PGMAT wrote:N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?

A. 29
B. 49
C. 58
D. 113
E. 131

To MINIMIZE N-M:
The difference between the HUNDREDS digits must be 1.
The TENS digit of N must be as SMALL as possible (1) and the tens digit of M must be as GREAT as possible (8).
Thus, there are only two cases to consider:

Case 1: N = 31X and M = 28X.
Case 2: N = 71X and M = 68X.

Of the remaining 2 digits, the SMALLER must be assigned to N and the GREATER to M, so that the distance between N and M is minimized:
Case 1: N = 316 and M = 287, with the result that N-M = 316-287 = 29.
Case 2: N = 712 and M = 683, with the result that N-M = 712-683 = 29.

The correct answer is A.

Trial and error isn't necessary here.

Suppose you have the number 123. You can write this as 1*100 + 2*10 + 3*1.

Now suppose you have the three digit number abc. You can write this as a*100 + b*10 + c*1.

We want the difference between two three digit numbers, so

abc - def =

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

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

Since all the digits are different, we can't have a = d or b = e, etc. So let's have (a - d) = 1, for that to be as small as possible: we'll have a = 3 and d = 2. (We need the difference to be positive, so (a - d) must be > 0.)

From here, we want to make 10*(b - e) + (c - f) negative, if possible, and we want (b - e) to be smaller than (c - f). If we do this, we'll shrink 100*(a - d), since we're adding negative numbers to it.

If we make (b - e) = 1 - 8, we have (b - e) = -7, giving us

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

100*1 + 10*(-7) + (c - f) =>

30 + (c - f)

We used 1, 2, 3, and 8, so we're left with 6 and 7. We want (c - f) to be negative, so we'll have c = 6 and f = 7. That gives us

30 + -1

or

29

with abc = 316 and def = 287