N and M are each 3-digit integers. Each of the numbers 1, 2,

Responding to a pm:

You have 6 digits: 1, 2, 3, 6, 7, 8
Each digit needs to be used to make two 3 digit numbers. This means that we will use each of the digits only once and in only one of the numbers. The numbers need to be as close to each other as possible. The numbers cannot be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other.

The first digit (hundreds digit) of both numbers should be consecutive integers i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3**. This gives us lots of options e.g. (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**).

Now let's think about the next digit (the tens digit). To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1 is possible) and the tens digit of the smaller number should be as large as possible (8 if possible). So let's not use 1 and 8 in the hundreds places and reserve them for the tens places since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?
Try and make a pair with (2** and 3**). Make the 2** number as large as possible and make the 3** number as small as possible. We get 287 and 316 (difference is 29) or
Try and make a pair with (6** and 7**). Make the 6** number as large as possible and make the 7** number as small as possible. We get 683 and 712 (difference is 29)

The smallest of the given options is 29 so we need to think no more. Answer must be (A).

The question is not a hit and trial question. It is completely based on logic and hence do not ignore it.

Consider N = (100X1 + 10Y1+ Z1)
Consider M = (100X2 + 10Y2+ Z2)

N - M = 100(X1-X2) + 10(Y1-Y2) + (Z1-Z2)

Lets analyze these terms:-
100(X1-X2) = 100; We need to keep it minimum at 100 (i.e. X1-X2=1 with pair of consecutive numbers). We do not want it over 200 as it will increase the overall value.
10(Y1-Y2) = -70; To offset 100 from above, we should minimize this term to lowest possible negative value. Pick extreme numbers as 1 & 8 -> 10(1-8)= -70
(Z1-Z2) = -1; Excluding (1,8) taken by (Y2,Y2) and pair of consecutive numbers taken by (X1,X2) -> we are left with 1 pair of consecutive numbers -> Minimize it to -1;

Finally, \(N-M=100(X1-X2)+10(Y1-Y2)+(Z1-Z2) = 100-70-1=29.\)

Hence choice(A) is the answer.

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PS: Once you allocate (1,8) to (Y1,Y2), it doesn't matter which pair of consecutive numbers you choose for (X1,X2) and (Z1, Z2). Either of them can take (6,7) or (2,3)
Both these combinations are valid and give minimum difference of 29: (316-287=29) OR (712-683=29)

In a problem like that you have to play with the numbers untill you realize a strategy.

We need to minimize the difference between the two numbers so we need to make the larger number as small as possible and the smaller number as large as possible so their difference is smallest. Looking at the available digits, the smallest difference in the hundreds is 1. So choose the hundreds to be say 3 and 2. For the remaining digits of the larger number, choose the smallest remaining digits ordered to make the number the smallest. For the smaller number, order the remaining digits to make it largest.

So I got: 316 and 287 with difference of 29.

Another possibility is if you choose 7 and 6 as hundreds: 712 and 683 with difference of 29.

Since 29 is the smallest answer given, it must be the right one.

Note, you don't always get 29. For example if you go with 8 and 7 for hundreds, you get 813 and 762 with difference of 49.

Is there any other approach to solve this question, its very time consuming to think of a solution for this question!

GMAT rewards you for thinking. If you are taking too much time, it means you need to learn to focus and think faster (i.e. practice). Don't be surprised if you get such 'logic based' questions which don't have an 'algebra solution' at higher level.

I understood the explanations here but could not figure out a takeaway for this problem .. what is the take away here?

The question is testing your logic skills in number properties. How do you make two 3 digit numbers such that they use different digits but are as close as possible to each other. So you start out with consecutive hundreds digits and so on...
Not every question on GMAT needs to test a defined sub heading in the Quant book. Sometimes, it will require you to develop your own logic. Though admittedly, some questions don't appear very often.

This approach is fairly straightforward, derived from the GMATprep suggested answer:

To minimize the difference in the two numbers, we pick minimum difference in the hundreds digit which is 1. there are 4 combinations:

2-- | 3-- | 7-- | 8--
1-- | 2-- | 6-- | 7--

Next we write down the rest of the available digits for each combination in ascending order:

3,6,7,8 | 1,6,7,8 | 1,2,3,8 | 1,2,3,6

In each combination, our task is to minimize the difference between the two 2-digit numbers (tens and ones).
This can be achieved by choosing the first two available digits in ascending order for the greater number and last two available digits in reverse order for the smaller number.

For example, in the case 2-- , we put, 236 and in the case of 1--, we put 187.

Hope the reason is clear. this is because it will maximize the value of the smaller number and minimize the value of the greater number. hence, the difference is the minimum.

doing so, we get:

236 | 316 | 712 | 812
-187 | -287 | - 683 |-763
-------------------------------
49 | 29 | 29 | 49

Hence the answer is 29. Choice A.

Video solution from Quant Reasoning starts at 23:48
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Hi karishma, does going from the answer choices help? It did help me, but this strategy could be one off and not a good strategy. Thoughts?
chetan2u MartyMurray - Curious to know your approaches to this problem and opinions too. Thank you all.

Started with 113 and tried to find a valid N and M, 286-173 = 113; Got stuck at B - 58. So, I went to 29 and did 861 -732 = 29 and voila!

Hi
You have gone wrong in your calculations. 861-732 = 129.

I tried it and one could get the answer in 30 seconds if logic is used.
Even hit and trial has to be used with some logic to cut down number of trials otherwise you could waste some precious time.
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The question has to be done logically.
Certain observations.
Let the number be ABC and DEF
1) A and D should be consecutive.(say A>D)
2) The most important aspect now would be to make BC the least and EF the maximum. This would make ABC the least and DEF the maximum.
3) E should be highest and B lowest => E=8 and B=1.
ABC - DEF = ABC - (A-1)EF = A1C - (A-1)8F. This itself will give us the answer as it means that difference will be less than 30. Only 29 possible.

A

Let us find possible pairs
A and D can be (3,2) or (7,6)
(3,2): EF to be greatest would mean E>F>C>B, it would be 87. BC to be minimum means BC would be 16.
Thus, ABC is 316 and DEF is 287 => 316-287 = 29
(7,6): EF to be greatest would mean E>F>C>B, it would be 83. BC to be minimum means BC would be 12.
Thus, ABC is 712 and DEF is 683 => 712-683 = 29


A
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Thanks for pointing out. I didn’t realize.

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