The problem asks us to find the smallest value of $k$ such that $p$ and $q$ are not uniquely determined given the conditions:

1. $p$ and $q$ are positive integers.
2. $p < q$.
3. $p + q = k$.

Solution:

To determine when $p$ and $q$ are not uniquely determined, we analyze the possible values of $k$ step by step:

1. Case $k = 3$:
   $p + q = 3$, with $p < q$. The only solution is $p = 1, q = 2$.
   Unique solution.

2. Case $k = 4$:
   $p + q = 4$, with $p < q$. The only solution is $p = 1, q = 3$.
   Unique solution.

3. Case $k = 5$:
   $p + q = 5$, with $p < q$. The only solution is $p = 2, q = 3$.
   Unique solution.

4. Case $k = 6$:
   $p + q = 6$, with $p < q$. Possible pairs:

   • $p = 1, q = 5$.
   • $p = 2, q = 4$.

   Here, there are two solutions, so $p$ and $q$ are not uniquely determined.

Conclusion:

The smallest value of $k$ for which $p$ and $q$ are not uniquely determined is $k = 6$.

Answer: (d) 6

Would you like a detailed breakdown of each case or more information about the reasoning?

Here are 5 related questions for further exploration:

1. What happens if the condition $p < q$ is removed?
2. Can $k$ be a non-integer value for this problem?
3. How does the uniqueness of $p$ and $q$ depend on the parity (even/odd nature) of $k$?
4. What if $p$ and $q$ must both be prime numbers?
5. How would the problem change if $p + q = k$ and $p = q$ were allowed?

Tip: Always verify constraints step by step for mathematical problems to avoid skipping potential cases!