Finding Perfect Square Factors of 360

This problem asks us to identify how many of the factors of the number 360 are also perfect squares. To solve this, we first need to understand the factors of 360 and the properties of perfect squares.

Understanding Factors and Perfect Squares

• A factor of a number is a number that divides it evenly without leaving a remainder.
• A perfect square is an integer that is the square of an integer. For example, 1 ($1^2$), 4 ($2^2$), 9 ($3^2$), 16 ($4^2$), and so on, are perfect squares. In terms of prime factorization, a number is a perfect square if and only if all the exponents in its prime factorization are even.

Step 1: Prime Factorization of 360

The first step is to find the prime factorization of 360. We break 360 down into its prime number components:

So, the prime factorization of 360 is \(2^3 \times 3^2 \times 5^1\).

Step 2: Understanding Factors from Prime Factorization

Any factor of 360 will be of the form \(2^a \times 3^b \times 5^c\), where the exponents satisfy:

• \(0 \le a \le 3\) (Possible values for \(a\): 0, 1, 2, 3)
• \(0 \le b \le 2\) (Possible values for \(b\): 0, 1, 2)
• \(0 \le c \le 1\) (Possible values for \(c\): 0, 1)

The total number of factors of 360 is \((3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24\).

Step 3: Identifying Perfect Square Factors

For a factor \(2^a \times 3^b \times 5^c\) to be a perfect square, all the exponents \(a\), \(b\), and \(c\) must be even numbers.

Let's look at the constraints on the exponents again, but this time, we only consider the even possibilities:

• For \(a\), the possible values are 0, 1, 2, 3. The even values in this range are 0 and 2. (2 options)
• For \(b\), the possible values are 0, 1, 2. The even values in this range are 0 and 2. (2 options)
• For \(c\), the possible values are 0, 1. The only even value in this range is 0. (1 option)

Step 4: Counting the Perfect Square Factors

To find the total number of perfect square factors, we multiply the number of options for each exponent:

Thus, there are 4 factors of 360 that are perfect squares.

We can list these factors by choosing the even exponents:

• \(a=0, b=0, c=0 \Rightarrow 2^0 \times 3^0 \times 5^0 = 1 \times 1 \times 1 = 1\) (\(1^2\))
• \(a=2, b=0, c=0 \Rightarrow 2^2 \times 3^0 \times 5^0 = 4 \times 1 \times 1 = 4\) (\(2^2\))
• \(a=0, b=2, c=0 \Rightarrow 2^0 \times 3^2 \times 5^0 = 1 \times 9 \times 1 = 9\) (\(3^2\))
• \(a=2, b=2, c=0 \Rightarrow 2^2 \times 3^2 \times 5^0 = 4 \times 9 \times 1 = 36\) (\(6^2\))

The perfect square factors of 360 are 1, 4, 9, and 36. There are indeed 4 such factors.

Summary of Perfect Square Factors of 360

Looking at the table, only the factors 1, 4, 9, and 36 are perfect squares. There are 4 such factors.

Conclusion on Perfect Square Factors of 360

By finding the prime factorization of 360 as \(2^3 \times 3^2 \times 5^1\), we determined that a factor is a perfect square if its exponents in the prime factorization are even. The possible even exponents for 2 are {0, 2}, for 3 are {0, 2}, and for 5 is {0}. The total number of combinations of these even exponents is \(2 \times 2 \times 1 = 4\). Therefore, there are 4 perfect square factors of 360.

Revision Table: Factors and Perfect Squares

Additional Information on Number Properties

Understanding prime factorization is fundamental in number theory. It helps in finding the total number of factors, the sum of factors, and identifying properties like perfect squares, perfect cubes, etc.

• To find the total number of factors of a number \(N\) with prime factorization \(p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}\), the formula is \((e_1+1)(e_2+1)\dots(e_k+1)\).
• To find the number of perfect square factors, you find the prime factorization \(p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}\). For each prime \(p_i\), the exponent \(a_i\) in a perfect square factor \(p_1^{a_1} \dots p_k^{a_k}\) must be even and \(0 \le a_i \le e_i\). The possible values for \(a_i\) are \(0, 2, 4, \dots, 2m \le e_i\). The number of such even values is \(\lfloor e_i/2 \rfloor + 1\). The total number of perfect square factors is \((\lfloor e_1/2 \rfloor + 1)(\lfloor e_2/2 \rfloor + 1)\dots(\lfloor e_k/2 \rfloor + 1)\).

Let's apply this formula to 360 = \(2^3 \times 3^2 \times 5^1\):

• For prime 2, exponent is 3. Even possibilities for \(a\) are 0, 2. Number of options = \(\lfloor 3/2 \rfloor + 1 = 1 + 1 = 2\).
• For prime 3, exponent is 2. Even possibilities for \(b\) are 0, 2. Number of options = \(\lfloor 2/2 \rfloor + 1 = 1 + 1 = 2\).
• For prime 5, exponent is 1. Even possibility for \(c\) is 0. Number of options = \(\lfloor 1/2 \rfloor + 1 = 0 + 1 = 1\).

Total perfect square factors = \(2 \times 2 \times 1 = 4\). This confirms our earlier calculation.