Henry Ernest Dudeney/Puzzles and Curious Problems/110 - An Absolute Skeleton/Solution/Initial Deductions

Puzzles and Curious Problems by Henry Ernest Dudeney: $110$

An Absolute Skeleton

Here is a good skeleton puzzle.

The only conditions are:

$(1)$ No digit appears twice in any row of figures except the dividend.

$(2)$ If $2$ be added to the last figure in the quotient it equals the last but one,

and if $2$ be added to the third figure from the end it gives the last figure but $3$ in the quotient.

That is to say, the quotient might end in, say, $9742$, or in $3186$.

        ********
    ------------
 ***)***********
      ***
       ---
       ***
       ***
       ----
       ****
       ****
       -----
         ***
         ***
         ----
         ****
         ****
         -----
          ****
          ****
          -----
           ****
           ****
           -----
            ****
            ****
            ----

Initial Deductions

Declarations

Let $D$ denote the divisor.

Let $Q$ denote the quotient.

Let $N$ denote the dividend.

Let $q_1$ to $q_8$ denote the digits of $Q$ which are calculated at each stage of the long division process in turn.

Let $n_1$ to $n_8$ denote the partial dividends which are subject to the $1$st to $8$th division operations respectively.

Let $j_1$ to $j_8$ denote the least significant digits of $n_1$ to $n_8$ as they are brought down from $N$ at each stage of the long division process in turn.

Let $p_1$ to $p_8$ denote the partial products generated by the $1$st to $8$th division operations respectively: $p_k = q_k D$

Let $d_1$ to $d_8$ 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:

        ********  -->     Q
    ------------        ---
 ***)***********  --> D ) N
      ***         --> p_1
       ---
       ***        --> n_2
       ***        --> p_2
       ----
       ****       --> n_3
       ****       --> p_3
       -----
         ***      --> n_4
         ***      --> p_4
         ----
         ****     --> n_5
         ****     --> p_5
         -----
          ****    --> n_6
          ****    --> p_6
          -----
           ****   --> n_7
           ****   --> p_7
           -----
            ****  --> n_8
            ****  --> p_8
            ----

It is apparent from the structure of the long division sum presented that no digit of $Q$ may be zero.

We also have, from the additional constraint on $Q$, that:

$\ds q_7$

$=$

$\ds q_8 + 2$

If $2$ be added to the last figure in the quotient it equals the last but one,

$\ds q_5$

$=$

$\ds q_6 + 2$

and if $2$ be added to the third figure from the end it gives the last figure but $3$ in the quotient.

This also gives us that:

$\ds q_5$

$\ge$

$\ds 3$

as $q_6 \ne 0$

$\ds q_6$

$\le$

$\ds 7$

$\ds q_7$

$\ge$

$\ds 3$

as $q_8 \ne 0$

$\ds q_8$

$\le$

$\ds 7$

The rule about repeated digits offers similar constraints on $n_2$ to $n_8$ and $p_1$ to $p_8$, and will be used without further comment in the following.

However, note that this does not apply to $n_1$, which in this context is the first $4$ digits of $N$, which has no constraint on the uniqueness of its digits.

We have:

$\ds p_1$

$\le$

$\ds 987$

$\ds d_1$

$\le$

$\ds 98$

$\ds \leadsto \ \ $

$\ds n_1$

$=$

$\ds p_1 + d_1$

$\ds $

$\le$

$\ds 987 + 98$

$\ds $

$=$

$\ds 1085$

This gives us a firm upper bound on $N$, and we can immediately state:

$(1): \quad 10 \, 000 \, 000 \, 000 \le N \le 10 \, 859 \, 999 \, 999$

Hence also:

$\ds p_1$

$=$

$\ds n_1 - d_1$

$\ds p_1$

$\ge$

$\ds 1000 - 98$

the lower bound of $n_1$ minus the upper bound of $d_1$

$\ds $

$=$

$\ds 902$

Thus we have:

$902 \le p_1 \le 987$

We have that $p_1$, $p_2$ and $p_4$ each have $3$ digits.

But each of $q_1$, $q_2$ and $q_4$ are distinct.

Hence:

$\ds \max \set {q_1, q_2, q_4}$

$\ge$

$\ds 3$

$\ds \leadsto \ \ $

$\ds \max \set {q_1, q_2, q_4} \times D$

$=$

$\ds \max \set {p_1, p_2, p_4}$

$\ds $

$\le$

$\ds 987$

$\ds \leadsto \ \ $

$\ds D$

$\le$

$\ds \dfrac {987} 3$

$\text {(2)}: \quad$

$\ds \leadsto \ \ $

$\ds D$

$\le$

$\ds 329$

We have that $p_3$, $p_5$, $p_6$, $p_7$ and $p_8$ each have $4$ digits.

