The following problem was given in the Eureka journal (April 1978/University of Cambridge). This problem can be solved by hand without computers. There is a unique solution. The phrase, "exact long division", means the remainder is 0. Have fun solving!

Long division puzzle with no given digits

EDIT: I have found an earlier source of this puzzle in the book, “536 PUZZLES & CURIOUS PROBLEMS” by Henry Ernest Dudeney. This book credits this puzzle to Mr. A. Corrigan. In the book the puzzle is worded differently but it the same mathematically.


2 Answers 2



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Since the result has four decimal places, the divisor must be divisible by either $2^4$ or $5^4$. Trying $5^4$ first: the only multiple of $625$ with three digits is $625$, so the first four nonzero digits of the result are 1. The final digit is somewhere between $2$ and $9$, and trying them all out shows that only a final digit of $8$ yields an integer numerator. So if the result is $1011.1008$ and the denominator is $625$, the numerator must be $1011.1008 \cdot 625 = 631938$.

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    $\begingroup$ Could you please give more detail about how you solved this. Thanks! $\endgroup$ Commented Aug 29, 2022 at 9:12
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    $\begingroup$ Your edit explains your solution well except one part. Why must the divisor be divisible by either $2^4$ or $5^4$? $\endgroup$ Commented Aug 30, 2022 at 4:50
  • $\begingroup$ @WillOctagonGibson Well, one way is: Multiply numerator and result by 10k. How do you remove those 4 zeros at the end of numerator (last digit of result cannot be 0 - otherwise it wouldn't be there)? You need to divide by 2^4 or 5^4 (N*2^4 and whatnot is also an option). Also, if your denominator is 5^4, result (if it was multiplied by 10^4 to remove decimal places) needs to be N*2^4. Noticing that last 2 digits before the end are 0 (see other answer) immediately gives 8 as the last digit (1000/16 = 62.5; so we need 8/16 to make it integer) $\endgroup$ Commented Aug 30, 2022 at 7:51

Here is how I reached the solution with my elementary knowledge :P


we can find that the red coloured digits must be zero as they are brought down from the top row.
The red digit in the quotient is zero because when $***×\color{red}*=**$, where the first digit is non-zero (clearly, in this case), the red digit must be a zero, otherwise the result will be a three digit number.

Now consider,

Clearly $d\neq0$ and so is $c$. As $a\cdot b=..c\neq0$, none of $a$ and $b$ is equal to zero. Thus in order to get the product $a\cdot e=...0$, one of them must be $5$ with other being an even number.


... assume $e=5$. In order to get $**a×e=d000$, '$a$' has to be zero. (Eg:- consider $200×5=1000$, $400×5=2000,$ etc.), but then it's a contradiction. Therefore from the previous argument, $a$ must be $5$. Also then, $e$ is an even number but we can reduce the possiblity to $\{4,8\}$, because when multiplied, there should be a number with last two (three, actually) digits being $0$.


... since $c\neq0$, it should be $5$. So $d$ also becomes $5$. Then we have $**5×e=5000$. We see that $5000=2^3×5^4$. We need to fill the numbers in $\text{_ _} 5 × \text_ =5000$, where the one digit number is confirmed to be $4$ or $8$. If it is $4$ the other number in the product should be a 4-digit number, which leads to a contradiction. Thus we can easily find that it is $8$ as in $625×8=5000$.

Now we can easily deduce that..

... the other digits in the quotient are $1$s because the every product remaining (i.e. $625×*$) results in a 3-digit number. So we can see that $b$ should be a $1$, giving $625$ (where the last digit was $c$ previously) and the prior number $630$. From there working backwards we can find all the missing numbers. four

Hope I made it clear.

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    $\begingroup$ Excellent work. One comment: You used the fact that if a product of an even and odd number ends in "00" then the even number is a multiple of 4. You could skip a bit using instead that if a product of an even and odd number ends in "000" then the even number is a multiple of 8. (This continues for any number of zeroes and matching power of two, since divisibility by $10^n = 2^n 5^n$ implies divisibility by $2^n$.) $\endgroup$
    – aschepler
    Commented Aug 30, 2022 at 12:15

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