The total number of 6-digit numbers in which the sum of the digits is divisible by 5, is

Options

A. 180000
B. 540000
C. `5xx10^(5)`
D. none of these

Explanation

To solve the problem of finding the total number of 6-digit numbers in which the sum of the digits is divisible by 5, we can follow these steps:

### Step 1: Determine the structure of a 6-digit number
A 6-digit number can be represented as \(d_1d_2d_3d_4d_5d_6\), where \(d_1\) is the first digit and cannot be zero (since it is a 6-digit number). The remaining digits \(d_2, d_3, d_4, d_5, d_6\) can be any digit from 0 to 9.

### Step 2: Count the possible choices for each digit
- The first digit \(d_1\) can be any digit from 1 to 9 (9 options).
- Each of the remaining digits \(d_2, d_3, d_4, d_5, d_6\) can be any digit from 0 to 9 (10 options each).

Thus, the total number of ways to choose the first five digits is:
\[ 9 \times 10 \times 10 \times 10 \times 10 = 9 \times 10^4 \]

### Step 3: Analyze the sum of the digits
Let \(S = d_1 + d_2 + d_3 + d_4 + d_5 + d_6\). We want \(S\) to be divisible by 5.

The possible remainders when \(S\) is divided by 5 can be:
- \(5m\)
- \(5m + 1\)
- \(5m + 2\)
- \(5m + 3\)
- \(5m + 4\)

### Step 4: Determine the last digit
The last digit \(d_6\) will determine the total sum \(S\) modulo 5. Depending on the sum of the first five digits \(d_1 + d_2 + d_3 + d_4 + d_5\), we can choose \(d_6\) such that the total sum \(S\) is divisible by 5.

- If \(d_1 + d_2 + d_3 + d_4 + d_5 \equiv 0 \mod 5\), then \(d_6\) can be \(0, 5\) (2 options).
- If \(d_1 + d_2 + d_3 + d_4 + d_5 \equiv 1 \mod 5\), then \(d_6\) can be \(4, 9\) (2 options).
- If \(d_1 + d_2 + d_3 + d_4 + d_5 \equiv 2 \mod 5\), then \(d_6\) can be \(3, 8\) (2 options).
- If \(d_1 + d_2 + d_3 + d_4 + d_5 \equiv 3 \mod 5\), then \(d_6\) can be \(2, 7\) (2 options).
- If \(d_1 + d_2 + d_3 + d_4 + d_5 \equiv 4 \mod 5\), then \(d_6\) can be \(1, 6\) (2 options).

### Step 5: Calculate the total combinations
Since there are 2 valid choices for \(d_6\) for each case of the sum of the first five digits, the total number of valid 6-digit numbers is:
\[ \text{Total} = (9 \times 10^4) \times 2 = 18 \times 10^4 \]

### Final Answer
Thus, the total number of 6-digit numbers in which the sum of the digits is divisible by 5 is:
\[ \boxed{180000} \]