How many natural-number factors does $\textit{N}$ have if $\textit{N} = 2^3 \cdot 3^2 \cdot 5^1$?
Any positive integer divisor of $N$ must take the form $2^a \cdot 3^b \cdot 5^c$ where $0 \le a \le 3$, $0 \le b \le 2$ and $0 \le c \le 1$. In other words, there are 4 choices for $a$, 3 choices for $b$ and 2 choices for $c$. So there are $4 \cdot 3 \cdot 2 = \boxed{24}$ natural-number factors of $N$.
24
Level 3
Number Theory
Problem: How many natural-number factors does $\textit{N}$ have if $\textit{N} = 2^3 \cdot 3^2 \cdot 5^1$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of natural-number factors of a number expressed in its prime factorization form, we utilize the formula for finding the total number of factors. The formula states that if a number \( N \) can be expressed as \( p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_k^{e_k} \), where \( p_i \) are distinct prime factors and \( e_i \) are their respective positive integer exponents, then the total number of factors \( T(N) \) is given by:

\[ T(N) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]

In this case, the prime factorization of \( N \) is \( 2^3 \cdot 3^2 \cdot 5^1 \). Here, the exponents are \( e_1 = 3 \) for the prime \( 2 \), \( e_2 = 2 \) for the prime \( 3 \), and \( e_3 = 1 \) for the prime \( 5 \).

Applying the formula, we calculate each term:

1. For the prime \( 2 \): \( e_1 + 1 = 3 + 1 = 4 \)
2. For the prime \( 3 \): \( e_2 + 1 = 2 + 1 = 3 \)
3. For the prime \( 5 \): \( e_3 + 1 = 1 + 1 = 2 \)

Next, we multiply these results together to find the total number of factors:

\[ T(N) = 4 \cdot 3 \cdot 2 \]

Calculating this step-by-step:

1. First, multiply \( 4 \) and \( 3 \):
   \( 4 \cdot 3 = 12 \)

2. Then, multiply the result by \( 2 \):
   \( 12 \cdot 2 = 24 \)

Thus, the total number of natural-number factors of \( N \) is \( 24 \).<|END_OF_LATENT|> The final answer is $24$.