For `(P,Q)gt0` the number 340 PQQ is divisible by both 3 and 8 then the total possible values of (P,Q) is :

Options

A. 2
B. 3
C. 6
D. none of these

Answer

To solve the problem, we need to determine the values of \( P \) and \( Q \) such that the number \( 340PQQ \) is divisible by both 3 and 8.

### Step 1: Check divisibility by 8
A number is divisible by 8 if the last three digits form a number that is divisible by 8. In our case, the last three digits are \( PQQ \).

We can express \( PQQ \) as:
\[ PQQ = 100P + 10Q + Q = 100P + 11Q \]

Now, we need to find values of \( P \) and \( Q \) such that \( 100P + 11Q \) is divisible by 8.

Calculating \( 100 \mod 8 \) and \( 11 \mod 8 \):
\[ 100 \mod 8 = 4 \quad \text{and} \quad 11 \mod 8 = 3 \]
Thus,
\[ 100P + 11Q \equiv 4P + 3Q \mod 8 \]
We need \( 4P + 3Q \equiv 0 \mod 8 \).

### Step 2: Check divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. The digits of \( 340PQQ \) are \( 3, 4, 0, P, Q, Q \).

Calculating the sum of the digits:
\[ 3 + 4 + 0 + P + Q + Q = 7 + P + 2Q \]
We need \( 7 + P + 2Q \equiv 0 \mod 3 \).

### Step 3: Solve the congruences
**For divisibility by 8:**
We need to find pairs \( (P, Q) \) such that:
\[ 4P + 3Q \equiv 0 \mod 8 \]

**For divisibility by 3:**
We need:
\[ 7 + P + 2Q \equiv 0 \mod 3 \]
This simplifies to:
\[ P + 2Q \equiv -7 \equiv 2 \mod 3 \quad \text{(since -7 mod 3 is 2)} \]

### Step 4: Find possible values of \( P \) and \( Q \)
1. **Divisibility by 8:**
- For \( P = 0 \): \( 4(0) + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 0 \mod 8 \) → \( Q = 0, 8 \) (not valid since \( Q > 0 \))
- For \( P = 1 \): \( 4(1) + 3Q \equiv 0 \mod 8 \) → \( 4 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 4 \mod 8 \) → \( Q = 4 \)
- For \( P = 2 \): \( 4(2) + 3Q \equiv 0 \mod 8 \) → \( 8 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 0 \mod 8 \) → \( Q = 0, 8 \) (not valid since \( Q > 0 \))
- For \( P = 3 \): \( 4(3) + 3Q \equiv 0 \mod 8 \) → \( 12 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 4 \mod 8 \) → \( Q = 4 \)
- For \( P = 4 \): \( 4(4) + 3Q \equiv 0 \mod 8 \) → \( 16 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 0 \mod 8 \) → \( Q = 0, 8 \) (not valid since \( Q > 0 \))
- For \( P = 5 \): \( 4(5) + 3Q \equiv 0 \mod 8 \) → \( 20 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 4 \mod 8 \) → \( Q = 4 \)
- For \( P = 6 \): \( 4(6) + 3Q \equiv 0 \mod 8 \) → \( 24 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 0 \mod 8 \) → \( Q = 0, 8 \) (not valid since \( Q > 0 \))
- For \( P = 7 \): \( 4(7) + 3Q \equiv 0 \mod 8 \) → \( 28 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 4 \mod 8 \) → \( Q = 4 \)
- For \( P = 8 \): \( 4(8) + 3Q \equiv 0 \mod 8 \) → \( 32 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 0 \mod 8 \) → \( Q = 0, 8 \) (not valid since \( Q > 0 \))
- For \( P = 9 \): \( 4(9) + 3Q \equiv 0 \mod 8 \) → \( 36 + 3Q \equiv 0 \mod 8 \) → \( 3Q \equiv 4 \mod 8 \) → \( Q = 4 \)

2. **Divisibility by 3:**
- For \( P = 1, Q = 4 \): \( 1 + 2(4) = 9 \equiv 0 \mod 3 \) (valid)
- For \( P = 3, Q = 4 \): \( 3 + 2(4) = 11 \equiv 2 \mod 3 \) (not valid)
- For \( P = 5, Q = 4 \): \( 5 + 2(4) = 13 \equiv 1 \mod 3 \) (not valid)
- For \( P = 7, Q = 4 \): \( 7 + 2(4) = 15 \equiv 0 \mod 3 \) (valid)

### Conclusion
The valid pairs \( (P, Q) \) are:
1. \( (1, 4) \)
2. \( (7, 4) \)

Thus, the total possible values of \( (P, Q) \) is **2**.