N and M are each 3-digit integers. Each of the numbers 1, 2,

Responding to a pm:

You have 6 digits: 1, 2, 3, 6, 7, 8
Each digit needs to be used to make two 3 digit numbers. This means that we will use each of the digits only once and in only one of the numbers. The numbers need to be as close to each other as possible. The numbers cannot be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other.

The first digit (hundreds digit) of both numbers should be consecutive integers i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3**. This gives us lots of options e.g. (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**).

Now let's think about the next digit (the tens digit). To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1 is possible) and the tens digit of the smaller number should be as large as possible (8 if possible). So let's not use 1 and 8 in the hundreds places and reserve them for the tens places since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?
Try and make a pair with (2** and 3**). Make the 2** number as large as possible and make the 3** number as small as possible. We get 287 and 316 (difference is 29) or
Try and make a pair with (6** and 7**). Make the 6** number as large as possible and make the 7** number as small as possible. We get 683 and 712 (difference is 29)

The smallest of the given options is 29 so we need to think no more. Answer must be (A).

The question is not a hit and trial question. It is completely based on logic and hence do not ignore it.