Step 1: Understanding the Question:
The question asks for the count of distinct prime factors of the number 9900. To find this, we need to perform the prime factorization of 9900 and then count the unique prime bases used in the resulting expression.
Step 2: Key Formula or Approach:
Break down the number using a factor tree or continuous division by primes until only prime numbers remain. Express the number in the form $p_1^a \times p_2^b \times p_3^c \dots$
Step 3: Detailed Explanation:
We can break down 9900 into smaller, more manageable factors to save time.
9900 ends in two zeros, so it is clearly a multiple of 100. We can write $9900 = 99 \times 100$.
Now, let's factorize each part separately.
For 99: It is divisible by 9. So, $99 = 9 \times 11$. Since 9 is $3^2$, we have $99 = 3^2 \times 11^1$. Both 3 and 11 are prime numbers.
For 100: It is a perfect square, $10^2$. Since 10 is $2 \times 5$, we have $100 = (2 \times 5)^2 = 2^2 \times 5^2$. Both 2 and 5 are prime numbers.
Combine these parts back together to get the full prime factorization of 9900:
$9900 = (3^2 \times 11^1) \times (2^2 \times 5^2)$.
Rearranging them in ascending order of the prime bases, we get: $9900 = 2^2 \times 3^2 \times 5^2 \times 11^1$.
The distinct prime factors are the base prime numbers used in this factorization, regardless of their exponent powers.
The distinct prime bases are 2, 3, 5, and 11.
Counting them up, we have exactly 4 distinct prime factors.
Step 4: Final Answer:
There are 4 distinct prime factors in 9900.