# What are the hidden digits in this long division puzzle?

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!

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.

Solution:

Reasoning:

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 $$1011.1008 \cdot 625 = 631938$$.

• Could you please give more detail about how you solved this. Thanks! Commented Aug 29, 2022 at 9:12
• Your edit explains your solution well except one part. Why must the divisor be divisible by either $2^4$ or $5^4$? Commented Aug 30, 2022 at 4:50
• @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) Commented Aug 30, 2022 at 7:51

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

First,

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.

Now...

... 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 $$0$$.

Then...

... 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.

• 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 $2^n$.) Commented Aug 30, 2022 at 12:15