Since the cards are drawn one after the other without replacement from a well shuffled pack of cards untill an ace card appears. Then the probability that exactly 5 cards are drawn before the first ace card appears
$=\frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{45}{49} \times \frac{44}{48} \times \frac{4}{47}$
$=\frac{4}{49}\left[\frac{46 \times 45 \times 44}{52 \times 51 \times 50}\right]=\frac{4}{49}\left(\frac{11 \times 23 \times 3}{13 \times 17 \times 5}\right)$
$=\frac{4}{49}\left(\frac{P_1 \cdot P_2 \cdot P_3}{P_4 \cdot P_5 \cdot P_6}\right), \quad$ (given) where $P_i$ is prime
$\therefore \quad \max \left(p_i\right)=23$ and $\min \left\{P_i\right\}=3$
$\therefore \quad \max \left\{p_i\right\}-\min \left\{p_i\right\}=23-3=20$