The number of functions `f:{1,2,3,4}rarr{a in Z}|a|le8}` satisfying `f(n)+(1)/(n)f(n+1)=1,AA n in{1,2,3}` is

Options

A. `2`
B. `1`
C. `4`
D. `3`

Answer

To solve the problem, we need to find the number of functions \( f: \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} \mid |a| \leq 8\} \) satisfying the equation:

\[ f(n) + \frac{1}{n} f(n+1) = 1 \quad \text{for } n \in \{1, 2, 3\} \]

### Step 1: Rewrite the equation
We start with the given equation:

\[ f(n) + \frac{1}{n} f(n+1) = 1 \]

Multiplying through by \( n \) to eliminate the fraction gives:

\[ n f(n) + f(n+1) = n \]

### Step 2: Set up equations for \( n = 1, 2, 3 \)
Now we can set up equations for \( n = 1, 2, 3 \):

1. For \( n = 1 \):
\[ 1 f(1) + f(2) = 1 \quad \Rightarrow \quad f(1) + f(2) = 1 \quad \text{(Equation 1)} \]

2. For \( n = 2 \):
\[ 2 f(2) + f(3) = 2 \quad \Rightarrow \quad 2f(2) + f(3) = 2 \quad \text{(Equation 2)} \]

3. For \( n = 3 \):
\[ 3 f(3) + f(4) = 3 \quad \Rightarrow \quad 3f(3) + f(4) = 3 \quad \text{(Equation 3)} \]

### Step 3: Express \( f(2) \) and \( f(3) \) in terms of \( f(1) \)
From Equation 1, we can express \( f(2) \):

\[ f(2) = 1 - f(1) \]

Substituting \( f(2) \) into Equation 2:

\[ 2(1 - f(1)) + f(3) = 2 \]

This simplifies to:

\[ 2 - 2f(1) + f(3) = 2 \quad \Rightarrow \quad f(3) = 2f(1) \]

### Step 4: Express \( f(4) \) in terms of \( f(1) \)
Now substituting \( f(3) = 2f(1) \) into Equation 3:

\[ 3(2f(1)) + f(4) = 3 \]

This simplifies to:

\[ 6f(1) + f(4) = 3 \quad \Rightarrow \quad f(4) = 3 - 6f(1) \]

### Step 5: Determine the constraints on \( f(1) \)
We know that \( |f(n)| \leq 8 \) for \( n = 1, 2, 3, 4 \). Thus, we have:

1. For \( f(1) \):
\[ |f(1)| \leq 8 \]

2. For \( f(2) = 1 - f(1) \):
\[ |1 - f(1)| \leq 8 \quad \Rightarrow \quad -8 \leq 1 - f(1) \leq 8 \]
This gives:
\[ -9 \leq f(1) \leq 7 \]

3. For \( f(3) = 2f(1) \):
\[ |2f(1)| \leq 8 \quad \Rightarrow \quad -4 \leq f(1) \leq 4 \]

4. For \( f(4) = 3 - 6f(1) \):
\[ |3 - 6f(1)| \leq 8 \quad \Rightarrow \quad -8 \leq 3 - 6f(1) \leq 8 \]
This gives:
\[ -11 \leq -6f(1) \leq 5 \quad \Rightarrow \quad -\frac{11}{6} \leq f(1) \leq \frac{5}{6} \]

### Step 6: Combine the constraints
From the inequalities, we have:

\[ -\frac{11}{6} \leq f(1) \leq \frac{5}{6} \]

Since \( f(1) \) must be an integer, the possible values for \( f(1) \) are \( 0 \) and \( 1 \).

### Step 7: Calculate the corresponding values of \( f(2), f(3), f(4) \)
1. If \( f(1) = 0 \):
- \( f(2) = 1 \)
- \( f(3) = 0 \)
- \( f(4) = 3 \)

2. If \( f(1) = 1 \):
- \( f(2) = 0 \)
- \( f(3) = 2 \)
- \( f(4) = -3 \)

### Conclusion
Thus, there are **2 functions** that satisfy the given conditions.

### Final Answer
The number of functions \( f \) is **2**.