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.

I like the other solutions.

Here's how I approached it:

The number is "abcd" (where a,b,c,d are 0-9), then:

1000a+100b+10c+d+1000d+100c+10b+a = 10(1000a+100b+10c+d),

collecting terms:

8999a+890b-10c-991d = 0

since 991(9)=8919 < 8999, and by inspection:

a=1, d=9, and so, b=0 and c=8,

solution: 1089.

It's interesting to note that the constants of the equation (*) have the only those digits as in the answer. Conicidence or a hint to a general solution?

Posted by ryan smith on 2003-07-06 20:04:10