If 635P61 is completely divisible by 9, then what is the value of P?

Options

A. 3
B. 5
C. 6
D. 8

Answer

To determine the value of \( P \) in the number \( 635P61 \) such that it is completely divisible by 9, we can follow these steps:

### Step 1: Understand the Rule for Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9.

### Step 2: Calculate the Sum of the Digits
The digits of the number \( 635P61 \) are \( 6, 3, 5, P, 6, 1 \). We will sum these digits:
\[ 6 + 3 + 5 + P + 6 + 1 \]
Calculating the sum of the known digits:
\[ 6 + 3 = 9 \]
\[ 9 + 5 = 14 \]
\[ 14 + 6 = 20 \]
\[ 20 + 1 = 21 \]
Thus, the sum of the digits is:
\[ 21 + P \]

### Step 3: Set Up the Condition for Divisibility
We need \( 21 + P \) to be divisible by 9.

### Step 4: Check Possible Values of \( P \)
The possible values for \( P \) are \( 3, 5, 6, 8 \). We will check each value to see if \( 21 + P \) is divisible by 9.

- **If \( P = 3 \)**:
\[ 21 + 3 = 24 \quad (\text{Not divisible by 9}) \]

- **If \( P = 5 \)**:
\[ 21 + 5 = 26 \quad (\text{Not divisible by 9}) \]

- **If \( P = 6 \)**:
\[ 21 + 6 = 27 \quad (\text{Divisible by 9}) \]

- **If \( P = 8 \)**:
\[ 21 + 8 = 29 \quad (\text{Not divisible by 9}) \]

### Step 5: Conclusion
The only value of \( P \) that makes \( 21 + P \) divisible by 9 is \( P = 6 \).

Thus, the value of \( P \) is:
\[ \boxed{6} \]