Answer
(a) \[
\begin{array}{rcl}
N_1 &=& 3,\\[1mm]
N_2 &=& 7,\\[1mm]
N_3 &=& 31,\\[1mm]
N_4 &=& 211,\\[1mm]
N_5 &=& 2311,\\[1mm]
N_6 &=& 30031.
\end{array}
\]
(b) \[
\begin{array}{rcl}
q_1 &=& 3,\\[1mm]
q_2 &=& 7,\\[1mm]
q_3 &=& 31,\\[1mm]
q_4 &=& 211,\\[1mm]
q_5 &=& 2311,\\[1mm]
q_6 &=& 59.
\end{array}
\]
Work Step by Step
We begin with the list of primes in order:
\[
p_1 = 2,\quad p_2 = 3,\quad p_3 = 5,\quad p_4 = 7,\quad p_5 = 11,\quad p_6 = 13.
\]
For each \(i = 1,2,\dots,6\), define
\[
N_i = p_1 \cdot p_2 \cdots p_i + 1.
\]
### Part (a): Compute \(N_1, N_2, \dots, N_6\)
1. **\(N_1\):**
\[
N_1 = 2 + 1 = 3.
\]
2. **\(N_2\):**
\[
N_2 = 2\cdot3 + 1 = 6 + 1 = 7.
\]
3. **\(N_3\):**
\[
N_3 = 2\cdot3\cdot5 + 1 = 30 + 1 = 31.
\]
4. **\(N_4\):**
\[
N_4 = 2\cdot3\cdot5\cdot7 + 1 = 210 + 1 = 211.
\]
5. **\(N_5\):**
\[
N_5 = 2\cdot3\cdot5\cdot7\cdot11 + 1 = 2310 + 1 = 2311.
\]
6. **\(N_6\):**
\[
N_6 = 2\cdot3\cdot5\cdot7\cdot11\cdot13 + 1 = 30030 + 1 = 30031.
\]
### Part (b): Find the smallest prime divisor \(q_i\) of each \(N_i\)
1. **For \(N_1 = 3\):**
\(3\) is itself prime, so the smallest prime divisor is
\[
q_1 = 3.
\]
2. **For \(N_2 = 7\):**
\(7\) is prime, so
\[
q_2 = 7.
\]
3. **For \(N_3 = 31\):**
\(31\) is prime, so
\[
q_3 = 31.
\]
4. **For \(N_4 = 211\):**
\(211\) is prime (it is not divisible by any prime less than or equal to \(\sqrt{211} \approx 14.5\)), so
\[
q_4 = 211.
\]
5. **For \(N_5 = 2311\):**
We check divisibility by small primes:
- \(2311\) is odd (not divisible by 2).
- Sum of digits is \(2+3+1+1 = 7\), so not divisible by 3.
- It does not end in 0 or 5, so not divisible by 5.
- A quick check with 7, 11, 13, etc., shows no divisor among primes \(\le \sqrt{2311}\) (and indeed \(\sqrt{2311} \approx 48.1\)).
Thus, \(2311\) is prime, and
\[
q_5 = 2311.
\]
6. **For \(N_6 = 30031\):**
We check the small primes up to \(\sqrt{30031}\) (note that \(\sqrt{30031} \approx 173.3\)):
- Not divisible by 2, 3, 5, or 7 (since \(30030\) is divisible by these, and \(30031 = 30030+1\)).
- We test further: it turns out that when dividing by 59 we get
\[
59 \times 509 = 30031.
\]
Hence, the smallest prime divisor is
\[
q_6 = 59.
\]
