How many factors of 216 are perfect squares?

Answer: There are $4$ factors of $216$ that are perfect squares (namely $1, 4, 9,$ and $36$ ).

Step-by-step solution

Step 1: Find the prime factorisation of 216

To analyze the factors of $216$ , we first express $216$ as a product of prime powers using prime factorisation. Dividing repeatedly by prime numbers, $216 = 2 \times 108 = 2 \times 2 \times 54 = 2^3 \times 27 = 2^3 \times 3^3$ .

Step 2: Identify condition for a factor to be a perfect square

Any factor of $216$ must be of the form $2^a \times 3^b$ , where $0 \le a \le 3$ and $0 \le b \le 3$ . For a number to be a perfect square, the exponents of all prime factors in its prime factorisation must be even integers.

Step 3: Determine possible even powers for each prime base

The possible values for $a$ from the set $\{0, 1, 2, 3\}$ that are even are $0$ and $2$ . Similarly, the even values for $b$ from $\{0, 1, 2, 3\}$ are $0$ and $2$ . Thus, there are $2$ choices for the power of $2$ and $2$ choices for the power of $3$ .

Step 4: Calculate the total number of square factors and list them

By the fundamental principle of counting, the total number of factors that are perfect squares is $2 \times 2 = 4$ . These factors are $2^0 \times 3^0 = 1$ , $2^2 \times 3^0 = 4$ , $2^0 \times 3^2 = 9$ , and $2^2 \times 3^2 = 36$ .

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