What is the largest digit $N$ for which $2345N$ is divisible by 6?
To be divisible by 6, a number must be divisible by both 2 and 3. To be divisible by 2, the last digit must be even, so $N$ must be 0, 2, 4, 6, or 8. To be divisible by 3, the sum of the digits must be divisible by 3, so $2 + 3 + 4 + 5 + N$ must be divisible by 3. The sum of the first four digits is 14, which is not divisible by 3, so $N$ must be either 1, 4, or 7 to make the sum divisible by 3. The only digit that satisfies both conditions is 4, so $N = 4$. 