We are tasked with finding a 4-digit number that, when successively divided by 9, 8, and 6, leaves remainders 4, 3, and 2, respectively. Ultimately, we want to determine the remainder when this number is divided by 20.

Step-by-step solution:

Let's denote the number by $N$. We are given the following conditions:

1. $N \equiv 4 \ (\text{mod} \ 9)$
2. $N \equiv 3 \ (\text{mod} \ 8)$
3. $N \equiv 2 \ (\text{mod} \ 6)$

We will solve this system using the Chinese Remainder Theorem.

Step 1: Solve the first two congruences

We have: $N \equiv 4 \ (\text{mod} \ 9)$ $N \equiv 3 \ (\text{mod} \ 8)$

Let's assume $N = 9k + 4$ for some integer $k$ (from $N \equiv 4 \ (\text{mod} \ 9)$). Substituting this into the second congruence:

$9k + 4 \equiv 3 \ (\text{mod} \ 8)$ Simplifying: $9k \equiv -1 \equiv 7 \ (\text{mod} \ 8)$ Since $9 \equiv 1 \ (\text{mod} \ 8)$, we have: $k \equiv 7 \ (\text{mod} \ 8)$ Thus, $k = 8m + 7$ for some integer $m$.

Step 2: Substitute back into the equation for $N$

Now substitute $k = 8m + 7$ into $N = 9k + 4$: $N = 9(8m + 7) + 4 = 72m + 63 + 4 = 72m + 67$ Thus: $N \equiv 67 \ (\text{mod} \ 72)$

Step 3: Solve the third congruence

We now incorporate the third condition: $N \equiv 2 \ (\text{mod} \ 6)$ Substitute $N = 72m + 67$ into this congruence: $72m + 67 \equiv 2 \ (\text{mod} \ 6)$ Since $72 \equiv 0 \ (\text{mod} \ 6)$ and $67 \equiv 1 \ (\text{mod} \ 6)$, we have: $1 \equiv 2 \ (\text{mod} \ 6)$ This is clearly false, so let's correct the third modulus condition. Instead, we handle the modulus 6 step by looking for compatible simultaneous conditions.