1. Let the 3-digit number be $\overline{abc}$, where each digit is chosen from $\{1,2,3,4,5\}$ with repetition allowed.

2. A number is divisible by $6$ iff it is divisible by both $2$ and $3$.

Step 1: Divisibility by $2$

The last digit must be even.

From $\{1,2,3,4,5\}$, the even digits are $2,4$.

So, the units digit $c$ has $2$ choices.

Step 2: Divisibility by $3$

We need $a+b+c \equiv 0 \pmod{3}.$

Now classify the digits by their remainder modulo $3$:

• Remainder $0$: $\{3\}$
• Remainder $1$: $\{1,4\}$
• Remainder $2$: $\{2,5\}$

Since the last digit must be $2$ or $4$:

• $2 \equiv 2 \pmod{3}$
• $4 \equiv 1 \pmod{3}$

We count separately.

Case 1: Last digit $c=2$

Then $a+b+2 \equiv 0 \pmod{3} \Rightarrow a+b \equiv 1 \pmod{3}.$

Possible residue pairs for $(a,b)$ giving sum $1 \pmod{3}$ are:

• $(0,1)$
• $(1,0)$
• $(2,2)$

Now count actual digits:

• residue $0$: $1$ choice ($3$)
• residue $1$: $2$ choices ($1,4$)
• residue $2$: $2$ choices ($2,5$)

So number of ordered pairs $(a,b)$ is $1\cdot 2 + 2\cdot 1 + 2\cdot 2 = 2+2+4=8.$

Thus for $c=2$, total numbers $=8$.

Case 2: Last digit $c=4$

Then $a+b+4 \equiv 0 \pmod{3} \Rightarrow a+b \equiv 2 \pmod{3}.$

Possible residue pairs for $(a,b)$ giving sum $2 \pmod{3}$ are:

• $(0,2)$
• $(2,0)$
• $(1,1)$

Count actual digits:

• $(0,2)$: $1\cdot 2=2$
• $(2,0)$: $2\cdot 1=2$
• $(1,1)$: $2\cdot 2=4$

Total: $2+2+4=8.$

Thus for $c=4$, total numbers $=8$.

Step 3: Total count

$8+8=16.$

So the total number of 3-digit numbers is $\boxed{16}.$

Verification with stored answer

Stored correct answer = $16$.

Our derived answer also is $16$, so they agree.