 3600 divisible by 6? $\displaystyle 3600 \div 6 = 600$, yes.
• Is 3600 divisible by 8? $\displaystyle 3600 \div 8 = 450$, yes.
• Is 3600 divisible by 15? $\displaystyle 3600 \div 15 = 240$, yes.

Since 3600 is a multiple of LCM(6, 8, 15) = 120, and it is the smallest such multiple that is a perfect square, it is the smallest number satisfying the conditions.

Comparison with Options

Let's briefly check the given options:

From the comparison, only 3600 is both a perfect square and divisible by all three numbers.

Conclusion

The smallest number that is a perfect square and is divisible by each of the numbers 6, 8, and 15 is 3600.

Revision Table: Smallest Perfect Square Multiple

Additional Information: Properties of Perfect Squares

Understanding the properties of perfect squares is key to solving this type of problem. Here are some points:

• A number is a perfect square if and only if in its prime factorization, every prime factor has an even exponent. For example, $\displaystyle 100 = 2^2 \times 5^2$ is a perfect square because the exponents (2 and 2) are even. $\displaystyle 72 = 2^3 \times 3^2$ is not a perfect square because the exponent of 2 is 3 (odd).
• When you multiply two perfect squares, the result is a perfect square. For example, $\displaystyle 4 \times 9 = 36$, and $\displaystyle 2^2 \times 3^2 = (2 \times 3)^2 = 6^2$.
• To make any positive integer a perfect square by multiplying it by the smallest possible integer, you look at its prime factorization. For every prime factor with an odd exponent, you need to multiply by that prime raised to the power of 1. The product of these primes is the smallest multiplier.

In this problem, 120 has prime factorization $\displaystyle 2^3 \times 3^1 \times 5^1$. To make the exponents even (4, 2, 2), we need to multiply by $\displaystyle 2^1 \times 3^1 \times 5^1$.