Let a, b, c, d be non zero distinct digits. The number of 4 digit numbers abcd such that ab + cd is even is divisible by :

Options

A. 3
B. 4
C. 7
D. 11

Answer

To solve the problem, we need to find the number of 4-digit numbers \( abcd \) such that \( ab + cd \) is even, where \( a, b, c, d \) are distinct non-zero digits.

### Step-by-Step Solution:

1. **Understanding the Condition for Evenness**:
- The sum \( ab + cd \) is even if both \( ab \) and \( cd \) are either even or both are odd. This gives us two cases to consider:
- Case 1: Both \( ab \) and \( cd \) are even.
- Case 2: Both \( ab \) and \( cd \) are odd.

2. **Identifying Even and Odd Digits**:
- The non-zero digits available are \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \).
- The even digits are \( 2, 4, 6, 8 \) (4 even digits).
- The odd digits are \( 1, 3, 5, 7, 9 \) (5 odd digits).

3. **Calculating for Case 1 (Both \( ab \) and \( cd \) are even)**:
- We need to select 2 even digits from the 4 available.
- The number of ways to choose 2 even digits from 4 is given by \( \binom{4}{2} \).
- We can arrange these 2 even digits in \( 2! \) ways for \( ab \) and \( cd \).
- The total arrangements for this case:
\[ \text{Total for Case 1} = \binom{4}{2} \times 2! \times 2! = 6 \times 2 \times 2 = 24 \]

4. **Calculating for Case 2 (Both \( ab \) and \( cd \) are odd)**:
- We need to select 2 odd digits from the 5 available.
- The number of ways to choose 2 odd digits from 5 is given by \( \binom{5}{2} \).
- We can arrange these 2 odd digits in \( 2! \) ways for \( ab \) and \( cd \).
- The total arrangements for this case:
\[ \text{Total for Case 2} = \binom{5}{2} \times 2! \times 2! = 10 \times 2 \times 2 = 40 \]

5. **Combining Both Cases**:
- The total number of valid combinations is the sum of the totals from both cases:
\[ \text{Total Combinations} = 24 + 40 = 64 \]

6. **Finding Divisibility**:
- We need to determine if 64 is divisible by any specific numbers.
- The factors of 64 are \( 1, 2, 4, 8, 16, 32, 64 \).
- Therefore, 64 is divisible by \( 1, 2, 4, 8, 16, 32, \) and \( 64 \).

### Conclusion:
The number of 4-digit numbers \( abcd \) such that \( ab + cd \) is even is **64**, and it is divisible by **1, 2, 4, 8, 16, 32,** and **64**.