Henry Ernest Dudeney/Modern Puzzles/71 - A Complete Skeleton/Solution

by : $71$

 * A Complete Skeleton

Solution
1011.1008   - 625)631938     625     ---       693       625       ---        688        625        ---         63 0         62 5            5000            5000

Proof
Let $d$ denote the divisor.

Let $q$ denote the quotient.

Let $n$ denote the dividend.

Let $n_1$ to $n_5$ denote the partial dividends which are subect to the $1$st to $5$th division operations respectively.

Let $p_1$ to $p_5$ denote the partial products generated by the $1$st to $5$th division operations respectively.

As suggested, the $3$ zeroes are dropped down into $n_5$ and $p_5$.

Another zero is dropped into $n_4$.

This also gives us the last two but one digits of $q$.

Thus $d$ is a divisor of a multiple of $1000$.

The divisors of $d$ can only be $5$ or $2$ up to a multiplicity of $3$ each, or $x$ where $x < 10$.

Because $d$ has $3$ digits, one divisor must be $5$, and the last digit must therefore be $5$ or $0$.

The subtraction from the final $0$ in $n_4$ giving a digit in the of the difference means it must be $5$.

$d$ cannot have $2$ as a divisor or $d$ will end in $0$.

It follows immediately that $d = 5^4 = 625$, and that all of $p_1$ to $p_4$ are also $625$.

The remaining digits of $q$ are now obviously either $1$ or $0$: 1011.1008   - 625)******     625     ---       ***       625       ---        ***        625        ---         ** 0         62 5            5000            5000

and $n$ follows by multiplying $d$ by $q$.