Henry Ernest Dudeney/Modern Puzzles/70 - The Solitary Seven/Solution

by : $70$

 * The Solitary Seven

Solution
97809   - 124)12128316     1116       968       868       ---       1003        992         1116         1116

Proof
Let $d$ denote the divisor.

Let $q$ denote the quotient.

Let $n$ denote the dividend.

Let $n_1$ to $n_4$ denote the partial dividends which are subject to the $1$st to $4$th division operations respectively.

Let $p_1$ to $p_4$ denote the partial products generated by the $1$st to $4$th division operations respectively.

As suggested by :


 * $(1): \quad$ Because $7 d$ has $3$ digits, the first digit in $d$ is $1$.


 * $(2): \quad$ $p_1$ is a $4$-digit number, which can be no greater than $199 \times 9 = 1791$. Hence the first digit in $n$ is $1$, as is that of $p_1$.


 * $(3): \quad$ The last $2$ digits of $n$ come down together, which means the last but one digit of $q$ is $0$.

So far: *7*0*   - 1**)1*******     1***       ***       ***       ---       ****        ***         ****         ****


 * $(4): \quad$ The first and last digits of $q$ must be greater than $7$, because its product with $d$ has $4$ digits.


 * $(5): \quad$ $n_3$, $n_4$ and $p_4$ all have $4$ digits, and so begin with $1$, for the same reason as $(2)$.


 * $(6): \quad$ The $3$-digit $p_3$ must begin with $9$, as its difference from the $4$-digit $n_3$ has $2$ digits.


 * $(7): \quad$ Because $n_4$ begins with $1$:
 * the second digit of $n_3$ must be $0$.
 * the third digit of $n_3$ must be $0$ or $1$.
 * the second digit of $p_3$ must must be $8$ or $9$.


 * $(8): \quad$ $d \le 142$, otherwise $7 d$ would have $4$ digits.


 * $(9): \quad$ $d \ge 112$, otherwise $9 d$ would have $3$ digits.


 * $(10): \quad$ As $142 \times 9 = 1278$, the second digit of $n$ can be no greater than $2$.

So far: *7*0*   - 1**)1*******     1***       ***       ***       ---       10**        9**         1***         1***

As $142 \times 9 = 1278$, the difference $n_3 - p_3$ cannot exceed $12$.

Since $n_3 \ge 1000$, we have:
 * $988 \le p_3 \le 999$

We also have $p_3 \ge 7 d$ as $p_3$ must be the largest $3$-digit multiple of $d$,

and that $p_3 < 9 d$ since the latter has $4$ digits.

If $p_3 = 7 d$, then $p_2$ would also start with a $9$.

Then $n_2$, being at least $100$ larger, cannot be a $3$-digit number.

Hence we must have $p_3 = 8 d$.

The only number between $988$ and $999$ that is divisible by $8$ is $992$:
 * $992 = 8 \times 124$

Hence we arrive at $q = 97 \, 809$ and we can fill in all other numbers accordingly.