JEE Mathematics 2025 Question with Solution

Given: We need the number of $3$-digit numbers divisible by $2$ and $3$, but not divisible by $4$ and $9$.

Find: The required count.

A number divisible by $2$ and $3$ is divisible by $6$.

The first $3$-digit multiple of $6$ is $102$ and the last is $996$.

So the total number of $3$-digit numbers divisible by $6$ is

Now count those divisible by $36$. Since $36$ is divisible by both $4$ and $9$, these are excluded in the extracted solution.

The first $3$-digit multiple of $36$ is $108$ and the last is $972$.

Hence the number of $3$-digit multiples of $36$ is

Therefore, the required number is

So, the required number of $3$-digit numbers is $125$.

Given: Count $3$-digit numbers divisible by $2$ and $3$, but not divisible by $4$ and $9$.

Find: The required number.

One extracted approach notes that divisible by $2$ and $3$ means divisible by $6$. Thus total eligible numbers before exclusion are the $3$-digit multiples of $6$:

So their count is

Another extracted approach applies inclusion-exclusion to numbers divisible by $6$ and also divisible by $4$ or $9$:

• divisible by $6$ and $4$ means divisible by $12$
• divisible by $6$ and $9$ means divisible by $18$
• divisible by all three means divisible by $36$

That gives

Therefore,

However, the solution itself concludes with the final boxed answer $125$, and the primary solution also concludes $125$ using the subtraction of multiples of $36$ only. Following the provided solution conclusion as instructed, the answer is taken as $125$.

Therefore, the final answer is $125$.