Pick a four digit number, all digits different, such that when you add its reverse and divide it by 10, you get the number you started with.

For example: 1749+9471=11220, 11220/10 = 1122. Since 1749 is not equal to 1122, this is not the right number.

OK, pseudoalgebraically:

((1000A+100B+10C+D)+(1000D+100C+10B+A))/10 = 1000A+100B+10C+D

simplify the left, and multiply both sides by ten:

1001A+110B+110C+1001D = 10000A+1000B+100C+10D
Subtract the left:
0= 8999A+890B-10C-991D

If you look at the puzzle, we have one more equation at our disposal. Since the combined numbers must be divisible by ten, the sum must end in zero, so A + D = 10, and D = 10-A.

Substitute for D:
0 = 8999A+890B-10C-9910+991A
9910 = 9990A+890B-10C
We can divide by ten to simplify:
991 = 999A+89B-C

OK. Now lets use logic (thus the pseudoalgebra only).
We know A is a non-zero integer (otherwise we'd be working with a three digit number), so 89B-C must be negative. The only way this is possible would be for B to be zero (since even a positive 89 would make a negative impossible).

So B = 0, leaving 991 = 999A-C
Since C must be a single integer, A can't be greater than one. Substituting 1 for A:
-8 = -C
Thus far we have
A = 1
B = 0
C = 8
and, since A+D = 10
D=9

1089+9801 = 10890
10890/10 = 1089

Posted by Jim C on 2003-07-07 08:14:27