A mysterious fact related to digital root of a 'special incorrect division' and 'correct division'.Whats the reason behind it?

I observed that the digital root of an 'incorrect division' obtained by dividing a number in a 'special way'/'particular manner' and the digital root of the correct division are always same.

The Special way/Particular manner is as follows: While obtaining first digit of the incorrect division deduct '1' from the correct division digit. Just do this only for the first digit in the division, the remaining division must be carried according to rules of division. e.g. In 125/5 the correct division is 25 and the special incorrect division is: 125/5=196. If we observe the digital roots of correct and 'special incorrect division', we find that they are same that is '7' This fact holds true for any number. Some more examples are,
1] Special Incorrect division 11/2 = 41.5 has digital root 4+1+5=>10=>1+0=>1
Correct division 11/2 = 5.5 has digital root 5+5=>10=>1+0=>1
2] Special Incorrect division 81/3 = 198 has digital root 1+9+8=>18=>9 Correct division 81/3 = 27 has digital root 2+7=>9

How does this happen? Any mathematical proof?

• What?${}{}{}{}{}$ – Saucy O'Path Apr 7 at 13:42
• This is very hard to follow. All this talk about "incorrect division", "special way", "particular manner", is just confusing. If you have an operation you want to study, just define it. Don't worry about what to call it. To be specific: I can't follow your example which takes $125$ and $5$ and returns $196$. – lulu Apr 7 at 14:04
• Voting to close the question as it is not clear what you are asking. If you can, edit your post for clarity. In particular, I suggest clarifying your definition. Walk through an example (or two) carefully, step by step. – lulu Apr 7 at 14:22
• Yes...that makes sense, thanks for editing (I've retracted the close vote). Where did you encounter this operation? – lulu Apr 7 at 17:13
• It seems to work like carrying. With $\frac {125}{5}$ the second "digit" ought to be $15$ since $5\times 15=75$, but you mark it as $9$ because that's as high as you are allowed to go. Not immediately clear to me why that implies that your result is congruent to the standard quotient $\pmod 9$... – lulu Apr 7 at 17:18