Question

The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is.

Hint:

When solving divisibility problems, break down the conditions logically and apply the principle of inclusion-exclusion to ensure you don't double-count numbers.

Answer 1

We are asked to find the number of 3-digit numbers divisible by 6 but not divisible by 36.

Step 1: Find the total number of 3-digit numbers divisible by 6

A number is divisible by 6 if it is divisible by both 2 and 3. The first three-digit number divisible by 6 is 102, and the last three-digit number divisible by 6 is 996. To find the total number of 3-digit numbers divisible by 6, we use the formula for the number of terms in an arithmetic sequence: \[ \text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 \] Substituting the values: \[ \text{Number of terms} = \frac{996 - 102}{6} + 1 = \frac{894}{6} + 1 = 149 + 1 = 150. \] Thus, there are 150 numbers divisible by 6.

Step 2: Find the total number of 3-digit numbers divisible by 36

A number is divisible by 36 if it is divisible by both 4 and 9. The first three-digit number divisible by 36 is 108, and the last three-digit number divisible by 36 is 972. Using the same formula for the number of terms: \[ \text{Number of terms} = \frac{972 - 108}{36} + 1 = \frac{864}{36} + 1 = 24 + 1 = 25. \] Thus, there are 25 numbers divisible by 36.

Step 3: Subtract the numbers divisible by 36 from those divisible by 6

The required number of 3-digit numbers that are divisible by 6 but not divisible by 36 is: \[ \text{Required number} = \text{(Divisible by 6)} - \text{(Divisible by 36)} = 150 - 25 = 125. \]

Final Answer:

The required number of 3-digit numbers is \( \boxed{125} \).