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