Hmm, my last long division puzzle (Reconstruct a long division given less than a quarter of the digits, and all of those are wrong) didn't last the night. So let's try to make it harder. All of the digits of this puzzle are wrong except for one of them, which is correct. So you cannot deduce anything from just looking at a single digit! I found this makes it harder to solve, but it can still be done by hand.

Long Division sum

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

  • This division sum leaves a remainder - the two digits on the last row.
  • 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 in most cases the actual digit is different to the displayed digit. But for exactly one of the digits the digit shown and the digit in the solution are the same.
  • Treat each digit independently. The digits that replace the 2s (for example) might be the same or different to each other.
  • You may assume that the problem has a unique solution.

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.

  • $\begingroup$ @ACB I really hope that I have not made a transcription error. But I've triple checked against the output of the program I used to find the problem. The problem as shown here is as intended. I've compared the answer the program gave against the answer I got when I solved it by hand. They are the same. So I'm at least 90% confident that the problem I showed here has a solution and that it is unique. All I can say is that having one digit correct makes it much harder to solve than when all of the digits were wrong. $\endgroup$
    – user23087
    Commented Jun 1 at 18:08

1 Answer 1




Solve Path

It's easy to see that the last digit of the quotient is 0.

The key point is the last digit of the remainder. Note that for a unique solution, the tens digits of the divisor and the remainder should be the same, while the ones digits of each should be 1 and 0 respectively. Otherwise the last digit of the remainder (or dividend) is not constrained.

This leaves 11, 21, 31, ..., 91 as the only options for the divisor. 11 also can be immediately removed since it cannot produce a 3-digit multiple.

If the divisor is greater than 50, the second digit of the quotient (1) should be the correct digit. Consider divisors greater than 60. Then the minuend of second subtraction is ≤88 and the subtrahend of the same is ≥61. But that forces either of the leading digits in third subtraction also to be correct; contradiction. Therefore we can also remove divisors greater than 60. Now our list is reduced to 21, 31, 41 and 51.

21 and 31 can also be removed because they cannot produce a multiple greater than 290. (It is essential to have such a multiple to prevent both of the first and the third subtractions containing correct leading digits.)

41 can be dismissed beacuse it leads second digit of the quotient to be 2, which leads the second subtrahend to be 82, which leads its minuend to have a 9 in the tens place contradicting the condition of only one correct digit.

Thus, the divisor is 51 and the remainder is 50. The correct digit is the second digit of the quotient.

Now the third subtraction should contain numbers greater than 300 and less than 370. Since the third digit of the quotient cannot be 6, the only possible value is 51×7=357. Then following the obvious deductions we can complete the division from bottom upto the first subtrahend. Due to the second digit of the dividend we can conclude that the first subtrahend cannot be 306 but 357. Hence it's solved.

  • 1
    $\begingroup$ Yes - that's the same answer that I got and modulo trivial differences the same solve path. What I particularly liked about this problem is the "deduce from uniqueness" aspect. The path we both followed concluded that the dividend ended with '0' and the divisor ended with '1' because if they didn't then we would be able to find a forbidden second solution. Once we have slogged though the options for the first digit in the divisor we have enough that the rest of the solution follows without difficulty. $\endgroup$
    – user23087
    Commented Jun 3 at 11:36

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