1. Condition for divisibility by 15

A number is divisible by $15$ if and only if it is divisible by both $3$ and $5$.

So we need:

• last digit must make it divisible by $5$,
• sum of digits must be divisible by $3$.

2. Digits available

The digits allowed are $1,2,3,5$, and repetition is allowed.

For divisibility by $5$, the last digit must be $5$ (since $0$ is not available).

So every valid 4-digit number has the form $abcd$ with $d=5$.

Thus we only need to choose the first three digits from $\{1,2,3,5\}$.

3. Divisibility by 3 condition

The sum of digits must be divisible by $3$.

Since the last digit is $5$, we need $a+b+c+5 \equiv 0 \pmod{3}.$

Now, $5 \equiv 2 \pmod{3},$ so $a+b+c+2 \equiv 0 \pmod{3}$ which gives $a+b+c \equiv 1 \pmod{3}.$

4. Residues of allowed digits modulo 3

The digits $1,2,3,5$ give residues:

• $1 \equiv 1 \pmod{3}$
• $2 \equiv 2 \pmod{3}$
• $3 \equiv 0 \pmod{3}$
• $5 \equiv 2 \pmod{3}$

So for each of the first three positions, the possible residues are:

• one digit of residue $0$ : $\{3\}$
• one digit of residue $1$ : $\{1\}$
• two digits of residue $2$ : $\{2,5\}$

We need ordered triples $(a,b,c)$ such that $a+b+c \equiv 1 \pmod{3}.$

5. Count all ordered triples by residue cases

We count residue patterns of length $3$ whose sum is $1 \pmod{3}$.

Possible residue combinations are:

Case 1: $(1,0,0)$

Sum $=1$.

Number of arrangements of residues: $\frac{3!}{2!}=3.$ For each arrangement:

• residue $1$ digit can only be $1$ : $1$ choice,
• each residue $0$ digit can only be $3$ : $1$ choice each.

So total from this case: $3 \times 1 = 3.$

Case 2: $(2,2,0)$

Sum $=4 \equiv 1 \pmod{3}$.

Number of arrangements of residues: $\frac{3!}{2!}=3.$ For each arrangement:

• residue $0$ digit: $1$ choice ($3$),
• each residue $2$ digit: $2$ choices ($2$ or $5$).

So total from this case: $3 \times 2^2 = 12.$

Case 3: $(1,1,2)$

Sum $=4 \equiv 1 \pmod{3}$.

Number of arrangements of residues: $\frac{3!}{2!}=3.$ For each arrangement:

• each residue $1$ digit: only $1$ choice ($1$),
• residue $2$ digit: $2$ choices ($2$ or $5$).

So total from this case: $3 \times 2 = 6.$

6. Total count

Adding all valid cases: $3+12+6=21.$

Therefore, the number of such 4-digit numbers is $\boxed{21}.$

7. Comparison with stored answer

Stored correct answer = $21$.

Our derived answer is also $21$, so they agree.