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

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Modern Puzzles by Henry Ernest Dudeney: $70$

The Solitary Seven
        *7***
    ---------
 ***)********
     ****
     ----
       ***
       ***
       ---
       ****
        ***
       ----
         ****
         ****
         ----


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 Dudeney:

$(1): \quad$ 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.

$\blacksquare$


Historical Note

This puzzle was originally submitted to Henry Ernest Dudeney by a person identified merely as the Rev. E. F. O.

Dudeney then published it in The Strand Magazine.

Research is needed to determine which issue.

The uniqueness of the solution was proved at the time by Harold Revell.


Sources

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