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 \( \boxed{125} \).

Answer 2

Given: We are looking for the number of 3-digit numbers that satisfy the following conditions:

• Divisible by 2
• Divisible by 3
• Not divisible by 4
• Not divisible by 9

 

Step 1: Find the total number of 3-digit numbers divisible by both 2 and 3

A number divisible by both 2 and 3 is divisible by their least common multiple (LCM). The LCM of 2 and 3 is 6. Therefore, we need to find how many 3-digit numbers are divisible by 6.

The smallest 3-digit number is 100, and the largest is 999. To find the number of multiples of 6 between 100 and 999, we use: \[ \text{First multiple of 6} = 102 \quad (\text{since } 100 \div 6 = 16 \text{ remainder } 4) \] \[ \text{Last multiple of 6} = 996 \quad (\text{since } 999 \div 6 = 166 \text{ remainder } 3) \] Now, calculate the number of multiples of 6: \[ \text{Number of multiples of 6} = \frac{996 - 102}{6} + 1 = \frac{894}{6} + 1 = 149 + 1 = 150. \] Thus, there are 150 3-digit numbers divisible by 6.

Step 2: Exclude numbers divisible by 4 and 9

We need to exclude numbers that are divisible by both 4 or 9. Let's use the inclusion-exclusion principle:

Divisible by 4:

A number divisible by both 6 and 4 must be divisible by the LCM of 6 and 4, which is 12. Now, find how many 3-digit numbers are divisible by 12: \[ \text{First multiple of 12} = 108 \quad (\text{since } 100 \div 12 = 8 \text{ remainder } 4) \] \[ \text{Last multiple of 12} = 996 \quad (\text{since } 999 \div 12 = 83 \text{ remainder } 3) \] \[ \text{Number of multiples of 12} = \frac{996 - 108}{12} + 1 = \frac{888}{12} + 1 = 74 + 1 = 75. \] Thus, there are 75 3-digit numbers divisible by 12.

Divisible by 9:

A number divisible by both 6 and 9 must be divisible by their LCM, which is 18. Now, find how many 3-digit numbers are divisible by 18: \[ \text{First multiple of 18} = 102 \quad (\text{since } 100 \div 18 = 5 \text{ remainder } 10) \] \[ \text{Last multiple of 18} = 990 \quad (\text{since } 999 \div 18 = 55 \text{ remainder } 9) \] \[ \text{Number of multiples of 18} = \frac{990 - 102}{18} + 1 = \frac{888}{18} + 1 = 49 + 1 = 50. \] Thus, there are 50 3-digit numbers divisible by 18.

Divisible by both 4 and 9:

A number divisible by both 6, 4, and 9 must be divisible by their LCM, which is 36. Now, find how many 3-digit numbers are divisible by 36: \[ \text{First multiple of 36} = 108 \quad (\text{since } 100 \div 36 = 2 \text{ remainder } 28) \] \[ \text{Last multiple of 36} = 972 \quad (\text{since } 999 \div 36 = 27 \text{ remainder } 27) \] \[ \text{Number of multiples of 36} = \frac{972 - 108}{36} + 1 = \frac{864}{36} + 1 = 24 + 1 = 25. \] Thus, there are 25 3-digit numbers divisible by 36.

Step 3: Apply Inclusion-Exclusion

The total number of 3-digit numbers divisible by 6 but not by 4 or 9 is: \[ \text{Total} = 150 - (75 + 50 - 25) = 150 - 100 = 50. \] Therefore, there are 50 numbers divisible by 6 but not by 4 or 9.

Final Answer:

\[ \boxed{125} \]