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N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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07 Jul 2012, 20:28
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N and M are each 3digit 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
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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07 Jul 2012, 21:59
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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.



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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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24 Jul 2012, 23:30
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this is so time consuming. Is there a shorter way??
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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13 Oct 2012, 00:18
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nobelgirl777 wrote: N and M are each 3digit 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 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.
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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09 Jan 2013, 20:00
Is there any other approach to solve this question, its very time consuming to think of a solution for this question!
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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09 Jan 2013, 20:24
fozzzy wrote: 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.
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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09 Jan 2013, 22:16
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Consider N = (100X1 + 10Y1+ Z1) Consider M = (100X2 + 10Y2+ Z2) N  M = 100(X1X2) + 10(Y1Y2) + (Z1Z2) Lets analyze these terms:100(X1X2) = 100; We need to keep it minimum at 100 (i.e. X1X2=1 with pair of consecutive numbers). We do not want it over 200 as it will increase the overall value. 10(Y1Y2) = 70; To offset 100 from above, we should minimize this term to lowest possible negative value. Pick extreme numbers as 1 & 8 > 10(18)= 70(Z1Z2) = 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, \(NM=100(X1X2)+10(Y1Y2)+(Z1Z2) = 100701=29.\) Hence choice(A) is the answer. 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: (316287=29) OR (712683=29)
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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24 Jan 2013, 01:35
there is only one way. pick numbers first pick 1,and 2 as hundereds of the 2 number, then make the larger smallest, the smaller largest. then pick 2 and 3 as the hundreds then pick 3 and 4 as the undreds stop, 29 is smallest in the 5 choices. pick 29 and go this question will be appear at the late on the test. dont worry of this question. if we fail on basic question which is on the first of the test, we die. failing on this question is no problem.
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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30 Jan 2013, 03:32
hardest, if we see this on the test date, we are at 50/51 already the difference between hundreds must be 1. there are many couple. 1 and 2, or 7 and 8 the bigger number must be as smallas possible the smaller number must be as big as possibl we should choose 87 as 2 last digit in the smaller number. now, how to choose 1,2,3,6, if we choose 1 and 2 as hundreds of the 2 number we have 63 as the last digits in the smaller. if we choose 23 as hundered of the 2 numbers, we have 61 as the last digits in the smaller. this is worse than above case we choose 1,2 as hundreds of the 2 numbers. 312 287 is the result gmat is terrible when it make this question. but forget this question, we do not need to do this question to get 49/51
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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21 Apr 2013, 06:19
i dont think repetion is possible... hence it would 316 287 = 29
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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21 Apr 2013, 20:57
Since we are asked to find the smallest integer, I began with option A To get 9 as a unit digit we need 12 as the last 2 digits of one integer and 3 as the last digit of another integer we need 8 as the tenth digit of the smaller integer. so we have 712 as the 1st integer and 683 as the 2nd integer.
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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01 May 2013, 11:16
I understood the explanations here but could not figure out a takeaway for this problem .. what is the take away here?
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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02 May 2013, 09:34
TheNona wrote: 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.
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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This is one of the hardest questions I have seen from OG. Even OG marks it as 'Hard'. Should not it be tagged 700+ level instead of 600700? Thanks moderators!
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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03 Jun 2014, 07:38
PraPon wrote: Consider N = (100X1 + 10Y1+ Z1) Consider M = (100X2 + 10Y2+ Z2)
N  M = 100(X1X2) + 10(Y1Y2) + (Z1Z2)
Lets analyze these terms: 100(X1X2) = 100; We need to keep it minimum at 100 (i.e. X1X2=1 with pair of consecutive numbers). We do not want it over 200 as it will increase the overall value. 10(Y1Y2) = 70; To offset 100 from above, we should minimize this term to lowest possible negative value. Pick extreme numbers as 1 & 8 > 10(18)= 70 (Z1Z2) = 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, \(NM=100(X1X2)+10(Y1Y2)+(Z1Z2) = 100701=29.\)
Hence choice(A) is the answer.
 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: (316287=29) OR (712683=29) ProPan, you beauty! This one looks the most efficient solution to me out of all different solutions I have seen so far on different forums. Thanks for sharing
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N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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10 Nov 2014, 22:13
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 2digit 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.



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Re: N and M are each 3digit integers. Each of the numbers 1, 2, [#permalink]
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28 Feb 2015, 00:27
Interesting One!!! I picked up 283 and 176 and diff was 8..However ,that was not in answer choice.



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