N and M are each 3-digit integers

step 1: Look at the options.

The smallest one is A) 29

Now can we have two three digit integres whose difference is 3 ?

i.e N - M = 29 ?

Step 2: what can be last digit of N and M to get a result of 9 ?
n can be XY0 and M can be AB1

Step 2 : now can XY - (A(B+ 1 form Carry over) = 2

Step 3 X80 - X51
Step 4 X can be anything 2,3,6,7 ( As one digit is either N or M [not mutually exclusive])

hence A is the answer.
--- I am new to GMAT, hence for any better approach, please correct me.

We could do a bit of algebra here. Let's say that our numbers are abc and def. Then we have

abc - def =

(100a + 10b + c) - (100d + 10e + f) =

100*(a - d) + 10*(b - e) + (c - f)

Since all our numbers are distinct, we're going to get positive or negative numbers for (a - d), (b - e), and (c - f).

We want (a - d) to be as small as possible, so we'll have (a - d) = 1. That starts us off with 100*1, and we can use either a=3, b=2 or a=6, b=5. Let's use a = 3, b = 2.

Now we want (b - e) to be as NEGATIVE as possible, since we want to subtract as much as we can from the 100*1 we started with. The furthest we can go in our set is (1 - 8), or -7.

Now we've got 100 - 70. If we can make the last term negative, we're set. We're left with only 6 and 5, so we'll do (5 - 6), or -1.

This gives us 100 - 70 - 1, or 29. (Or, if we've lost track of the numbers, we've got 315 - 286.)