Since BEL has only 3 digits A cannot be greater than 3, in fact it cannotOriginally Posted by ferdo
even be 3 (since the last two digits of BEL are different AxN>9, and so
there ill be a carry and ANNxA would have four digits).
Also A cannot be 1, since then AxANN=ANN, so we may conclude that A=2.
Finnaly trial and error will alow us to find that: