Problem 1542

To determine all possible values for $n$ and $k$ given that three out of the four statements are true and one is false, we will analyze each case systematically.

### Case 1: Statements $i, ii, iii$ are true, and $iv$ is false
1. **Statement $i$: $n + 1$ is divisible by $k$**
$$n + 1 = mk \quad \text{for some integer } m$$
2. **Statement $ii$: $n = 2k + 5$**
$$n = 2k + 5$$
3. **Statement $iii$: $n + k$ is divisible by $3$**
$$n + k = 3j \quad \text{for some integer } j$$
4. **Statement $iv$: $n + 7k$ is prime (assumed false in this case)**

From statement $ii$, substitute $n$ in statement $i$:
$$2k + 5 + 1 = mk \implies 2k + 6 = mk \implies k(m - 2) = 6$$
The possible values for $k$ are the divisors of 6: $k = 1, 2, 3, 6$.

Next, check if $n + k$ is divisible by 3:
- For $k = 1$:
$$n = 2(1) + 5 = 7 \quad \text{and} \quad n + k = 7 + 1 = 8 \quad (\text{not divisible by 3})$$
- For $k = 2$:
$$n = 2(2) + 5 = 9 \quad \text{and} \quad n + k = 9 + 2 = 11 \quad (\text{not divisible by 3})$$
- For $k = 3$:
$$n = 2(3) + 5 = 11 \quad \text{and} \quad n + k = 11 + 3 = 14 \quad (\text{not divisible by 3})$$
- For $k = 6$:
$$n = 2(6) + 5 = 17 \quad \text{and} \quad n + k = 17 + 6 = 23 \quad (\text{not divisible by 3})$$

None of these values satisfy all three statements $i, ii, iii$ simultaneously. Therefore, this case is not possible.

### Case 2: Statements $i, ii, iv$ are true, and $iii$ is false
1. **Statement $i$: $n + 1$ is divisible by $k$**
$$n + 1 = mk$$
2. **Statement $ii$: $n = 2k + 5$**
$$n = 2k + 5$$
3. **Statement $iv$: $n + 7k$ is prime**
$$n + 7k = 2k + 5 + 7k = 9k + 5$$

From statement $ii$, substitute $n$ in statement $i$:
$$2k + 5 + 1 = mk \implies 2k + 6 = mk \implies k(m - 2) = 6$$
The possible values for $k$ are the divisors of 6: $k = 1, 2, 3, 6$.

Next, check if $9k + 5$ is prime:
- For $k = 1$:
$$n = 2(1) + 5 = 7 \quad \text{and} \quad 9(1) + 5 = 14 \quad (\text{not prime})$$
- For $k = 2$:
$$n = 2(2) + 5 = 9 \quad \text{and} \quad 9(2) + 5 = 23 \quad (\text{prime})$$
- For $k = 3$:
$$n = 2(3) + 5 = 11 \quad \text{and} \quad 9(3) + 5 = 32 \quad (\text{not prime})$$
- For $k = 6$:
$$n = 2(6) + 5 = 17 \quad \text{and} \quad 9(6) + 5 = 59 \quad (\text{prime})$$

Thus, the possible solutions are $(n, k) = (9, 2)$ and $(17, 6)$.

### Case 3: Statements $i, iii, iv$ are true, and $ii$ is false
1. **Statement $i$: $n + 1$ is divisible by $k$**
$$n + 1 = mk$$
2. **Statement $iii$: $n + k$ is divisible by $3$**
$$n + k = 3j$$
3. **Statement $iv$: $n + 7k$ is prime**
$$n + 7k = p \quad (\text{prime})$$

From statement $i$:
$$n + 1 = mk \implies n = mk - 1$$

From statement $iii$:
$$mk - 1 + k = 3j \implies k(m + 1) - 1 = 3j \implies k(m + 1) = 3j + 1$$

Since $k(m + 1)$ must be of the form $3j + 1$, we need to check if $n + 7k$ is prime for possible values of $k$.

However, if $n + k$ is divisible by 3, then $n + 7k$ will also be divisible by 3 (since $n + 7k = n + k + 6k$). This contradicts the statement that $n + 7k$ is prime unless $n + 7k = 3$, which is not possible for positive integers $n$ and $k$. Therefore, this case is not possible.

### Case 4: Statements $ii, iii, iv$ are true, and $i$ is false
1. **Statement $ii$: $n = 2k + 5$**
$$n = 2k + 5$$
2. **Statement $iii$: $n + k$ is divisible by $3$**
$$n + k = 3j$$
3. **Statement $iv$: $n + 7k$ is prime**
$$n + 7k = 2k + 5 + 7k = 9k + 5$$

From statement $ii$:
$$n = 2k + 5$$

From statement $iii$:
$$2k + 5 + k = 3j \implies 3k + 5 = 3j \implies 3k = 3j - 5 \implies k = j - \frac{5}{3}$$

Since $k$ must be an integer, $j$ must be of the form $3m + 2$ for some integer $m$:
$$j = 3m + 2 \implies k = 3m + 2 - \frac{5}{3} = 3m - 1$$

Next, check if $9k + 5$ is prime:
- For $k = 1$:
$$n = 2(1) + 5 = 7 \quad \text{and} \quad 9(1) + 5 = 14 \quad (\text{not prime})$$
- For $k = 2$:
$$n = 2(2) + 5 = 9 \quad \text{and} \quad 9(2) + 5 = 23 \quad (\text{prime})$$
- For $k = 3$:
$$n = 2(3) + 5 = 11 \quad \text{and} \quad 9(3) + 5 = 32 \quad (\text{not prime})$$
- For $k = 6$:
$$n = 2(6) + 5 = 17 \quad \text{and} \quad 9(6) + 5 = 59 \quad (\text{prime})$$

Thus, the possible solutions are $(n, k) = (9, 2)$ and $(17, 6)$.

### Conclusion
The only possible values for $n$ and $k$ that satisfy three out of the four statements are:
$$(n, k) = (9, 2) \quad \text{and} \quad (17, 6)$$