Henry Ernest Dudeney/Puzzles and Curious Problems/111 - Odds and Evens/Solution/Initial Deductions
Puzzles and Curious Problems by Henry Ernest Dudeney: $111$
- Odds and Evens
- Every asterisk and letter represents a figure,
- Can you construct an arrangement complying with these conditions?
- There are $6$ solutions.
- Can you find one, or all of them?
******* ---------- ***)********* OE* ----- **** OO** ----- *** EE* ---- *** EO* ---- **** EE** ----- *** OO* ---
Initial Deductions
Declarations
Let $D$ denote the divisor.
Let $Q$ denote the quotient.
Let $N$ denote the dividend.
Let $q_1$ to $q_7$ denote the digits of $Q$ which are calculated at each stage of the long division process in turn.
Let $n_1$ to $n_7$ denote the partial dividends which are subject to the $1$st to $7$th division operations respectively.
- Note that $n_2 = 0$, but has been retained for consistency of numbering.
Let $j_1$ to $j_7$ denote the least significant digits of $n_1$ to $n_7$ as they are brought down from $N$ at each stage of the long division process in turn.
Let $p_1$ to $p_7$ denote the partial products generated by the $1$st to $7$th division operations respectively: $p_k = q_k D$
- Again note that $p_2 = 0$, but has been retained for consistency of numbering.
Let $d_1$ to $d_7$ 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 OE* --> p_1 ----- **** --> n_3 OO** --> p_3 ----- *** --> n_4 EE* --> p_4 ---- *** --> n_5 EO* --> p_5 ---- **** --> n_6 EE** --> p_6 ----- *** --> n_7 OO* --> p_7 ---
First we note that the partial products are structured as follows:
- $p_1: \mathtt {O E} *$
- $p_3: \mathtt {O O} * *$
- $p_4: \mathtt {E E} *$
- $p_5: \mathtt {E O} *$
- $p_6: \mathtt {E E} * *$
- $p_7: \mathtt {O O} *$
and so are all distinct.
Hence $q_1$ to $q_7$ are also distinct.
We also know that $q_2 = 0$ as $2$ digits were pulled down into $n_3$.
Thus the first $3$ digits of $n_3$ form a number strictly smaller than $D$.
We have that $p_1$, $p_4$, $p_5$ and $p_7$ each have $3$ digits.
But each of $q_1$, $q_4$, $q_5$ and $q_7$ are distinct.
Hence:
\(\ds \max \set {q_1, q_4, q_5, q_7}\) | \(\ge\) | \(\ds 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \max \set {q_1, q_4, q_5, q_7} \times D\) | \(=\) | \(\ds \max \set {p_1, p_4, p_5, p_7}\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds 999\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds D\) | \(\le\) | \(\ds \dfrac {999} 4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds D\) | \(\le\) | \(\ds 249\) |
Notice that $p_6$ is a $4$-digit number that starts with an even number.
Therefore $p_6 \ge 2000$.
Hence:
\(\ds q_6\) | \(\le\) | \(\ds 9\) | ||||||||||||
\(\ds q_6 \times D\) | \(=\) | \(\ds p_6\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds 2000\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds D\) | \(\ge\) | \(\ds \dfrac {2000} 9\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds D\) | \(\ge\) | \(\ds 223\) |
That is:
- $223 \le D \le 249$
The four $3$-digit multiples of $D$ are of the patterns:
- $p_1: \mathtt {O E} *$, $p_4: \mathtt {E E} *$, $p_5: \mathtt {E O} *$, $p_7: \mathtt {O O} *$
If $3 D < 700$, the first three multiples of $D$ all begin with the digits $2$, $4$ and $6$, which causes the candidate $D$ to fail the criteria above.
Thus:
- $3 D \ge 700$
or equivalently:
- $D \ge 234$
We check the rest of candidates for $D$ to see whether they have multiples that are of each of the six forms.
First we check the $3$-digit multiples.
