If the number `2345p60q` is exactly divisible by 3 and 5, then the maximum value of `p+q` is

Options

A. 12
B. 13
C. 14
D. 15

Explanation

To solve the problem of finding the maximum value of \( p + q \) for the number \( 2345p60q \) to be exactly divisible by 3 and 5, we will follow these steps:

### Step 1: Check divisibility by 5
For a number to be divisible by 5, its last digit must be either 0 or 5. In our case, the last digit is \( q \). Therefore, \( q \) can be either 0 or 5.

### 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 our number are \( 2, 3, 4, 5, p, 6, 0, q \).

Calculating the sum of the known digits:
\[ 2 + 3 + 4 + 5 + 6 + 0 = 20 \]
Thus, the total sum of the digits is:
\[ 20 + p + q \]

### Step 3: Analyze possible values for \( q \)
1. **If \( q = 0 \)**:
\[ \text{Sum} = 20 + p + 0 = 20 + p \]
We need \( 20 + p \) to be divisible by 3. The possible values of \( p \) that make \( 20 + p \) divisible by 3 can be found by checking:
- \( p = 0 \): \( 20 + 0 = 20 \) (not divisible by 3)
- \( p = 1 \): \( 20 + 1 = 21 \) (divisible by 3)
- \( p = 2 \): \( 20 + 2 = 22 \) (not divisible by 3)
- \( p = 3 \): \( 20 + 3 = 23 \) (not divisible by 3)
- \( p = 4 \): \( 20 + 4 = 24 \) (divisible by 3)
- \( p = 5 \): \( 20 + 5 = 25 \) (not divisible by 3)
- \( p = 6 \): \( 20 + 6 = 26 \) (not divisible by 3)
- \( p = 7 \): \( 20 + 7 = 27 \) (divisible by 3)
- \( p = 8 \): \( 20 + 8 = 28 \) (not divisible by 3)
- \( p = 9 \): \( 20 + 9 = 29 \) (not divisible by 3)

Valid pairs for \( (p, q) \) when \( q = 0 \) are:
- \( (1, 0) \)
- \( (4, 0) \)
- \( (7, 0) \)

2. **If \( q = 5 \)**:
\[ \text{Sum} = 20 + p + 5 = 25 + p \]
We need \( 25 + p \) to be divisible by 3. The possible values of \( p \):
- \( p = 0 \): \( 25 + 0 = 25 \) (not divisible by 3)
- \( p = 1 \): \( 25 + 1 = 26 \) (not divisible by 3)
- \( p = 2 \): \( 25 + 2 = 27 \) (divisible by 3)
- \( p = 3 \): \( 25 + 3 = 28 \) (not divisible by 3)
- \( p = 4 \): \( 25 + 4 = 29 \) (not divisible by 3)
- \( p = 5 \): \( 25 + 5 = 30 \) (divisible by 3)
- \( p = 6 \): \( 25 + 6 = 31 \) (not divisible by 3)
- \( p = 7 \): \( 25 + 7 = 32 \) (not divisible by 3)
- \( p = 8 \): \( 25 + 8 = 33 \) (divisible by 3)
- \( p = 9 \): \( 25 + 9 = 34 \) (not divisible by 3)

Valid pairs for \( (p, q) \) when \( q = 5 \) are:
- \( (2, 5) \)
- \( (5, 5) \)
- \( (8, 5) \)

### Step 4: Calculate \( p + q \)
Now we calculate \( p + q \) for all valid pairs:
- For \( (1, 0) \): \( p + q = 1 + 0 = 1 \)
- For \( (4, 0) \): \( p + q = 4 + 0 = 4 \)
- For \( (7, 0) \): \( p + q = 7 + 0 = 7 \)
- For \( (2, 5) \): \( p + q = 2 + 5 = 7 \)
- For \( (5, 5) \): \( p + q = 5 + 5 = 10 \)
- For \( (8, 5) \): \( p + q = 8 + 5 = 13 \)

### Conclusion
The maximum value of \( p + q \) is \( 13 \).