1. We need to count all 3-digit numbers divisible by either 2 or 3 but not divisible by 7.

A 3-digit number ranges from $100$ to $999$.

So total numbers under consideration are in the interval $[100,999]$.

2. First, count 3-digit numbers divisible by $2$ or $3$.

Let

• $A =$ set of 3-digit numbers divisible by $2$
• $B =$ set of 3-digit numbers divisible by $3$

We need $|A \cup B|$.

Using inclusion-exclusion, $|A\cup B|=|A|+|B|-|A\cap B|$

Count $|A|$

3-digit multiples of $2$ go from $100$ to $998$.

Number of terms: $\frac{998-100}{2}+1=\frac{898}{2}+1=449+1=450$

So, $|A|=450$

Count $|B|$

Smallest 3-digit multiple of $3$ is $102$, largest is $999$.

Number of terms: $\frac{999-102}{3}+1=\frac{897}{3}+1=299+1=300$

So, $|B|=300$

Count $|A\cap B|$

Numbers divisible by both $2$ and $3$ are divisible by $6$.

Smallest 3-digit multiple of $6$ is $102$, largest is $996$.

Number of terms: $\frac{996-102}{6}+1=\frac{894}{6}+1=149+1=150$

So, $|A\cap B|=150$

Hence, $|A\cup B|=450+300-150=600$

So, there are $600$ three-digit numbers divisible by $2$ or $3$.

3. From these, remove those which are also divisible by $7$.

So now count 3-digit numbers divisible by $(2 \text{ or } 3)$ and divisible by $7$.

That means numbers divisible by:

• $14$ (for divisible by $2$ and $7$), or
• $21$ (for divisible by $3$ and $7$)

Let

• $C =$ multiples of $14$ among 3-digit numbers
• $D =$ multiples of $21$ among 3-digit numbers

We need $|C\cup D|$.

Again by inclusion-exclusion, $|C\cup D|=|C|+|D|-|C\cap D|$

Count $|C|$

Smallest 3-digit multiple of $14$ is $112$, largest is $994$.

Number of terms: $\frac{994-112}{14}+1=\frac{882}{14}+1=63+1=64$

So, $|C|=64$

Count $|D|$

Smallest 3-digit multiple of $21$ is $105$, largest is $987$.

Number of terms: $\frac{987-105}{21}+1=\frac{882}{21}+1=42+1=43$

So, $|D|=43$

Count $|C\cap D|$

Numbers divisible by both $14$ and $21$ are divisible by $\operatorname{lcm}(14,21)=42$

Smallest 3-digit multiple of $42$ is $126$, largest is $966$.

Number of terms: $\frac{966-126}{42}+1=\frac{840}{42}+1=20+1=21$

So, $|C\cap D|=21$

Thus, $|C\cup D|=64+43-21=86$

So, among those divisible by $2$ or $3$, exactly $86$ are also divisible by $7$.

4. Therefore required count is $600-86=514$

5. Final Answer: $\boxed{514}$

6. Comparison with stored correct answer:

Stored correct answer = $514$

Our derived answer = $514$

So the answer agrees with the stored correct answer.