Since BEL has only 3 digits A cannot be greater than 3, in fact it cannotOriginally Posted byferdo

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:

288

X 2

_____

576

RonL