What is the largest two digit number which when divided by 6 and 5 leaves remainder 1 in each case ?

Options

A. 61
B. 93
C. 91
D. 97

Answer

To solve the problem of finding the largest two-digit number that leaves a remainder of 1 when divided by both 6 and 5, we can follow these steps:

### Step 1: Understand the problem
We need to find a two-digit number \( N \) such that:
- \( N \mod 6 = 1 \)
- \( N \mod 5 = 1 \)

### Step 2: Find the least common multiple (LCM)
To solve the problem, we first find the least common multiple (LCM) of 6 and 5.
- The LCM of 6 and 5 is \( 30 \).

### Step 3: Set up the equation
Since \( N \) leaves a remainder of 1 when divided by both 6 and 5, we can express \( N \) in terms of the LCM:
- \( N = 30k + 1 \) for some integer \( k \).

### Step 4: Determine the largest two-digit number
The largest two-digit number is 99. We need to find the largest \( k \) such that \( N \) remains a two-digit number:
- \( 30k + 1 \leq 99 \)
- \( 30k \leq 98 \)
- \( k \leq \frac{98}{30} \)
- \( k \leq 3.2667 \)

Since \( k \) must be an integer, the largest possible value for \( k \) is 3.

### Step 5: Calculate \( N \) for \( k = 3 \)
Now we substitute \( k = 3 \) back into the equation to find \( N \):
- \( N = 30 \times 3 + 1 = 90 + 1 = 91 \)

### Step 6: Verify the result
We need to check if 91 satisfies the original conditions:
- \( 91 \div 6 = 15 \) remainder \( 1 \) (since \( 91 = 6 \times 15 + 1 \))
- \( 91 \div 5 = 18 \) remainder \( 1 \) (since \( 91 = 5 \times 18 + 1 \))

Both conditions are satisfied.

### Conclusion
The largest two-digit number which when divided by 6 and 5 leaves a remainder of 1 in each case is **91**.

---