Question

Using each of the digits 2, 3, 4, 6, 7, and 8 exactly once, construct two 3-digit integers M and N so that M-N is positive and is a small as possible. Compute M-N.

The smallest M-N I found was 39. Anyone?

Answer 1

39 was the lowest I found as well

Answer 2

I bet there's a mathematical way to figure out the smallest M-N without knowing M or N, or something. >:( I never know with this questions if there's something wrong with my answer.

Answer 3

234-243=9

Answer 4

Or cant we repeat digit for both M and N

Answer 5

Sorry, you can't repeat digits.

Answer 6

Well Yes it is 39

Answer 7

And there is a mathematical way to do it.

Answer 8

Let the first number "M" be = abc =a*100+b*10+c
And the other number "N"be =a'b'c' = a'*100+b'*100+c'
Then,
M-N=(a*100+b*10+c)-(a'*100+b'*10+c')
=(a-a')*100+(b-b')*100+(c-c')

Answer 9

So, M-N to be minimum......
a-a' must be positive smallest number (FIRST priority)
b-b' must be smallest (Second priority)
c-c' must be smallest (Third priority)

Answer 10

So, a-a'=1 when a=4 and b=3
b-b'=-6 when b=2 and b'=6
c-c'=-1 when c=6 and c'=7

Answer 11

Thus, Smallest M-N=1*100+(-6)*10+(-1)=39