Reconstruct a long division given less than a quarter of the digits, and all of those are wrong

Question

Inspired by "The Puffin Book of Brainteasers" I decided to try my hand at creating a long division missing digits problem where all of the displayed digits are wrong. On searching I found that it has been done before - Someone has vandalised this long division. Can you reconstruct it? - but that was nearly ten years ago. In honour of the previous problem, I am using a sum with the same shape, but I am only providing 7 digits, where the previous sum provided 22 digits, so I think my one is harder.

The task is the same. Can you construct a well formed long-division problem by replacing each digit or star with a digit?

• The division sum goes exactly - there is no remainder.
• You may assume that none of the numbers begin with a leading zero.
• Whenever the problem shows a star you have no information about the actual digit (other than that there is one)
• Whenever the problem shows a digit then the actual digit is different to the displayed digit
• Treat each digit independently. The digits that replace the $2$s (for example) might be the same or different to each other. All you know is that they will not be $2$s.
• You may assume that the solution is unique (and for what its worth that it is not the same as the solution of the old problem)

The problem can be solved entirely by logical deduction. There is no need to use a computer to find the solution by brute force. A valid answer should show at least some of the working.

Answer 1

The Answer

     3993
   ______
33|131769
    99
   ---
    327
    297
    ---
     306
     297
     ---
       99
       99
       ==

Solve Path

Consider the last digit of the quotient. Since it cannot be 1, it must be at least 2; and the last multiple of the divisor is 2-digit, which implies that the divisor is less than 50. Now consider the first digit of the quotient. It cannot clearly be 1, because the divisor is 2-digit while the dividend (being considered at the moment) is 3-digit^(*). Since it also can't be 2, it must be at least 3. As there are 2-digit multiples of the divisor, the divisor must at most be 33.

Now look at the third subtraction. Due to the given constraints, the subtrahend is greater than or equal to 200, which is followed by the minuend being greater than or equal to 300. But the difference of them is a 1-digit number. Therefore the subtrahend lies between 290 and 299. Since ⌈290÷9⌉=33, the divisor must indeed be 33. And the aforementioned subtrahend is 33×9=297.

Now consider the last 2-digit number. It can be 66 or 99. But 66 will lead to the last digit of the aforementioned minuend (which was taken from the dividend above) being 3, contradicting the given condition. Therefore it is 99.

Then consider the first subtraction. Since the leading digit of the difference cannot be 1, it is 2 or 3. Either way implies that the subtrahend is 99. Then the only possible values for the difference and minuend are (20, 119), (31, 130) and (32, 131). The first pair is rejected because the smallest 3-digit multiple of 33 is 132, which should be in the immediate step, if that's the case, leading to a contradiction. Afterall we can see that the minuend of the second subtraction is 310+, while the difference is 30, which will again lock 297 in the place of the subtrahend. Thus the division is completed.