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cathyangs
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?
39 was the lowest I found as well
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.
Or cant we repeat digit for both M and N
Sorry, you can't repeat digits.
And there is a mathematical way to do it.
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')
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)
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
Thus, Smallest M-N=1*100+(-6)*10+(-1)=39