1. a must be 1 or 2, as if it were any larger abcde*4 would have 6 digitsOriginally Posted byluckyvan

2. 4*abcde is even so a must be 2.

3. if a=2 then e is 8 or 9, as the leading digit on the RHS is either 4*a or 4*a+1.

4. if a is 2, then consider the trailing digit on the RHS, this a the solution of: 4*x=2 mod 10, which has solutions modulo 10 of 3 and 8, but we know from 3. above that e is either 9 or 8, so it must be 8.

5. We now know there is no carry from 4*b - otherwise e would be 9 its not so there is no carry. This means that b is either 0, 1 or 2, but 2 is taken so it is either 0 or 1.

6. Now trial and error can be used, and we find that abcde=21678.

RonL