The mysterious 6174 revisited
From my earlier post on 6174 comes a simpler 3 digit variant. Select 3 integers which are not all the same. Arrange these digits to form the largest and smallest numbers and subtract them. Repeat this process with the result. For instance,
You will end up with 495!
The following gives a proof that all 3 digit numbers which are not 0,111,222,333,...,999 will end up with 495. This proof can be adapted to 4 digits as well, as in the case of 6174.
Let x,y,z be 3 digits such that z is the largest while x is the smallest, and x,y and z are not all equal. Then if we subtract the largest number formed from the smallest number formed, we have
Since x < z
Hence B = 9 = z since z is the largest digit. We want A and C to be either x or y. Solving, we see that x=4 and y=5. Hence 495 is the only number that repeats itself after an iteration.
But do any set of 3 randomly selected integers not all equal will lead to 495 after a finite number of iterations? First we see that for x,y and z,
Hence the resulting number after any iteration is a 3-digit multiple of 99.
Hence we only need to check 9 possible numbers: 99,198,297,296,495,594,693,792 and 891 (10(99)=990 which is the same as 99)
Since all the above numbers lead to 495, QED.
Surprisingly, 495=5(99) which is in the middle of the pack of multiples of 99!
Can you see if you can generalise this result to 5 digit numbers and above?