Whoever finds the pattern i'll give you a cookie

58 -> 377
34 -> 183
27 -> 106
62 -> 461
83 -> ?
I will refer to the left number as X and right as Y
1) ones digit
X's ones digit is 1 larger than Y's ones digit (8-1 = 7, 4-1 = 3 and on)
so for 83, it will be 2 --> ??2
fairly straightforward
2) tens digit
first X's tens digit(5) + first Y's tens digit(3) = 8, thus 34 --> 183
5+3+2 = 10, 106 and on
so for 83, it will be 5+3+2+6+ 8 = 24, ?42
3) hundreds digit
genearlly, Y's hundreds digit is 2 less than X's.
but we see an exception: 27 --> 106
if we go back to how we got the tens digit, it was 5+3+2, 10. here, we only used the 0 for our solution, and dumped 1. I though that 1 back there should come back here, making it 2-2+1 = 1, 27 --> 106
so for 83, it should be 8-2 = 6, but remember, we dumped the 2 from 24 in tens digit, so 6+2 =8
THEREFORE, ANSWER = 842
Now, there are some faulty reasoning in my solution

1. my solution fails to explain the 7 in tens digit for 58 --> 377. i just said that the first value was given a random number
2. My solution only works if the order matters; for most of these problems, order often doesn't account for the answer, but that was the only way i saw the answer
3. the number becomes invalid if ones digit is 0 , because 0-1 = -1, we can't have that. similar errors could occur with hundreds digit
   BUT, for the given numbers, i didn't come across any error solutions, so i speculate im right? maybe idk