Henry Ernest Dudeney/Puzzles and Curious Problems/112 - Simple Division/Solution

by : $112$

 * Simple Division

Solution
The solution given by is as follows:

Solution $1$
There are further solutions:

Solution $4$
There may be more.

Proof
This section declares the variables which are to be used during the deduction of the solution to this skeleton puzzle.

Let $D$ denote the divisor.

Let $Q$ denote the quotient.

Let $N$ denote the dividend.

Let $q_1$ to $q_5$ denote the digits of $Q$ which are calculated at each stage of the long division process in turn.

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

Let $j_1$ to $j_5$ denote the least significant digits of $n_1$ to $n_5$ as they are brought down from $N$ at each stage of the long division process in turn.

Let $p_1$ to $p_5$ denote the partial products generated by the $1$st to $5$th division operations respectively: $p_k = q_k D$

Let $d_1$ to $d_5$ denote the differences between the partial dividends and partial products: $d_k = n_k - p_k$.

By the mechanics of a long division, we have throughout that:


 * $n_k = 10 d_{k - 1} + j_k$

for $k \ge 2$.

Hence we can refer to elements of the structure of this long division as follows: **7**  -->     Q       ---         --- ****7*)**7*******  --> D ) N        ******       --> p_1 ---       *****7*      --> n_2 *******     --> p_2 ---         *7****     --> n_3 *7****    --> p_3 --         *******    --> n_4 ***7**   --> p_4 ---           ******   --> n_5 ******  --> p_5 --

First we attempt to bound the divisor $D$.

Note that $n_4$, $p_4$ and $d_4$ are $7$, $6$ and $5$-digit numbers respectively.

Thus the first digit of $p_4$ must be $9$.

Since each of $n_3$, $p_3$ and $d_3$ are $6$-digit numbers, the first digit of $p_3$ cannot be $9$.

Also note that $p_2$ has $7$ digits.

These give:
 * $p_2 > p_4 > p_3 = 7 D$

Hence we must have $q_2 = 9$ and $q_4 = 8$.

Therefore:
 * $D \le \dfrac {999 \, 799} 8 = 124 \, 974 \tfrac 7 8$

Now we determine the first digit of $p_3$.

If it does not exceed $7$:
 * $p_4 = 8 D \le 8 \times \dfrac {780 \, 000} 7 \approx 891 \, 429$

which contradicts the fact that the first digit of $p_4$ must be $9$.

Since the first digit of $p_3$ cannot be $9$, it must be $8$.

This gives:
 * $D \ge \dfrac {870 \, 000} 7 = 124 \, 285 \tfrac 5 7$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 370$
 * $p_3 = 8 D \ge 8 \times 124 \, 370 = 994 \, 960$

Since the third to last digit of $p_3$ is $7$:
 * $p_3 \ge 995 \, 700$
 * $D \ge \dfrac {995 \, 700} 8 = 124 \, 462.5$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 470$
 * $p_3 \ge 8 \times 124 \, 470 = 995 \, 760$

As $124 \, 475 \times 8 = 995 \, 800$, we will need to check:
 * $D = 124 \, 470, 124 \, 471, 124 \, 472, 124 \, 473, 124 \, 474$

All the possible values of $N$ that can be generated by the values of $D$ above are of the form:
 * $N = D Q = 124 \, 47x \times y9 \, 78z$

where:
 * $0 \le x \le 4$
 * $1 \le y \le 7$
 * $1 \le z \le 8$

First we check that:

Now notice that:

and we see that none of the products above is within a $1 \, 500 \, 000$ range of a number with $7$ as its third digit.

Thus these values of $D$ are eliminated.

Now we consider $D \ge 124 \, 475$.

Since the third to last digit of $p_3$ is $7$:
 * $p_3 \ge 996 \, 700$
 * $D \ge \dfrac {996 \, 700} 8 = 124 \, 587.5$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 670$
 * $p_3 \ge 8 \times 124 \, 670 = 997 \, 360$

Since the third to last digit of $p_3$ is $7$:
 * $p_3 \ge 997 \, 700$
 * $D \ge \dfrac {997 \, 700} 8 = 124 \, 712.5$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 770$
 * $p_3 \ge 8 \times 124 \, 770 = 998 \, 160$

Since the third to last digit of $p_3$ is $7$:
 * $p_3 \ge 998 \, 700$
 * $D \ge \dfrac {998 \, 700} 8 = 124 \, 837.5$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 870$
 * $p_3 \ge 8 \times 124 \, 870 = 998 \, 960$

Since the third to last digit of $p_3$ is $7$:
 * $p_3 \ge 999 \, 700$
 * $D \ge \dfrac {999 \, 700} 8 = 124 \, 962.5$

Since the second to last digit of $D$ is $7$:
 * $D \ge 124 \, 970$

Thus we just need to check the these remaining values:
 * $D = 124 \, 970, 124 \, 971, 124 \, 972, 124 \, 973, 124 \, 974$

These values give:
 * $1 \, 124 \, 730 \le p_2 \le 1 \, 124 \, 766$
 * $874 \, 790 \le p_3 \le 874 \, 818$
 * $999 \, 760 \le p_4 \le 999 \, 792$

With this information, we can fill in parts of the puzzle: *978*      --- 12497*)**7*******        ******        ---        *****7*        11247**        ---          *7****          874***          ---          1******           9997**            ******            ******            --

Note that all values that $p_2$ can take contains a digit $7$.

Thus this puzzle cannot be completed unless more $7$'s are used than those given at the start.

Since $n_3 < 980 \, 000$, we have:
 * $D \times \sqbrk {78q_5} < 98 \, 000 \, 000$

which gives:
 * $\sqbrk {78q_5} < \dfrac {98 \, 000 \, 000} {124 \, 974} \approx 784.2$

This shows that $q_5 \le 4$.

Using the $7$ given in $n_2$, we need to find the pairs of $D$ and $q_5$ such that the sixth digit of the product $D \times \sqbrk {978q_5}$ is $7$.

Going through all $20$ possibities:
 * $\begin{array}{r|rrrr}

\times & 9781 & 9782 & 9783 & 9784 \\ \hline 124 \, 970 & 1 \, 222 \, 331 \, 570 & 1 \, 222 \, 456 \, 540 & 1 \, 222 \, 581 \, 510 & 1 \, 222 \, 706 \, 480 \\ 124 \, 971 & 1 \, 222 \, 341 \, 351 & 1 \, 222 \, 466 \, 322 & 1 \, 222 \, 591 \, 293 & 1 \, 222 \, 716 \, 264 \\ 124 \, 972 & 1 \, 222 \, 351 \, 132 & 1 \, 222 \, 4 \color{red} 76 \, 104 & 1 \, 222 \, 601 \, 076 & 1 \, 222 \, 726 \, 048 \\ 124 \, 973 & 1 \, 222 \, 360 \, 913 & 1 \, 222 \, 485 \, 886 & 1 \, 222 \, 610 \, 859 & 1 \, 222 \, 735 \, 832 \\ 124 \, 974 & 1 \, 222 \, 3 \color{red} 70 \, 694 & 1 \, 222 \, 495 \, 668 & 1 \, 222 \, 620 \, 642 & 1 \, 222 \, 745 \, 616 \\ \end{array} $

We see that the only possible pairs are:
 * $D = 124 \, 972$ and $q_5 = 2$
 * $D = 124 \, 974$ and $q_5 = 1$