How many pairs of digits A,B such that ABA+BAB is divisible by 74? ABA and BAB are not multiplication, they just denote digits. I realized that ABA plus BAB will result in the last 3 digits being repeated, so do I find all multiple of 74 that have 3 repeating digits?

ABA+BAB=111A+111B=111(A+B)=3x37(A+B) which divides 74 when A+B is even

## where A,B is an element of {1,2,3,4,5,6,7,8,9}

so (A,B)=(1,1) (1,3) (1,5) (1,7) (1,9) (3,...) (5,...) (7,...) (9,...) so there are 25 pairs where both A and B are odd

There is also (2,2) (2,4) (2,6) (2,8) (4,..) (6,..) (8,..) which is another 16 solutions

Total =41 pairs