Really tough. I started with 8** and 7** and found out the difference as 49. Since 49 was already present as an answer choice, I marked it.

to get minimum difference we need the following

the 2 hundred numbers must be as close as possible, for example 1 and 2, 2 and 3, 6 and 7 or 7 and 8
second. the 2 ten number must be as far as possible, there is only 1 couple which is 1 and 8. from this point now ther are only 2 case for the 2 hundred numbers which are 2 and 3 , and 6 and 7

suppose it is 2 and 3. we plug the number, 287 and 316, of course, for digit numbers, put the bigger into the smaller number
supoose, it is 6 and 7, we plug the number , 683 and 782,

we get the result .

to get the minimum difference, we find minim difference in hundred digits, which are
1 and 2
2 and 3
6 and 7
7 and 8

now we need to find the greatest 2-digit number and smallest 2-digit number to add the greatest to the number with smaller hundred digit and add the smallest to the number with greater hundred digit

for the first couple, we have 3,6,7,8, and we can find 87 and 36
for the second couple, we have 1,6,7,8 and we can find 87 and 16
for the third couple, we have 1,2,3,8 and we can find 83 and 12
for the forth couple, we have , 1,2,3,6 and we can find: 63 and 12

now we do
smaller 2 digit number-greater 2 digit number, adding 1 to be hundred digit of the smaller
1 36-87=59
1 16-87=29
1 12-83=29
1 12-83=29

answer is 29

The big thing is to remember that to get the smallest number we need to carry 1 over from hundreds and 1 from tens digit.
(so hundreds column should be equal to 1, and in tens/ones column of the second number should be more than the first number) to have 2 carry overs.
The smallest answers, A and B both have 9. Can we get 9 in the units digit? Yes... in 4 ways with carrying 1 over (7-8, 6-7, 2-3, 1-2)

_ _ 7
_ _ 8

Add the other digits, following the rules outlined:
2 3 7
1 6 8
=6 9

The others possibilities are:
3 1 6
2 8 7
=2 9

7 1 2
6 8 3
=2 9

7 3 1
6 8 2
=4 9

So A is definitely possible.

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?

Let N be n1n2n3 and let M be m1m2m3

min (|100(n1-m1) + 10(n2-m2) + (n3-m3)|) = ?

Let N be 287
Let M be 316
316 - 287 = 29

IMO A
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I guess a good way to tackle this question would be to plug in numbers

We know that N and M are composed of either 1,2,3,6,7,8 only

Notice that (8 _ _ // 7 _ _) and (7 _ _ // 6 _ _) are the only two pairs that can give you the smallest difference because the other numbers are 1,2,3 and 6 _ _ - 3 _ _ will be a bigger difference.

So, we only have to check for these two pairs.

Check for:
8_ _ and 7_ _
Now to find the smallest difference you would need to put smaller values in 8 _ _ i.e. minimize this number and larger values in 7 _ _ i.e maximize this number.
Thus, the number can be 812 and 763, the difference is 49

Now, check the options, 29 is also there, so we have to check for the other pair now.
You dont need to check for any other 8 _ _ and 7 _ _ combination because any such combination's difference would be > 49 since we have already minimized 8 _ _ and maximized 7 _ _

Check for:
7 _ _ and 6 _ _
Similar approach, minimize 7 _ _ and maximize 6 _ _
We get, 712 and 683, the difference is 29

ANSWER : A

Hi,

Because we want the smallest possible difference between N and M, we will minimize N's tens and maximize M's tens.
Our min is 1 and our max is 8. Now we have two options:

713-682 = 31
or
316-287=29

Answer A)

Hmm, I didnt think it was so hard compared to what Ive seen on here. To me, a bunch of lines forming angles is the hardest possible question.

Here you instantly see that the hundreds digit must be 1 apart. Then simply maximize the difference between the tens digits and you are home.

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