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

Step 1: Numbers divisible by 2 and 3 (i.e., divisible by 6). 
- First 3-digit number divisible by 6: 102 
- Last 3-digit number divisible by 6: 996 
- Common difference: 6 The total number of terms is: $$\frac{996 - 102}{6} + 1 = 150.$$ 
Step 2: Exclude numbers divisible by 4 or 9.
- Numbers divisible by 6 and 4 are divisible by 12. 
The number of terms divisible by 12: $$\frac{996 - 108}{12} + 1 = 75.$$ - Numbers divisible by 6 and 9 are divisible by 18. The number of terms divisible by 18: $$\frac{990 - 108}{18} + 1 = 50.$$ - Numbers divisible by 6, 4, and 9 are divisible by 36. The number of terms divisible by 36: $$\frac{972 - 108}{36} + 1 = 25.$$ Step 3: Apply inclusion-exclusion. $$\text{Required number of terms} = 150 - (75 + 50 - 25) = 150 - 100 = 150.$$ Thus, the correct answer is option (3) 150.