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N and M are each 3-digit integers

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N and M are each 3-digit integers

by Needgmat » Tue Sep 06, 2016 8:12 am
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

OAA

Please explain
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by cubicle_bound_misfit » Tue Sep 06, 2016 8:40 am
step 1: Look at the options.

The smallest one is A) 29

Now can we have two three digit integres whose difference is 3 ?

i.e N - M = 29 ?

Step 2: what can be last digit of N and M to get a result of 9 ?
n can be XY0 and M can be AB1

Step 2 : now can XY - (A(B+ 1 form Carry over) = 2

Step 3 X80 - X51
Step 4 X can be anything 2,3,6,7 ( As one digit is either N or M [not mutually exclusive])

hence A is the answer.
--- I am new to GMAT, hence for any better approach, please correct me.
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by Brent@GMATPrepNow » Tue Sep 06, 2016 3:24 pm
cubicle_bound_misfit wrote:step 1: Look at the options.

The smallest one is A) 29

Now can we have two three digit integres whose difference is 3 ?

i.e N - M = 29 ?

Step 2: what can be last digit of N and M to get a result of 9 ?
n can be XY0 and M can be AB1

Step 2 : now can XY - (A(B+ 1 form Carry over) = 2

Step 3 X80 - X51
Step 4 X can be anything 2,3,6,7 ( As one digit is either N or M [not mutually exclusive])

hence A is the answer.
--- I am new to GMAT, hence for any better approach, please correct me.
The question could be worded more clearly (and more mathematically), but I believe the idea here is that each of the 6 digits (1,2,3,6,7, and 8) must be used exactly once in the creation of numbers N and M.

"Each [digit] 1,2,3,6,7, and 8 is a digit of either N or M. So, for example the digit 1 must be in either N or M. Same with the digit 2 etc.

Cheers,
Brent
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by Brent@GMATPrepNow » Tue Sep 06, 2016 3:27 pm
Needgmat wrote: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

OAA

Please explain
I started with a line of reasoning that was similar to that used by cubicle_bound_misfit, and then started fiddling with the tens and hundreds digits.

I got N = 712 and M = 683, in which case N - M = 712 - 683 = 29

Answer: A
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by Matt@VeritasPrep » Thu Sep 15, 2016 8:23 pm
We could do a bit of algebra here. Let's say that our numbers are abc and def. Then we have

abc - def =

(100a + 10b + c) - (100d + 10e + f) =

100*(a - d) + 10*(b - e) + (c - f)

Since all our numbers are distinct, we're going to get positive or negative numbers for (a - d), (b - e), and (c - f).

We want (a - d) to be as small as possible, so we'll have (a - d) = 1. That starts us off with 100*1, and we can use either a=3, b=2 or a=6, b=5. Let's use a = 3, b = 2.

Now we want (b - e) to be as NEGATIVE as possible, since we want to subtract as much as we can from the 100*1 we started with. The furthest we can go in our set is (1 - 8), or -7.

Now we've got 100 - 70. If we can make the last term negative, we're set. We're left with only 6 and 5, so we'll do (5 - 6), or -1.

This gives us 100 - 70 - 1, or 29. (Or, if we've lost track of the numbers, we've got 315 - 286.)
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