Consider a number N=21 P 5 3 Q 4.
The number of ordered pairs (P,Q) so that the number' N' is divisible by 9, is

Options

A. 11
B. 12
C. 10
D. 8

Answer

To solve the problem of finding the number of ordered pairs (P, Q) such that the number \( N = 21P53Q4 \) is divisible by 9, we follow these steps:

### Step 1: Understand the divisibility rule for 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 of \( N \)
The digits of \( N \) are 2, 1, P, 5, 3, Q, and 4. We can express the sum of these digits as:
\[ \text{Sum} = 2 + 1 + P + 5 + 3 + Q + 4 \]
Simplifying this, we get:
\[ \text{Sum} = 15 + P + Q \]

### Step 3: Set up the equation for divisibility by 9
For \( N \) to be divisible by 9, the sum \( 15 + P + Q \) must be a multiple of 9. We can express this as:
\[ 15 + P + Q \equiv 0 \mod 9 \]
This simplifies to:
\[ P + Q \equiv -15 \mod 9 \]
Calculating \(-15 \mod 9\):
\[ -15 \equiv 3 \mod 9 \]
Thus, we have:
\[ P + Q \equiv 3 \mod 9 \]

### Step 4: Determine possible values for \( P + Q \)
The possible values for \( P + Q \) that satisfy \( P + Q \equiv 3 \mod 9 \) are:
- \( P + Q = 3 \)
- \( P + Q = 12 \)
- \( P + Q = 21 \)

### Step 5: Analyze each case for ordered pairs (P, Q)

#### Case 1: \( P + Q = 3 \)
The possible pairs (P, Q) are:
- (0, 3)
- (1, 2)
- (2, 1)
- (3, 0)

Total pairs = 4

#### Case 2: \( P + Q = 12 \)
The possible pairs (P, Q) are:
- (3, 9)
- (4, 8)
- (5, 7)
- (6, 6)
- (7, 5)
- (8, 4)
- (9, 3)

Total pairs = 7

#### Case 3: \( P + Q = 21 \)
This case is not possible because both P and Q must be single-digit numbers (0-9), and their sum cannot exceed 18.

### Step 6: Total the number of ordered pairs
Adding the total pairs from the valid cases:
\[ \text{Total pairs} = 4 + 7 = 11 \]

### Final Answer
The number of ordered pairs (P, Q) such that the number \( N \) is divisible by 9 is \( \boxed{11} \).