F=5 so B=1. [F>4 because we get a 2-digit result when we add F to itself. The only double-digit for BB is 11, because the doubling of a single digit cannot exceed 18, which starts with 1. This means that B must be 1, so BB=11=2*5 plus carryover.] So we have 5C5+5C1=11JC; C=6 because 5+1=6, therefore 565+561=1126. J=2.
(B,C,F,J)=(1,6,5,2).