How many 4 digit numbers are there, without repetition of digits, If each number is divisible by 5 ?

Answer

To find the number of 4-digit numbers that are divisible by 5 and do not have any repeated digits, we can break the problem down into two cases based on the last digit of the number.

### Step 1: Identify the cases for divisibility by 5
A number is divisible by 5 if its last digit is either 0 or 5. Therefore, we will consider two cases:
- Case 1: The last digit is 0.
- Case 2: The last digit is 5.

### Step 2: Case 1 - Last digit is 0
If the last digit is 0, the first digit can be any digit from 1 to 9 (since it cannot be 0). This gives us 9 options for the first digit. After choosing the first digit, we have 8 remaining digits to choose from for the second digit, and then 7 remaining digits for the third digit.

So, the total number of combinations for this case is:
\[ 9 \times 8 \times 7 \]

### Step 3: Case 2 - Last digit is 5
If the last digit is 5, the first digit can be any digit from 1 to 9 (but not 5). This gives us 8 options for the first digit (1-9 excluding 5). The second digit can be any digit from 0 to 9, excluding the first digit and 5, which gives us 8 options. The third digit can be any digit from 0 to 9, excluding the first digit, the second digit, and 5, which gives us 7 options.

So, the total number of combinations for this case is:
\[ 8 \times 8 \times 7 \]

### Step 4: Calculate the total combinations
Now we add the combinations from both cases:
\[ \text{Total} = (9 \times 8 \times 7) + (8 \times 8 \times 7) \]

Calculating each part:
- For Case 1:
\[ 9 \times 8 \times 7 = 504 \]
- For Case 2:
\[ 8 \times 8 \times 7 = 448 \]

Adding these together:
\[ 504 + 448 = 952 \]

### Final Answer
Thus, the total number of 4-digit numbers that are divisible by 5 and do not have repeated digits is **952**.

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