N is the largest two digit number, which when divided by 3, 4 and 6 leaves the remainder 1, 2 and 4 respectively. What is the remainder when N is divided by 5?

Options

A. 4
B. 2
C. 0
D. 1

Answer

To solve the problem step by step, we need to find the largest two-digit number \( N \) that satisfies the given conditions when divided by 3, 4, and 6, and then determine the remainder when \( N \) is divided by 5.

### Step 1: Understand the conditions
We need to find \( N \) such that:
- \( N \mod 3 = 1 \)
- \( N \mod 4 = 2 \)
- \( N \mod 6 = 4 \)

### Step 2: Rewrite the conditions
We can rewrite the conditions in terms of \( N \):
- \( N = 3k + 1 \) for some integer \( k \)
- \( N = 4m + 2 \) for some integer \( m \)
- \( N = 6p + 4 \) for some integer \( p \)

### Step 3: Find the least common multiple (LCM)
To solve these congruences, we can find the least common multiple (LCM) of the divisors (3, 4, and 6):
- The LCM of 3, 4, and 6 is 12.

### Step 4: Adjust the conditions
Since we need \( N \) to leave specific remainders, we can adjust our conditions:
- For \( N \mod 3 = 1 \): \( N + 2 \equiv 0 \mod 3 \)
- For \( N \mod 4 = 2 \): \( N + 2 \equiv 0 \mod 4 \)
- For \( N \mod 6 = 4 \): \( N + 2 \equiv 0 \mod 6 \)

This means \( N + 2 \) must be a multiple of 12.

### Step 5: Find the largest two-digit number
The largest two-digit number is 99. We need to find the largest two-digit number less than or equal to 99 that is a multiple of 12:
- The largest multiple of 12 less than or equal to 99 is \( 12 \times 8 = 96 \).

### Step 6: Calculate \( N \)
Now, since \( N + 2 \) is a multiple of 12, we have:
- \( N + 2 = 96 + 2 = 98 \)
- Therefore, \( N = 96 \).

### Step 7: Find the remainder when \( N \) is divided by 5
Now we need to find \( N \mod 5 \):
- \( 96 \div 5 = 19 \) remainder \( 1 \) (since \( 5 \times 19 = 95 \))
- Thus, the remainder when \( 96 \) is divided by \( 5 \) is \( 1 \).

### Final Answer
The remainder when \( N \) is divided by 5 is **1**.

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