Let $n$ and $k$ be integers, $1\le k\le n$. Find an integer $b$ and a set $A$ of $n$ integers satisfying the following conditions:
(i) No product of $k-1$ distinct elements of $A$ is divisible by $b$.
(ii) Every product of $k$ distinct elements of $A$ is divisible by $b$.
(iii) For all distinct $a,a'$ in $A$, $a$ does not divide $a'$.
**
   - Let \( b = 2^k \).
   - Let \( A = \{2p_1, 2p_2, \ldots, 2p_n\} \), where \( p_1, p_2, \ldots, p_n \) are distinct odd primes.

2. **Verification of Condition (i):**
   - Condition (i) states that no product of \( k-1 \) distinct elements of \( A \) is divisible by \( b \).
   - Each element of \( A \) is of the form \( 2p_i \), where \( p_i \) is an odd prime.
   - The product of \( k-1 \) distinct elements of \( A \) will be:
     \[
     (2p_{i_1})(2p_{i_2}) \cdots (2p_{i_{k-1}}) = 2^{k-1} (p_{i_1} p_{i_2} \cdots p_{i_{k-1}})
     \]
   - Since \( 2^{k-1} \) is not divisible by \( 2^k \), the product of \( k-1 \) distinct elements of \( A \) is not divisible by \( b = 2^k \).
   - Therefore, condition (i) is satisfied.

3. **Verification of Condition (ii):**
   - Condition (ii) states that every product of \( k \) distinct elements of \( A \) is divisible by \( b \).
   - The product of \( k \) distinct elements of \( A \) will be:
     \[
     (2p_{i_1})(2p_{i_2}) \cdots (2p_{i_k}) = 2^k (p_{i_1} p_{i_2} \cdots p_{i_k})
     \]
   - Since \( 2^k \) is divisible by \( 2^k \), the product of \( k \) distinct elements of \( A \) is divisible by \( b = 2^k \).
   - Therefore, condition (ii) is satisfied.

4. **Verification of Condition (iii):**
   - Condition (iii) states that for all distinct \( a, a' \) in \( A \), \( a \) does not divide \( a' \).
   - Each element of \( A \) is of the form \( 2p_i \), where \( p_i \) is an odd prime.
   - Since \( p_i \) and \( p_j \) are distinct odd primes, \( 2p_i \) does not divide \( 2p_j \) for \( i \neq j \).
   - Therefore, condition (iii) is satisfied.

Since all the conditions are satisfied, the proposed solution is correct.

The final answer is \( \boxed{ b = 2^k } \) and \( A = \{2p_1, 2p_2, \ldots, 2p_n\} \), where \( p_1, p_2, \ldots, p_n \) are distinct odd primes.