What cou ld be the possible values of X , if 2345691X32 is divisible by 3 (where X is a digit)?

Options

A. 1
B. 4
C. 7
D. All of these

Answer

To find the possible values of \( X \) such that the number \( 2345691X32 \) is divisible by 3, we can follow these steps:

### Step 1: Understand the divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.

### Step 2: Calculate the sum of the known digits
We will first sum the digits of the number \( 2345691X32 \) excluding \( X \):
\[ 2 + 3 + 4 + 5 + 6 + 9 + 1 + 3 + 2 \]

Calculating this step-by-step:
- \( 2 + 3 = 5 \)
- \( 5 + 4 = 9 \)
- \( 9 + 5 = 14 \)
- \( 14 + 6 = 20 \)
- \( 20 + 9 = 29 \)
- \( 29 + 1 = 30 \)
- \( 30 + 3 = 33 \)
- \( 33 + 2 = 35 \)

Thus, the sum of the digits without \( X \) is \( 35 \).

### Step 3: Include \( X \) in the sum
Now, we include \( X \) in the sum:
\[ 35 + X \]

### Step 4: Determine conditions for divisibility by 3
We need \( 35 + X \) to be divisible by 3. To find the possible values of \( X \), we can calculate \( 35 \mod 3 \):
\[ 35 \div 3 = 11 \quad \text{remainder } 2 \]
Thus, \( 35 \equiv 2 \mod 3 \).

For \( 35 + X \) to be divisible by 3, we need:
\[ (35 + X) \mod 3 \equiv 0 \]
This means:
\[ (2 + X) \mod 3 \equiv 0 \]
Which simplifies to:
\[ X \mod 3 \equiv 1 \]

### Step 5: Find possible values of \( X \)
The digits \( X \) can take values from 0 to 9. The values of \( X \) that satisfy \( X \mod 3 \equiv 1 \) are:
- \( X = 1 \) (since \( 1 \mod 3 = 1 \))
- \( X = 4 \) (since \( 4 \mod 3 = 1 \))
- \( X = 7 \) (since \( 7 \mod 3 = 1 \))

### Conclusion
The possible values of \( X \) that make \( 2345691X32 \) divisible by 3 are \( 1, 4, \) and \( 7 \).