For visual clarity, the first two digits are marked $\color { red } {\text {red} }$ when odd and $\color { blue } {\text {blue} }$ when even:
- $\begin{array} {r|rrrrrrrr} \times & 2 & 3 & 4 \\ \hline \color {blue} 2 \color {red} 3 4 & \color {blue} 4 \color {blue} 68 & \color {red} 7 \color {blue} 02 & \color {red} 9 \color {red} 3 6 & \\ \color {blue} 2 \color {red} 3 5 & \color {blue} 4 \color {red} 70 & \color {red} 7 \color {blue} 05 & \color {red} 9 \color {blue} 40 & \\ \color {blue} 2 \color {red} 3 6 & \color {blue} 4 \color {red} 72 & \color {red} 7 \color {blue} 08 & \color {red} 9 \color {blue} 44 & \\ \color {blue} 2 \color {red} 3 7 & \color {blue} 4 \color {red} 74 & \color {red} 7 \color {red} 11 & \color {red} 9 \color {blue} 48 & \\ \color {blue} 2 \color {red} 3 8 & \color {blue} 4 \color {red} 76 & \color {red} 7 \color {red} 14 & \color {red} 9 \color {red} 52 & \\ \color {blue} 2 \color {red} 3 9 & \color {blue} 4 \color {red} 78 & \color {red} 7 \color {red} 17 & \color {red} 9 \color {red} 56 & \\ \color {blue} 2 \color {blue} 40 & \color {blue} 4 \color {blue} 80 & \color {red} 7 \color {blue} 20 & \color {red} 9 \color {blue} 60 & \\ \color {blue} 2 \color {blue} 41 & \color {blue} 4 \color {blue} 82 & \color {red} 7 \color {blue} 23 & \color {red} 9 \color {blue} 64 & \\ \color {blue} 2 \color {blue} 42 & \color {blue} 4 \color {blue} 84 & \color {red} 7 \color {blue} 26 & \color {red} 9 \color {blue} 68 & \\ \color {blue} 2 \color {blue} 43 & \color {blue} 4 \color {blue} 86 & \color {red} 7 \color {blue} 29 & \color {red} 9 \color {red} 72 & \\ \color {blue} 2 \color {blue} 44 & \color {blue} 4 \color {blue} 88 & \color {red} 7 \color {red} 32 & \color {red} 9 \color {red} 76 & \\ \color {blue} 2 \color {blue} 45 & \color {blue} 4 \color {red} 90 & \color {red} 7 \color {red} 35 & \color {red} 9 \color {blue} 80 & \\ \color {blue} 2 \color {blue} 46 & \color {blue} 4 \color {red} 92 & \color {red} 7 \color {red} 38 & \color {red} 9 \color {blue} 84 & \\ \color {blue} 2 \color {blue} 47 & \color {blue} 4 \color {red} 94 & \color {red} 7 \color {blue} 41 & \color {red} 9 \color {blue} 88 & \\ \color {blue} 2 \color {blue} 48 & \color {blue} 4 \color {red} 96 & \color {red} 7 \color {blue} 44 & \color {red} 9 \color {red} 92 & \\ \color {blue} 2 \color {blue} 49 & \color {blue} 4 \color {red} 98 & \color {red} 7 \color {blue} 47 & \color {red} 9 \color {red} 96 & \end{array}$
Of the above, $234$, $245$, $246$, $248$ and $249$ have the appropriate $\mathtt {O E}$, $\mathtt {E E}$, $\mathtt {E O}$ and $\mathtt {O O}$ forms, in some order.
We now check the $4$-digit multiples of those $5$ numbers in the same way:
- $\begin{array} {r|rrrrrrrr} \times & 5 & 6 & 7 & 8 & 9 \\ \hline 234 & \color {red} {1170} & 1404 & 1638 & 1872 & 2106 \\ 245 & 1225 & 1470 & \color {red} {1715} & \color {red} {1960} & \color {blue} {2205} \\ 246 & 1230 & 1476 & \color {red} {1722} & \color {red} {1968} & \color {blue} {2214} \\ 248 & 1240 & 1488 & \color {red} {1736} & \color {red} {1984} & \color {blue} {2232} \\ 249 & 1245 & 1494 & \color {red} {1743} & \color {red} {1992} & \color {blue} {2241} \end{array}$
The numbers of $\mathtt {O O}$ form are marked $\color { red } {\text {red} }$, while those of $\mathtt {E E}$ form are marked $\color { blue } {\text {blue} }$.
It can be seen that of the above, $234$ is eliminated as a candidate as it has no $4$-digit multiple of $\mathtt {E E}$ form.
It remains to explore the individual long divisions which are composed of the above candidate values of $D$ and their multiples.
Since we know the form of each multiple, we can match each form to its corresponding digit of the quotient.
For $D = 246$, the quotient is either $4 \, 071 \, 293$ or $4 \, 081 \, 293$.
However, we would have:
- $N = D Q \ge 246 \times 4 \, 071 \, 293 = 1 \, 001 \, 538 \,078$
but $N$ only has $9$ digits.
Therefore $D \ne 246$.
The rest of the possibilities all yield solutions, and will be explored in their respective solution pages.