22
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Oh dear, someone has vandalised this long division.

They've replaced every digit either with a wrong digit or a hyphen.

Your task: reconstruct the entire thing. There's no remainder.

Please show all work. Use of a computer is not allowed.

long division

Please note the following :

  1. None of the digits in the puzzle, is correct.

  2. None of the numbers has a leading 0. This means that the actual numbers that replace 29, 86,96 and 35 are all two digit numbers as well, the actual numbers that replace 2_ _, 125, 3 _ _ and 248 are all three digit numbers themselves, the actual number that is there in place of 16_2 is a 4 digit number and the actual number that replaces 00 _8 is a 6 digit number.

  3. Let's say that 2 in the divisor, 29, is actually supposed to be 7. Then it is not necessary that all the 2s present in the image above will be replaced by 7. It can be that one of the 2s gets replaced by 7 and another 2 gets replaced by 3. Same is true for the other digits in the image : 0,1, 3…9. They each can be replaced by different digits at different places.

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  • 4
    $\begingroup$ Wow. So just to confirm, none of the digits that are in the puzzle are correct? $\endgroup$
    – LeppyR64
    Jun 14, 2015 at 0:31
  • 1
    $\begingroup$ Are there multiple answers to this? $\endgroup$
    – Quark
    Jun 14, 2015 at 0:37
  • $\begingroup$ @LeppyR64 thanks for mentioning that. I missed the wrong digit part and was wondering what this was all about. $\endgroup$
    – Bob
    Jun 14, 2015 at 0:38
  • 1
    $\begingroup$ Quark and JLee - There's only one correct answer. Leppy - Yes. $\endgroup$
    – h34
    Jun 14, 2015 at 0:43
  • 2
    $\begingroup$ This is a cool question. $\endgroup$
    – LeppyR64
    Jun 14, 2015 at 0:50

2 Answers 2

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Solution

In the second subtraction, the number being subtracted is three digits, and the first digit isn't 1. So it must be at least 200, and the divisor is at least 21. Since the first digit of the divisor isn't 2, the divisor must be at least 30.

In the first subtraction, the number being subtracted is at least twice the divisor, but is only two digits. So the divisor is at most 49.

At the end, there is a two digit number, where the first digit is not 9. It is at most 89. This means that the last digit of the quotient cannot be 3 or more, so it must be 1. Then the divisor cannot start with 3, so it is 40-49.

The first digit of the quotient must be 2. Twice the divisor cannot start with 8, so the divisor is 45-49. The last digit of the divisor is not 5, 6, 8, or 9, so the divisor is 47.

The second digit of the quotient must be at least 5, because 4*47 is less than 200. But it can't be 5 because the 5 is wrong, it can't be 7 (7*47=329) or 9 (9*47=423) because the 2 is wrong, and it can't be 6. So it is 8.

Now the third digit of the quotient is the only one missing. It can't be 1 or 2 because those are too small. It can't be 3 because the 4 is wrong, it can't be 4 because the 8 is wrong, and it can't be 5 or 6 because the 2 is wrong. It also can't be 7 or 8 because the 3 on the preceding line is wrong, and the numbers subtract to only 4. So the missing digit is 9.

The complete division is 135877/47=2891. The subtractions are 135-94, 418-376, 427-423, and 47-47.

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  • $\begingroup$ This is the right answer. Nicely done! :-) $\endgroup$
    – h34
    Jun 14, 2015 at 12:57
  • $\begingroup$ Well done. Here's a pic. (I can't seem to make it display here in the comment.) $\endgroup$
    – r.e.s.
    Jun 14, 2015 at 13:25
  • $\begingroup$ "The last digit of the divisor is not 5": why not? $\endgroup$
    – msh210
    Jun 7, 2023 at 21:21
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I think I found the answer.

Step one to find the divisor.

By looking at the number we know that the first three digits of dividend is quite close to 100 because the divisor is dividing it with 2 digits. And second things is the reminder of the first division is smaller than divisor and thats why the number has to be between 21-25. I thought odd number has more possibility to get this answer.

I started with

25 as divisor but it is hard to get 2 digit reminder in second step of division. Then I tried 23.

Now this number start making sense in my mind. Here's how

23x4 =92 which is very close to 100 and 23x5=115 which more quiet far from 100. so i chose 113 number between 100 and 115 Now if you divide 113 with 23 you will get 21 as remainder. Now according to given division it has to be small than divisor because we are going to drop one more number from dividend. Now that Number has to be close to a number which is close divisible by 23 and get 2 digit number as remainder which is also has to be smaller than 23.

Second step of division.

So far we have first three digits of dividend 113 and 21 as remainder and now we are going to decide the fourth digit of dividend. I started with number 9 because its the biggest single digit number. so now we got number 1139 and our remainder become 219.Now 23 x 9 =207 which is the closest to 219. so now we got two digit remainder 12 which is smaller that 23.

Third step of division.

So far we got first four digits of dividend which is 1139 and quotient 49. Now its time to find fifth digit of dividend. lets say the smallest number we can use is 0 which makes remainder as 120 and when you divide it with 23 you will get 5 as remainder and there is no two digit number which start with 5 is divisible by 23. so I change that fifth number to 1. Now remainder become 121 and when you divide 121 with 23 you will get 6 as remainder.

Fourth/Final step of division

Now we have first five digits of dividend which are 11391 and first three digits of quotient which are 495. We have 6 as a remainder and there is one number which start with 6 and also divisible by 23 and that is 69. ;-) (23 x 3=69) So our last digit of dividend is 9. and our last digit of quotient is 3.

Here is my actual work

Here is my calculations

Answer

Screen Shot of Final answer

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  • $\begingroup$ The 2 in the 219 is not allowed. $\endgroup$
    – r.e.s.
    Jun 14, 2015 at 1:55
  • $\begingroup$ The 2 in the 23 is not correct. Keep grinding though :) $\endgroup$
    – LeppyR64
    Jun 14, 2015 at 1:55
  • $\begingroup$ @LeppyR64 You couldn't stop yourself from brute-forcing it, could you? hahaha :) $\endgroup$
    – JLee
    Jun 14, 2015 at 2:47
  • $\begingroup$ @JLee I'll be honest, I couldn't :) My comment here was based on the description in the question though. $\endgroup$
    – LeppyR64
    Jun 14, 2015 at 2:49
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    $\begingroup$ +1 for effort, but I'm sorry to say that after all this work the answer is not right: as others have said, the 2s in 23 and 219 are not allowed. @HradayJoshi - No number in the right answer has first digit 0. $\endgroup$
    – h34
    Jun 14, 2015 at 10:08

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