Largest four digit number which when divided by 15 leaves a remainder of 12 and if the same number is divided by 8 it leaves the remainder 5. Such a greatest possible number is :

Options

A. 9963
B. 9957
C. 9945
D. 9999

Answer

To find the largest four-digit number that meets the specified conditions, we can follow these steps:

### Step 1: Understand the conditions
We need to find a four-digit number \( N \) such that:
- When \( N \) is divided by 15, it leaves a remainder of 12.
- When \( N \) is divided by 8, it leaves a remainder of 5.

### Step 2: Set up the equations
From the conditions, we can express \( N \) in terms of the divisors and remainders:
1. \( N \equiv 12 \mod 15 \)
2. \( N \equiv 5 \mod 8 \)

### Step 3: Rewrite the equations
We can rewrite these congruences:
1. \( N = 15k + 12 \) for some integer \( k \)
2. \( N = 8m + 5 \) for some integer \( m \)

### Step 4: Find a common solution
We need to find a common \( N \) that satisfies both equations. We can substitute the expression for \( N \) from the first equation into the second:
\[ 15k + 12 \equiv 5 \mod 8 \]
Now, simplify \( 15k + 12 \mod 8 \):
\[ 15 \mod 8 = 7 \quad \text{and} \quad 12 \mod 8 = 4 \]
Thus, we have:
\[ 7k + 4 \equiv 5 \mod 8 \]
Subtract 4 from both sides:
\[ 7k \equiv 1 \mod 8 \]

### Step 5: Solve for \( k \)
To solve \( 7k \equiv 1 \mod 8 \), we can find the multiplicative inverse of 7 modulo 8. Testing values:
- \( k = 1 \): \( 7 \times 1 = 7 \mod 8 \)
- \( k = 2 \): \( 7 \times 2 = 14 \equiv 6 \mod 8 \)
- \( k = 3 \): \( 7 \times 3 = 21 \equiv 5 \mod 8 \)
- \( k = 4 \): \( 7 \times 4 = 28 \equiv 4 \mod 8 \)
- \( k = 5 \): \( 7 \times 5 = 35 \equiv 3 \mod 8 \)
- \( k = 6 \): \( 7 \times 6 = 42 \equiv 2 \mod 8 \)
- \( k = 7 \): \( 7 \times 7 = 49 \equiv 1 \mod 8 \)

Thus, \( k \equiv 7 \mod 8 \), which means:
\[ k = 8j + 7 \quad \text{for some integer } j \]

### Step 6: Substitute back to find \( N \)
Substituting \( k \) back into the equation for \( N \):
\[ N = 15(8j + 7) + 12 = 120j + 105 + 12 = 120j + 117 \]

### Step 7: Find the largest four-digit number
We want \( N \) to be a four-digit number:
\[ 120j + 117 \leq 9999 \]
Subtracting 117 from both sides:
\[ 120j \leq 9882 \]
Dividing by 120:
\[ j \leq \frac{9882}{120} \approx 82.35 \]
Thus, the largest integer \( j \) is 82.

### Step 8: Calculate \( N \)
Substituting \( j = 82 \):
\[ N = 120 \times 82 + 117 = 9840 + 117 = 9957 \]

### Step 9: Verify the solution
1. **Check \( N \mod 15 \)**:
\[ 9957 \div 15 = 663 \quad \text{remainder } 12 \]
2. **Check \( N \mod 8 \)**:
\[ 9957 \div 8 = 1244 \quad \text{remainder } 5 \]

Both conditions are satisfied.

### Final Answer
The largest four-digit number that meets the conditions is **9957**.

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