To solve the problem of finding the number of six-digit numbers between 100,000 and 300,000 that are divisible by 4 and formed by rearranging the digits of 112233, we can follow these steps:
### Step 1: Identify the Range
We need to find six-digit numbers that lie between 100,000 and 300,000. This means that the first digit can only be 1 or 2. Since the number cannot start with 3 (as it would exceed 300,000), we will consider cases where the first digit is either 1 or 2.
### Step 2: Check Divisibility by 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Therefore, we need to check the possible combinations of the last two digits from the digits 1, 1, 2, 2, 3, 3.
### Step 3: List Possible Last Two Digits
The possible pairs of last two digits that can be formed from the digits 1, 1, 2, 2, 3, 3 are:
- 12
- 32
- 22
- 13
- 33
Now, we need to check which of these pairs are divisible by 4:
- 12 ÷ 4 = 3 (divisible)
- 32 ÷ 4 = 8 (divisible)
- 22 ÷ 4 = 5.5 (not divisible)
- 13 ÷ 4 = 3.25 (not divisible)
- 33 ÷ 4 = 8.25 (not divisible)
Thus, the valid pairs for the last two digits that are divisible by 4 are **12** and **32**.
### Step 4: Count Valid Combinations
Now, we will count how many six-digit numbers can be formed with each valid pair of last two digits while ensuring the first digit is either 1 or 2.
#### Case 1: Last two digits are 12
The remaining digits to arrange are 1, 2, 3, 3.
- The first digit can be 1 or 2.
- If the first digit is 1, the remaining digits are 2, 3, 3.
- If the first digit is 2, the remaining digits are 1, 3, 3.
Calculating the arrangements:
- For first digit 1: The arrangement of 2, 3, 3 is given by \( \frac{3!}{1! \cdot 2!} = 3 \).
- For first digit 2: The arrangement of 1, 3, 3 is given by \( \frac{3!}{1! \cdot 2!} = 3 \).
Total for last two digits 12 = 3 + 3 = 6.
#### Case 2: Last two digits are 32
The remaining digits to arrange are 1, 1, 2, 3.
- The first digit can be 1 or 2.
- If the first digit is 1, the remaining digits are 1, 2, 3.
- If the first digit is 2, the remaining digits are 1, 1, 3.
Calculating the arrangements:
- For first digit 1: The arrangement of 1, 2, 3 is given by \( 3! = 6 \).
- For first digit 2: The arrangement of 1, 1, 3 is given by \( \frac{3!}{2!} = 3 \).
Total for last two digits 32 = 6 + 3 = 9.
### Step 5: Final Count
Now, we add the totals from both cases:
- Total from last two digits 12 = 6
- Total from last two digits 32 = 9
Total valid six-digit numbers = 6 + 9 = **15**.
### Conclusion
The total number of six-digit numbers between 100,000 and 300,000 that are divisible by 4 and formed by rearranging the digits of 112233 is **15**.
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