Henry Ernest Dudeney/Modern Puzzles/73 - Alphabetical Arithmetic/Solution

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Modern Puzzles by Henry Ernest Dudeney: $73$

Alphabetical Arithmetic
                            F G
 Less A B multiplied by C = D E
                            ---
                   Leaving  H I
                            ---


Solution

                            9 3
 Less 1 7 multiplied by 4 = 6 8
                            ---
                   Leaving  2 5
                            ---

Proof

Immediately:

$F > D$
$A \le 4$
$B \ne 1, 5$
$C \ne 1, 5$


Suppose $A = 4$.

Since $3 \times 40 > 100$, $C = 2$ and $D \ge 8$.

Since $9 \ge F > D$, we must have $D = 8$, leaving $H = 1$.

$E$ is even, so $E = 6$ and $B = 3$.

So far we have: 

                            9 G
 Less 4 3 multiplied by 2 = 8 6
                            ---
                   Leaving  1 I
                            ---

The two remaining digits are $5$ and $7$, and cannot fit in the positions of $G$ and $I$.

Therefore $A \ne 4$.

$\Box$


Suppose $A = 3$.

Since $4 \times 30 > 100$, $C = 2$ and $D \ge 6$.

Since $F \le 9$, $H \le F - D \le 9 - 6 = 3$.

This forces $H = 1$.

We check the case $B = 4$.

So far we have: 

                            F G
 Less 3 4 multiplied by 2 = 6 8
                            ---
                   Leaving  1 I
                            ---

The remaining digits are $5, 7$ and $9$.

This forces $F = 7$.

But then $5$ and $9$ cannot fit in the positions of $G$ and $I$.

Thus $B \ne 4$.


For $B > 5$, we have $D = 7$.

$B \ne 6$ since $E \ne 2$.

We check the cases $B = 8, 9$.

For $B = 8$ we have: 

                            F G
 Less 3 8 multiplied by 2 = 7 6
                            ---
                   Leaving  1 I
                            ---

The remaining digits are $4, 5$ and $9$.

This forces $F = 9$.

But then $4$ and $5$ cannot fit in the positions of $G$ and $I$.

Thus $B \ne 8$.


For $B = 9$ we have: 

                            F G
 Less 3 9 multiplied by 2 = 7 8
                            ---
                   Leaving  1 I
                            ---

The remaining digits are $4, 5$ and $6$.

None of the digits can fit $F$.

Therefore $B \ne 9$, and thus $A \ne 3$.

$\Box$


Suppose $A = 2$.

We have $C = 3$ or $4$.

If $C = 4$, since $B \ge 3$:

$D E \ge 23 \times 4 = 92$

But then $F > 9$ is impossible.

Thus $C = 3$.

Similar to the above, we have $H = 1$.

If $B = 9$, $DE = 87$.

But $FG > 87$ is impossible as $9$ is taken.

If $B = 8$, $DE = 84$ which repeats $8$.

If $B = 7$, $DE = 81$ which repeats $1$.

If $B = 4$, $DE = 72$ which repeats $2$.

Thus $B = 6$.

So far we have: 

                            F G
 Less 2 6 multiplied by 3 = 7 8
                            ---
                   Leaving  1 I
                            ---

The remaining digits are $4, 5$ and $9$.

Thus $F = 9$.

But then $4$ and $5$ cannot fit in the positions of $G$ and $I$.

Therefore $A \ne 2$.

$\Box$


We arrive at $A = 1$.

Then $DE = FG - HI \le 98 - 23 = 75$.

This leaves the following cases for $AB \times C = DE$ to check:

$13 \times 4 = 52$
$13 \times 2 = 26$ repeats $2$
$14 \times 3 = 42$ repeats $4$
$14 \times 2 = 28$ repeats $2$
$16 \times 4 = 64$ repeats $4$ and $6$
$16 \times 3 = 48$
$16 \times 2 = 32$ repeats $2$
$17 \times 4 = 68$
$17 \times 3 = 51$ repeats $1$
$17 \times 2 = 34$
$18 \times 4 = 72$
$18 \times 3 = 54$
$18 \times 2 = 36$
$19 \times 4 = 76$
$19 \times 3 = 57$
$19 \times 2 = 38$


This needs considerable tedious hard slog to complete it.
In particular: While checking the final $10$ cases is certainly trivial, it might be worth it to first check whether there is an alternative way to solve this. There is no obvious grouping of these $10$ cases that could reduce the amount of work needed.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

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