Write down any three-digit number. Now write it down backwards, so you have two numbers before you. Subtract the smaller from the larger one. Take the resulting number and add it to its own reversal. The result of this addition is the title of this node.

Why this works for any starting number is left as an exercise for the interested reader.

Proof: A 3-digit number abc can be expressed as 100a+10b+c, and its reversal as 100c+10b+a. Then |abc-cba| = 99|a-c|. Let |a-c| = d. Since a and c are 1-digit numbers, d has 1 digit.

Now 99d = (100-1)d = 100d - d = 100d - 100 + 90 + 10 - d = 100(d-1) + 10*9 + (10-d)
Thus, provided d>0 (i.e. provided abc is not a palindrome), the digits of |abc-cba| are, from left to right, (d-1), 9, and (10-d), so it's reversal is 100(10-d) + 10*9 + (d-1), and the final sum is

100((d-1)+(10-d)) + 10*(9+9) + ((10-d)+(d-1))
= 100(9) + 10*18 + 9
= 1089.

Note that this only works if we take the result of the first subtraction as a 3-digit number even if it has 2 digits. For example, if the initial number is 504, the result of the subtraction is 99, which must be read as 099.

Also, note that this doesn't work at all for palindromes. For example, 505 gives a final result of .

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