1. Total number of 3-digit numbers

We have the digits $\{2,3,4,5,7\}$ and repetition is not allowed.

The total number of 3-digit numbers that can be formed is:

$^5P_3 = 5\times 4\times 3 = 60$

2. Condition for divisibility by 3

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

Let us classify the given digits modulo $3$:

• $2 \equiv 2 \pmod{3}$
• $3 \equiv 0 \pmod{3}$
• $4 \equiv 1 \pmod{3}$
• $5 \equiv 2 \pmod{3}$
• $7 \equiv 1 \pmod{3}$

So the residue classes are:

• Remainder $0$: $\{3\}$
• Remainder $1$: $\{4,7\}$
• Remainder $2$: $\{2,5\}$

For divisibility by $3$, the sum of the residues must be $0 \pmod{3}$.

Possible residue combinations for 3 digits are:

• $(0,1,2)$
• $(1,1,1)$ — not possible since only two digits with remainder $1$
• $(2,2,2)$ — not possible since only two digits with remainder $2$
• $(0,0,0)$ — not possible since only one digit with remainder $0$

Thus only the combination $(0,1,2)$ works.

3. Count numbers divisible by 3

• Choose the digit with remainder $0$: only $3$ → $1$ way
• Choose one digit with remainder $1$: from $\{4,7\}$ → $2$ ways
• Choose one digit with remainder $2$: from $\{2,5\}$ → $2$ ways

So the number of valid sets of digits is:

$1\times 2\times 2 = 4$

Each set of 3 distinct digits can be arranged in:

$3! = 6$

Hence numbers divisible by $3$:

$4\times 6 = 24$

4. Count numbers not divisible by 3

$60 - 24 = 36$

5. Final Answer

The required number is:

$\boxed{36}$

6. Comparison with stored answer

Stored correct answer = $36$

Our derived answer = $36$

So, the answer agrees with the stored correct answer.