But each of $q_3$, $q_5$, $q_6$, $q_7$ and $q_8$ are distinct.

Hence:

$\ds \min \set {q_3, q_5, q_6, q_7, q_8}$

$\le$

$\ds 5$

$\ds \leadsto \ \ $

$\ds \min \set {q_3, q_5, q_6, q_7, q_8} \times D$

$=$

$\ds \min \set {p_3, p_5, p_6, p_7, p_8}$

$\ds $

$\ge$

$\ds 1023$

$\ds \leadsto \ \ $

$\ds D$

$\ge$

$\ds \dfrac {1023} 5$

$\text {(3)}: \quad$

$\ds \leadsto \ \ $

$\ds D$

$\ge$

$\ds 205$

That is:

$205 \le D \le 329$

Thus we have established bounds on $D$ and $N$.

Hence we can now establish bounds on $Q$:

$\ds Q$

$\le$

$\ds \dfrac {\map \max N} {\map \min D}$

abusing notation

$\ds $

$=$

$\ds \dfrac {10 \, 859 \, 999 \, 999} {205}$

from $(1)$ and $(3)$

$\ds $

$=$

$\ds 52 \, 975 \, 609$

$\ds Q$

$\ge$

$\ds \dfrac {\map \min N} {\map \max D}$

again abusing notation

$\ds $

$=$

$\ds \dfrac {10 \, 000 \, 000 \, 000} {329}$

from $(1)$ and $(2)$

$\ds $

$=$

$\ds 30 \, 395 \, 136$

So we immediately see that $3 \le q_1 \le 5$.

Suppose $q_1 = 5$.

Then:

$\ds D$

$\ge$

$\ds 205$

from $(3)$

$\ds \leadsto \ \ $

$\ds p_1$

$\ge$

$\ds 5 \times 205$

as $p_1 = q_1 D$

$\ds \leadsto \ \ $

$\ds p_1$

$\ge$

$\ds 1025$

But we already have that $p_1 \le 987$.

So $q_1 \ne 5$.

Suppose $q_1 = 4$.

Then:

$\ds D$

$\ge$

$\ds 205$

from $(3)$

$\ds \leadsto \ \ $

$\ds p_1$

$\ge$

$\ds 4 \times 205$

as $p_1 = q_1 D$

$\ds \leadsto \ \ $

$\ds p_1$

$\ge$

$\ds 820$

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But we already have that $p_1 \le 987$.

So $q_1 \ne 4$.

This directly gives us that $q_1 = 3$.

We also have that:

$\ds p_2$

$\le$

$\ds 987$

$\ds p_4$

$\le$

$\ds 987$

Hence by the same reasoning:

$q_2 < 4$ and $q_4 < 4$

and so either:

$q_2 = 1$ and $q_4 = 2$

or:

$q_2 = 2$ and $q_4 = 1$

Now let us consider the lower bound and upper bound for $Q$.

Recall:

$\ds q_7$

$=$

$\ds q_8 + 2$

$\ds q_5$

$=$

$\ds q_6 + 2$

and that $q_1$ to $q_8$ are unique.

We also have that:

$\ds q_1$

$=$

$\ds 3$

$\ds q_2$

$=$

$\ds 1 \text { or } 2$

$\ds q_4$

$=$

$\ds 2 \text { or } 1$

This gives us the limits on $Q$:

$31 \, 427 \, 586 \le Q \le 32 \, 918 \, 675$

Now:

$\ds Q$

$\le$

$\ds 32 \, 918 \, 675$

from above

$\ds D$

$\le$

$\ds 329$

from above

$\ds \leadsto \ \ $

$\ds N$

$=$

$\ds D \times Q$

$\ds $

$\le$

$\ds 10 \, 830 \, 244 \, 075$

Similarly:

$\ds Q$

$\le$

$\ds 32 \, 918 \, 675$

from above

$\ds N$

$\ge$

$\ds 10 \, 000 \, 000 \, 000$

still no better lower bound on $N$

$\ds \leadsto \ \ $

$\ds D$

$=$

$\ds \dfrac N Q$

$\ds $

$\ge$

$\ds 304$

Thus we have revised our bounds on $D$:

$304 \le D \le 329$

Between $304$ and $329$ are $26$ numbers.

Of these, $4$ have duplicated digits:

$311$, $313$, $322$ and $323$

so these cannot equal $D$.

There are sufficiently few of these to compute their multiples to investigate whether they have duplicated digits.

