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
Can some one show how this can be solved under 2 minutes? It can be solved using various combinations but takes too long.
Thanks.
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Smallest possible difference
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GMATGuruNY
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To MINIMIZE N-M:PGMAT 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 difference between the HUNDREDS digits must be 1.
The TENS digit of N must be as SMALL as possible (1) and the tens digit of M must be as GREAT as possible (8).
Thus, there are only two cases to consider:
Case 1: N = 31X and M = 28X.
Case 2: N = 71X and M = 68X.
Of the remaining 2 digits, the SMALLER must be assigned to N and the GREATER to M, so that the distance between N and M is minimized:
Case 1: N = 316 and M = 287, with the result that N-M = 316-287 = 29.
Case 2: N = 712 and M = 683, with the result that N-M = 712-683 = 29.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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joealam1
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Hey Mitch,
Is there any way to avoid the trial an error process?
I knew from the start that i must choose as below:
Case 1: N = 31X and M = 28X.
Case 2: N = 71X and M = 68X.
Of the remaining 2 digits, the SMALLER must be assigned to N and the GREATER to M, so that the distance between N and M is minimized:
Case 1: N = 316 and M = 287, with the result that N-M = 316-287 = 29.
Case 2: N = 712 and M = 683, with the result that N-M = 712-683 = 29.
However, i've chosen N= 314 and M=287 and the difference was 27 ( i thought i made a mistake somewhere and i repeated the calculation to double check and at the end i got stuck in the problem and exceeded the 2 min limit)
Is there any way to avoid the trial an error process?
I knew from the start that i must choose as below:
Case 1: N = 31X and M = 28X.
Case 2: N = 71X and M = 68X.
Of the remaining 2 digits, the SMALLER must be assigned to N and the GREATER to M, so that the distance between N and M is minimized:
Case 1: N = 316 and M = 287, with the result that N-M = 316-287 = 29.
Case 2: N = 712 and M = 683, with the result that N-M = 712-683 = 29.
However, i've chosen N= 314 and M=287 and the difference was 27 ( i thought i made a mistake somewhere and i repeated the calculation to double check and at the end i got stuck in the problem and exceeded the 2 min limit)
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DavidG@VeritasPrep
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Don't forget to glance at the answer choices! No matter what units digits you select for 31X and 28X, the difference has to be less than 49. So you know the answer must be A.Is there any way to avoid the trial an error process?
I knew from the start that i must choose as below:
Case 1: N = 31X and M = 28X.
Case 2: N = 71X and M = 68X.
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Matt@VeritasPrep
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Trial and error isn't necessary here.
Suppose you have the number 123. You can write this as 1*100 + 2*10 + 3*1.
Now suppose you have the three digit number abc. You can write this as a*100 + b*10 + c*1.
We want the difference between two three digit numbers, so
abc - def =
(100a + 10b + c) - (100d + 10e + f) =
100*(a - d) + 10*(b - e) + (c - f)
Since all the digits are different, we can't have a = d or b = e, etc. So let's have (a - d) = 1, for that to be as small as possible: we'll have a = 3 and d = 2. (We need the difference to be positive, so (a - d) must be > 0.)
From here, we want to make 10*(b - e) + (c - f) negative, if possible, and we want (b - e) to be smaller than (c - f). If we do this, we'll shrink 100*(a - d), since we're adding negative numbers to it.
If we make (b - e) = 1 - 8, we have (b - e) = -7, giving us
100*(a - d) + 10*(b - e) + (c - f) =>
100*1 + 10*(-7) + (c - f) =>
30 + (c - f)
We used 1, 2, 3, and 8, so we're left with 6 and 7. We want (c - f) to be negative, so we'll have c = 6 and f = 7. That gives us
30 + -1
or
29
with abc = 316 and def = 287
Suppose you have the number 123. You can write this as 1*100 + 2*10 + 3*1.
Now suppose you have the three digit number abc. You can write this as a*100 + b*10 + c*1.
We want the difference between two three digit numbers, so
abc - def =
(100a + 10b + c) - (100d + 10e + f) =
100*(a - d) + 10*(b - e) + (c - f)
Since all the digits are different, we can't have a = d or b = e, etc. So let's have (a - d) = 1, for that to be as small as possible: we'll have a = 3 and d = 2. (We need the difference to be positive, so (a - d) must be > 0.)
From here, we want to make 10*(b - e) + (c - f) negative, if possible, and we want (b - e) to be smaller than (c - f). If we do this, we'll shrink 100*(a - d), since we're adding negative numbers to it.
If we make (b - e) = 1 - 8, we have (b - e) = -7, giving us
100*(a - d) + 10*(b - e) + (c - f) =>
100*1 + 10*(-7) + (c - f) =>
30 + (c - f)
We used 1, 2, 3, and 8, so we're left with 6 and 7. We want (c - f) to be negative, so we'll have c = 6 and f = 7. That gives us
30 + -1
or
29
with abc = 316 and def = 287
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elisagrace
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Answering these questions requires a lot of practice. Once you're trained, it is the play of your left hand.
