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Using each of the digits 2, 3, 4, 6, 7, and 8 exactly once, construct two 3digit integers M and N so that MN is positive and is a small as possible. Compute MN.
The smallest MN I found was 39. Anyone?
 one year ago
 one year ago
Using each of the digits 2, 3, 4, 6, 7, and 8 exactly once, construct two 3digit integers M and N so that MN is positive and is a small as possible. Compute MN. The smallest MN I found was 39. Anyone?
 one year ago
 one year ago

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liza2Best ResponseYou've already chosen the best response.0
39 was the lowest I found as well
 one year ago

cathyangsBest ResponseYou've already chosen the best response.0
I bet there's a mathematical way to figure out the smallest MN without knowing M or N, or something. >:( I never know with this questions if there's something wrong with my answer.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Or cant we repeat digit for both M and N
 one year ago

cathyangsBest ResponseYou've already chosen the best response.0
Sorry, you can't repeat digits.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Well Yes it is 39
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
And there is a mathematical way to do it.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
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, MN=(a*100+b*10+c)(a'*100+b'*10+c') =(aa')*100+(bb')*100+(cc')
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So, MN to be minimum...... aa' must be positive smallest number (FIRST priority) bb' must be smallest (Second priority) cc' must be smallest (Third priority)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So, aa'=1 when a=4 and b=3 bb'=6 when b=2 and b'=6 cc'=1 when c=6 and c'=7
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Thus, Smallest MN=1*100+(6)*10+(1)=39
 one year ago
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