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

## Modern Puzzles by Henry Ernest Dudeney: $73$

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

                            9 3
Less 1 7 multiplied by 4 = 6 8
---
Leaving  2 5
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## 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
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Leaving  1 I
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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
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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
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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
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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
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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 $1$
$17 \times 2 = 34$
$18 \times 4 = 72$
$18 \times 3 = 54$
$18 \times 2 = 36$
$19 \times 4 = 76$
$19 \times 3 = 57$
$19 \times 2 = 38$