ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer that can take any value from 0 to 9.

As Victor points out, A has to be 2 and E has to be 8. This, by substitution, yields
133B+10C-32D+1=0
Since the result begins with an 8, no carry goes through the 4th digit mutliplication, so whatever is carried from the 3rd digit, 4B must be less than 10. Therefore B is either 0, 1 or 2. However, 10C-32D is even, forcing 133B (and B) to be odd. So B=1 and the equation becomes
16D-5C=67
The product's second digit is 1, so 4D has to end in an 8 (3 is carried over by 4E=32). This means that D is either 2 or 7. Substituting D with 2, C becomes -7 which is impossible. Therefore D=7 and C=9, which completes the proof.
I agree with Victor's second post, only adding that NO other 5-digit number has this property, as my proof shows ;)

Posted by zaphod on 2003-11-09 10:23:24