19 May '08 00:34>
start with any 3 digit number, with a stipulation that the digits must be descending
321
reverse the order of the digits and subtract it from the original number
321 - 123 = 198
reverse the order of the new number and add it to the number first produced
198 + 891 = 1089
this will always process always produces 1089
algebraiclly
100a + 10b + c -(100c + 10b + a )
99 ( a - c ) is the first produced number, but how do we show that this number reversed added to itself always produces 1089
I understand that the multiples of 99 are the key, but is there algebraically a insight into this peculiar property?
321
reverse the order of the digits and subtract it from the original number
321 - 123 = 198
reverse the order of the new number and add it to the number first produced
198 + 891 = 1089
this will always process always produces 1089
algebraiclly
100a + 10b + c -(100c + 10b + a )
99 ( a - c ) is the first produced number, but how do we show that this number reversed added to itself always produces 1089
I understand that the multiples of 99 are the key, but is there algebraically a insight into this peculiar property?