A number 'x' is completely divisible by 9 but leaves remainder 3 when divided by 6, 5, 7 and 8. Find the sum of digits of 'x' ?

Options

A. 21
B. 22
C. 18
D. 24

Answer

To solve the problem step by step, we need to find a number \( x \) that meets the following conditions:

1. \( x \) is completely divisible by 9.
2. \( x \) leaves a remainder of 3 when divided by 6, 5, 7, and 8.

### Step 1: Understand the conditions
Since \( x \) leaves a remainder of 3 when divided by 6, 5, 7, and 8, we can express \( x \) in the form:
\[ x = 840k + 3 \]
where \( 840 \) is the least common multiple (LCM) of 6, 5, 7, and 8.

### Step 2: Calculate the LCM
To find the LCM of 6, 5, 7, and 8:
- The prime factorization of each number is:
- \( 6 = 2 \times 3 \)
- \( 5 = 5 \)
- \( 7 = 7 \)
- \( 8 = 2^3 \)

The LCM is found by taking the highest power of each prime:
- \( 2^3 \) from 8
- \( 3^1 \) from 6
- \( 5^1 \) from 5
- \( 7^1 \) from 7

Thus, the LCM is:
\[ LCM = 2^3 \times 3^1 \times 5^1 \times 7^1 = 8 \times 3 \times 5 \times 7 = 840 \]

### Step 3: Set up the equation for \( x \)
Now we can express \( x \) as:
\[ x = 840k + 3 \]

### Step 4: Ensure \( x \) is divisible by 9
Next, we need \( x \) to be divisible by 9. Therefore, we can set up the equation:
\[ 840k + 3 \equiv 0 \, (\text{mod} \, 9) \]

Calculating \( 840 \mod 9 \):
- \( 840 \div 9 = 93 \) remainder \( 3 \)
- Thus, \( 840 \equiv 3 \, (\text{mod} \, 9) \)

Substituting this back into our equation:
\[ 3k + 3 \equiv 0 \, (\text{mod} \, 9) \]
This simplifies to:
\[ 3(k + 1) \equiv 0 \, (\text{mod} \, 9) \]

### Step 5: Solve for \( k \)
This means \( k + 1 \) must be divisible by 3:
\[ k + 1 \equiv 0 \, (\text{mod} \, 3) \]
Thus, \( k \equiv 2 \, (\text{mod} \, 3) \). The possible values for \( k \) can be \( 2, 5, 8, \ldots \)

### Step 6: Try values for \( k \)
Let’s start with \( k = 2 \):
\[ x = 840 \times 2 + 3 = 1680 + 3 = 1683 \]

### Step 7: Check divisibility by 9
Now, we need to check if \( 1683 \) is divisible by 9:
- Sum of the digits of \( 1683 = 1 + 6 + 8 + 3 = 18 \)
- Since \( 18 \) is divisible by 9, \( 1683 \) is also divisible by 9.

### Step 8: Find the sum of the digits of \( x \)
The sum of the digits of \( x = 1683 \) is:
\[ 1 + 6 + 8 + 3 = 18 \]

### Final Answer
Thus, the sum of the digits of \( x \) is \( 18 \).

---