A number when divided successively by 13 and 3 gives respective remainder 5 and 1. What will the remainder when the largest such two-digit number is divided by 19.

Answer

To solve the problem step by step, we need to find a number that meets the conditions given in the question and then determine the remainder when that number is divided by 19.

### Step 1: Understand the conditions
The number we are looking for, let's call it \( N \), when divided by 13 gives a remainder of 5, and when divided by 3 gives a remainder of 1. This can be expressed mathematically as:
- \( N \equiv 5 \, (\text{mod} \, 13) \)
- \( N \equiv 1 \, (\text{mod} \, 3) \)

### Step 2: Express \( N \) in terms of \( k \)
From the second condition, we can express \( N \) in terms of a quotient \( k \):
\[ N = 3k + 1 \]
This means that \( N \) can be any number that is 1 more than a multiple of 3.

### Step 3: Substitute \( N \) into the first condition
Now, we substitute \( N \) into the first condition:
\[ 3k + 1 \equiv 5 \, (\text{mod} \, 13) \]
Subtracting 1 from both sides gives:
\[ 3k \equiv 4 \, (\text{mod} \, 13) \]

### Step 4: Solve for \( k \)
To solve for \( k \), we need to find the multiplicative inverse of 3 modulo 13. The inverse is a number \( x \) such that:
\[ 3x \equiv 1 \, (\text{mod} \, 13) \]
By testing values, we find that \( x = 9 \) works because:
\[ 3 \times 9 = 27 \equiv 1 \, (\text{mod} \, 13) \]
Now, we multiply both sides of \( 3k \equiv 4 \, (\text{mod} \, 13) \) by 9:
\[ k \equiv 36 \, (\text{mod} \, 13) \]
Calculating \( 36 \mod 13 \):
\[ 36 \div 13 = 2 \quad \text{(remainder 10)} \]
So,
\[ k \equiv 10 \, (\text{mod} \, 13) \]

### Step 5: Express \( k \)
This means \( k \) can be expressed as:
\[ k = 13m + 10 \quad \text{for some integer } m \]

### Step 6: Substitute back to find \( N \)
Substituting \( k \) back into the equation for \( N \):
\[ N = 3(13m + 10) + 1 = 39m + 30 + 1 = 39m + 31 \]

### Step 7: Find the largest two-digit number
To find the largest two-digit number, we set \( N \leq 99 \):
\[ 39m + 31 \leq 99 \]
Subtracting 31 from both sides gives:
\[ 39m \leq 68 \]
Dividing by 39:
\[ m \leq \frac{68}{39} \approx 1.74 \]
Thus, the largest integer \( m \) can be is 1.

### Step 8: Calculate \( N \) for \( m = 1 \)
Substituting \( m = 1 \):
\[ N = 39(1) + 31 = 70 \]

### Step 9: Check if \( N \) meets the conditions
- \( 70 \div 13 = 5 \) remainder \( 5 \) (Condition 1 satisfied)
- \( 70 \div 3 = 23 \) remainder \( 1 \) (Condition 2 satisfied)

### Step 10: Find the remainder when \( N \) is divided by 19
Now we need to find the remainder when \( 70 \) is divided by \( 19 \):
\[ 70 \div 19 = 3 \quad \text{(remainder } 13\text{)} \]

### Final Answer
The remainder when the largest two-digit number \( N \) is divided by 19 is **13**.