Q2014/008) Solve in positive integers for 577(bcd+b+c)=520(abcd+ab+ac+cd+1)

577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)

(577−520a)(bcd+b+c)=+520(cd+1)

First conclusion:  a=1

57(bcd+b+c)=+520(cd+1)

57b(cd+1)+57c=+520(cd+1)

57c=(520−57b)(cd+1) →b<10

Further

57c=(520−57b)(cd+1) 
=> 520 - 57b < 57
or b >= 9
so b = 9
hence

57c=7(cd+1)

c(57−7d)=7 →d=8 →c=7
refer to http://mathhelpboards.com/challenge-questions-puzzles-28/solve-positive-integers-8202.html#post37808