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

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Manager
Joined: 22 Jan 2018
Posts: 52
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,  [#permalink]

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19 Dec 2018, 09:57
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.
Director
Joined: 29 Jun 2017
Posts: 927
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,  [#permalink]

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26 Feb 2019, 06:57
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 .
Director
Joined: 29 Jun 2017
Posts: 927
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,  [#permalink]

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16 Apr 2019, 03:59
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

Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 440
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,  [#permalink]

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26 Jun 2019, 15:55
nobelgirl777 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

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.
SVP
Joined: 03 Jun 2019
Posts: 1880
Location: India
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,  [#permalink]

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24 Sep 2019, 09:48
nobelgirl777 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

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
Re: N and M are each 3-digit integers. Each of the numbers 1, 2,   [#permalink] 24 Sep 2019, 09:48

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