We require that $D k$ have no duplicate digits where $1 \le k \le 3$.

If none of them do, then we require that $D k$ have no duplicate digits for at least $5$ values of $k$ where $4 \le k \le 9$.

By the Pigeonhole Principle, in order to eliminate a candidate for $D$, we need to find just $2$ such multiples with duplicated digits where $4 \le k \le 9$.

We will present them as an array, where the numbers in $\color { red } {\text {red} }$ have duplicated digits.

We bother to go no further with a candidate $D$ once the criteria have been breached.

$\begin{array} {r|rrrrrrrr} \times & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline

304 & 608 & 912 & \color {red} {1216} & 1520 & 1824 & \color {red} {2128} & \\ 305 & 610 & 915 & \color {red} {1220} & \color {red} {1525} \\ 306 & 612 & 918 & \color {red} {1224} & 1530 & 1836 & \color {red} {2142} \\ 307 & 614 & 921 & \color {red} {1228} & \color {red} {1353} & \\ 308 & \color {red} {616} & \\ 309 & 618 & 927 & 1236 & \color {red} {1545} & 1854 & 2163 & \color {red} {2472} \\ 310 & 620 & 930 & 1240 & \color {red} {1550} & 1860 & 2170 & 2480 & 2790 \\ 312 & 624 & 936 & 1248 & 1560 & 1872 & 2184 & 2496 & \color {red} {2808} \\ 314 & 628 & 942 & 1256 & 1570 & \color {red} {1884} & 2198 & \color {red} {2512} & \\ 315 & 630 & 945 & 1260 & \color {red} {1575} & 1890 & \color {red} {2205} & \\ 316 & 632 & 948 & 1264 & 1580 & 1896 & \color {red} {2212} & \color {red} {2528} & \\ 317 & 634 & 951 & 1268 & \color {red} {1585} & 1902 & \color {red} {2219} & \\ 318 & \color {red} {636} & \\ 319 & 638 & 957 & 1276 & \color {red} {1595} & \color {red} {1914} & \\ 320 & 640 & 960 & 1280 & \color {red} {1600} & 1920 & \color {red} {2240} & \\ 321 & 642 & 963 & 1284 & 1605 & 1926 & \color {red} {2247} & 2568 & \color {red} {2889} \\ 324 & 648 & 972 & 1296 & 1620 & \color {red} {1944} & \color {red} {2268} & \\ 325 & 650 & 975 & \color {red} {1300} & 1625 & 1950 & \color {red} {2275} & \\ 326 & 652 & 978 & 1304 & 1630 & 1956 & \color {red} {2282} & 2608 & 2934 \\ 327 & 654 & 981 & 1308 & 1635 & 1962 & \color {red} {2289} & \color {red} {2616} & \\ 328 & \color {red} {656} & \\ 329 & 658 & 987 & \color {red} {1316} & 1645 & 1974 & \color {red} {2303} & \end{array}$

As can be seen, there are three candidates for $D$:

$310$, $312$, $326$

Suppose $D = 326$.

Then $Q$ is such that it has no $7$.

So $Q$ is such that $q_8, q_8 + 2, q_6, q_6 + 2$ come from $\set {4, 5, 6, 8, 9}$.

It is seen that such a $Q$ is not possible.

Hence $D \ne 326$.

It remains to investigate $310$ and $312$.

Let $D = 310$.

Then $Q$ is made from $\set {1, 2, 3, 4, 6, 7, 8, 9}$.

Hence $Q$ can be:

$31428697$

$31429786$

$31826497$

$31829764$

$32418697$

$32419786$

$32816497$

$32819764$

Let $D = 312$.

Then $Q$ is made from $\set {1, 2, 3, 4, 5, 6, 7, 8}$.

Hence $Q$ can be:

$31427586$

$31428675$

$31826475$

$31827564$

$32417586$

$32418675$

$32816475$

$32817564$

But:

$10000000000 = 32051282 \times 312 + 16$

and:

$10000000000 = 32258064 \times 310 + 160$

and so it is seen that the candidate values for $Q$ which are smaller than $32000000$ are too small.

Hence we are left with the following to investigate.

For $D = 310$:

$32418697$

$32419786$

$32816497$

$32819764$

For $D = 312$:

$32417586$

$32418675$

$32816475$

$32817564$

These are to be tested one by one.

Henry Ernest Dudeney/Puzzles and Curious Problems/110 - An Absolute Skeleton/Solution/Initial Deductions

Puzzles and Curious Problems by Henry Ernest Dudeney: $110$

Initial Deductions

Declarations

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