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Question Pack 1 = N and M are each 3-digit integers. Each of

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Question Pack 1 = N and M are each 3-digit integers. Each of

by richachampion » Thu Dec 08, 2016 2:58 pm
N and M are each 3-digit integers. Each of the number 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
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by Matt@VeritasPrep » Thu Dec 08, 2016 7:35 pm
Let's start with a number to get the idea, then move to algebra.

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
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by Matt@VeritasPrep » Thu Dec 08, 2016 7:38 pm
That isn't the only way to do it, of course!

One fast way is to work with the answers. Since you want the smallest difference, start with the smallest answer, and see if you can make it. If you can, great, you're done! If you can't, OK, shoot for the next one.
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