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If a b c are three positive intejers{I}.Such that ab+bc+ac+a+b+c=989
then find a+b+c=?
Dear student,
abc+ab+ac+bc+a+b+c=989, abc+ab+ac+bc+a+b+c+1=990 (a+1)(b+1)(c+1)=990 Since a,b,c >0, then (a+b+c+3)/3= ((a+1)+(b+1)+(c+1))/3>= ((a+1)(b+1)(c+1))^(1/3)=(990)^(1/3) >(729)^(1/3)=9 So a+b+c+3>27, a+b+c>24. 990=2*(3^2)*5*11 Since a,b,c are all positive integers, hence >0, then (a+1), (b+1), (c+1) are all >=2. We are not given that the sought numbers are consecutive, this makes the problem a little lengthy. Then to mention a few, the following options are possible: a+1=2, b+1=9, c+1=55 => a=1, b=8, c=54. Then a+b+c=63 a+1=2, b+1=3, c+1=165 => a=1, b=2, c=164. Then a+b+c=167, a+1=9, b+1=10, c+1=11 => a=8, b=9, c=10, Then a+b+c=27 a+1=2, b+1=5, c+1=99 => a=1, b=4, c=98 =>a+b+c=103 and so forth in a finite number of steps all relevant a,b,c with a+b+c>24 can be listed
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