Question

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function such that \( f(m + n) = f(m) + f(n) \) for every \( m, n \in \mathbb{N} \). If \( f(6) = 18 \), then \( f(2) \cdot f(3) \) is equal to :

• A. 6
• B. 18
• C. 36
• D. 54

Hint:

For any functional equation of the form \( f(x+y) = f(x) + f(y) \), the solution is always a linear function passing through the origin, i.e., \( f(x) = kx \).
This saves time during the exam instead of deriving \( f(1), f(2), \dots \) iteratively.

Answer 1

The Correct Option is D

To solve this problem, we are given a function \( f : \mathbb{N} \to \mathbb{N} \) satisfying the functional equation \( f(m + n) = f(m) + f(n) \) for every \( m, n \in \mathbb{N} \). This type of function is known as an additive function. Our goal is to find \( f(2) \cdot f(3) \), given that \( f(6) = 18 \).

The functional equation provided is a well-known property of linear functions. An additive function from \( \mathbb{N} \) to \( \mathbb{N} \) is commonly of the form \( f(x) = cx \), where \( c \) is a constant. Let's determine the value of this constant using the given information:

1. We know \( f(6) = 18 \). Assuming \( f(x) = cx \), for \( x = 6 \), the function becomes: f(6) = c \cdot 6 = 6c.
2. We set this equal to the provided \( f(6) = 18 \): 6c = 18.
3. Solving for \( c \): c = \frac{18}{6} = 3.

Therefore, the function is \( f(x) = 3x \).

Now, we calculate \( f(2) \cdot f(3) \):

1. Find \( f(2) \): f(2) = 3 \cdot 2 = 6.
2. Find \( f(3) \): f(3) = 3 \cdot 3 = 9.

Thus, \( f(2) \cdot f(3) = 6 \cdot 9 = 54 \).

The correct answer is 54.