N and M are each 3-digit integers

lheiannie07 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.

In each case, the difference will clearly be LESS THAN 49, so the correct answer is A.

lheiannie07 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

Since we want the positive difference, we can assume N is the larger of the two numbers.

Therefore, in order to find the smallest possible difference between N and M, we want their hundreds digits to be as close as possible. For example, N can be in the 200s and M can be in the 100s. Then we want, from the remaining digits, the tens digit of N to be as small as possible and the tens digit of M to be as large as possible. Therefore, N can be in the 230s and M can be in the 180s. Finally we assign the smaller of the remaining two digits as N’s units digit and the larger of the two remaining digits as M’s units digits. Thus, N = 236 and M = 187 and their difference is N - M = 236 - 187 = 49.

Of course, one might wonder: is it possible to have N and M as other numbers and perhaps find an even smaller difference between the two numbers? The answer is yes. Let’s use the same strategy as above, but now we will have N in the 300s and M in the 200s. In this case, N must be 316 and M must be 287 and N - M = 316 - 287 = 29. Since 29 is the smallest number in the answer choices, 29 must be the smallest difference between N and M.

A