1. Interpret the question

We need to count all 3-digit numbers between $100$ and $1000$ formed from digits $1,2,3,4,5$ without repetition, such that the number is divisible by 3 or 5.

So we count:

• numbers divisible by $3$
• numbers divisible by $5$
• subtract numbers divisible by both $3$ and $5$

This is the inclusion-exclusion principle.

2. Total possible 3-digit numbers from $\{1,2,3,4,5\}$ without repetition

This is: ${}^5P_3 = 5\times 4\times 3 = 60$

But we only need those divisible by $3$ or $5$.

3. Count numbers divisible by 5

A number is divisible by $5$ only if its unit digit is $5$ (since $0$ is not available).

Fix the last digit as $5$.

Now the first two places are filled from $\{1,2,3,4\}$ without repetition: ${}^4P_2 = 4\times 3 = 12$

So, $N(5)=12$

4. Count numbers divisible by 3

A number is divisible by $3$ if the sum of its digits is divisible by $3$.

We must choose 3 distinct digits from $\{1,2,3,4,5\}$ whose sum is divisible by $3$.

All 3-digit selections:

• $\{1,2,3\}$, sum $=6$ ✓
• $\{1,2,4\}$, sum $=7$ ✗
• $\{1,2,5\}$, sum $=8$ ✗
• $\{1,3,4\}$, sum $=8$ ✗
• $\{1,3,5\}$, sum $=9$ ✓
• $\{1,4,5\}$, sum $=10$ ✗
• $\{2,3,4\}$, sum $=9$ ✓
• $\{2,3,5\}$, sum $=10$ ✗
• $\{2,4,5\}$, sum $=11$ ✗
• $\{3,4,5\}$, sum $=12$ ✓

So there are $4$ valid sets.

For each set, the number of 3-digit arrangements is: $3! = 6$

Hence, $N(3)=4\times 6=24$

5. Count numbers divisible by both 3 and 5

Such numbers must be divisible by $15$.

So:

• last digit must be $5$
• sum of digits must be divisible by $3$

Let the number be of form $\_\_5$. Choose 2 digits from $\{1,2,3,4\}$ such that their sum plus $5$ is divisible by $3$.

That means: $(a+b+5) \equiv 0 \pmod 3$ Since $5\equiv 2 \pmod 3$, $a+b \equiv 1 \pmod 3$

Check pairs from $\{1,2,3,4\}$:

• $1+2=3 \equiv 0$ ✗
• $1+3=4 \equiv 1$ ✓
• $1+4=5 \equiv 2$ ✗
• $2+3=5 \equiv 2$ ✗
• $2+4=6 \equiv 0$ ✗
• $3+4=7 \equiv 1$ ✓

Valid pairs: $\{1,3\}$ and $\{3,4\}$.

Each pair can be arranged in the first two places in $2! = 2$ ways:

• with $\{1,3\}$: $135, 315$
• with $\{3,4\}$: $345, 435$

So, $N(3\cap 5)=4$

6. Apply inclusion-exclusion

$N(3\cup 5)=N(3)+N(5)-N(3\cap 5)$ $=24+12-4=32$

7. Final answer

The total number of such numbers is: $\boxed{32}$

8. Comparison with stored answer

Stored correct answer = $32$

Our derived answer is also $32$, so they agree.