1 1 1, because any number can have factor of 1. also, only 1 can be true if any random number multiple plus has to be divided by the remaining digit. So, 1 1 1 is my answer.
They would be 111 or 123 . \& g; j) k v; c. uFor 111 would be 1x1+1=2 that is 2 times of 3rd 1' A! A$ u( W5 q' x" k5 R
For 123 # l: b; \* A# b3 K: [tvb now,tvbnow,bttvb1x2+1=3 that is 1 time of 31 q/ w8 Q) v6 h
1x3+1=4 that is 2 times of 2 ( X4 e# F% f# l6 Ptvb now,tvbnow,bttvb2x3+1=7 thta is 7 times of 1