1. Conditions for divisibility by 15

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

So the 6-digit number must satisfy:

• last digit is $5$ (for divisibility by $5$),
• sum of digits is divisible by $3$ (for divisibility by $3$).

2. Fix the last digit

Since the digits available are $1,2,3,4,5,6,7$ and there is no $0$, the only way for the number to be divisible by $5$ is that its last digit must be $5$.

So we must form a 6-digit number using digits from $\{1,2,3,4,5,6,7\}$ without repetition, with $5$ fixed at the units place.

Thus we need to choose the other $5$ digits from $\{1,2,3,4,6,7\}.$

3. Use divisibility by 3

The sum of all digits $1$ to $7$ is $1+2+3+4+5+6+7=28.$

A 6-digit number is formed by omitting exactly one digit from these 7 digits.

If the omitted digit is $x$, then the sum of the 6 used digits is $28-x.$

For divisibility by $3$, we need $28-x \equiv 0 \pmod{3}.$ Since $28 \equiv 1 \pmod{3},$ we get $1-x \equiv 0 \pmod{3} \implies x \equiv 1 \pmod{3}.$

Among the digits $1,2,3,4,5,6,7$, those congruent to $1 \pmod{3}$ are $1,4,7.$

But digit $5$ must be present as the last digit, so the omitted digit cannot be $5$ anyway. Hence valid omitted digits are: $1,4,7.$

So there are $3$ valid choices for the omitted digit.

4. Arrange the remaining digits

After omitting one valid digit, we have exactly 6 digits including $5$, with $5$ fixed in the last place.

The remaining $5$ digits can be arranged in the first $5$ positions in $5! = 120$ ways.

For each of the $3$ valid omitted digits, number of such arrangements is $120$.

Therefore total number of required 6-digit numbers is $3 \times 5! = 3 \times 120 = 360.$

5. Final answer

$\boxed{360}$

6. Comparison with stored correct answer

Stored correct answer: $360$

Our derived answer is also $360$, so they agree.