Henry Ernest Dudeney/Puzzles and Curious Problems/104 - Letter-Figure Puzzle/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $104$

Letter-Figure Puzzle
\(\text {(0)}: \quad\) \(\ds A \times B\) \(=\) \(\ds B\)
\(\text {(1)}: \quad\) \(\ds B \times C\) \(=\) \(\ds A C\)
\(\text {(2)}: \quad\) \(\ds C \times D\) \(=\) \(\ds B C\)
\(\text {(3)}: \quad\) \(\ds D \times E\) \(=\) \(\ds C H\)
\(\text {(4)}: \quad\) \(\ds E \times F\) \(=\) \(\ds D K\)
\(\text {(5)}: \quad\) \(\ds F \times H\) \(=\) \(\ds C J\)
\(\text {(6)}: \quad\) \(\ds H \times J\) \(=\) \(\ds K J\)
\(\text {(7)}: \quad\) \(\ds J \times K\) \(=\) \(\ds E\)
\(\text {(8)}: \quad\) \(\ds K \times L\) \(=\) \(\ds L\)
\(\text {(9)}: \quad\) \(\ds A \times L\) \(=\) \(\ds L\)
Every letter represents a different digit, and, of course, $A C$, $B C$ etc., are two-figure numbers.
Can you find the values in figures of all the letters?


Solution

 A B C D E F H J K L
---------------------
 1 3 5 7 8 9 6 4 2 0

Proof

Trivially from $(8)$ and $(9)$

$L = 0$

Hence immediately from $(1)$:

$A = 1$

Hence:

$10 < A C < 20$

From $(2)$ we inspect pairs ending in the same digit whose divisor is that same digit twice, and this leads us to:

$C = 5$

Hence:

\(\ds C\) \(=\) \(\ds 5\) by inspection of possibilities
\(\ds \leadsto \ \ \) \(\ds B \times C\) \(=\) \(\ds 3 \times 5 = 15\) \(\ds = A C\) $A = 1$ constrains $B$ to $3$
\(\ds \leadsto \ \ \) \(\ds C \times D\) \(=\) \(\ds 5 \times 7 = 35\) \(\ds = B C\) We know both $B = 3$ and $C = 5$, which gives us $D$
\(\ds \leadsto \ \ \) \(\ds D \times E\) \(=\) \(\ds 7 \times 8 = 56\) \(\ds = C H\) We know both $D = 7$ and $C = 5$, which constrains $E$ and $H$
\(\ds \leadsto \ \ \) \(\ds E \times F\) \(=\) \(\ds 8 \times 9 = 72\) \(\ds = D K\) We know both $E = 8$ and $D = 7$, which constrains $D$ and $K$
\(\ds \leadsto \ \ \) \(\ds F \times H\) \(=\) \(\ds 9 \times 6 = 54\) \(\ds = C J\) We know both $F = 9$ and $C = 5$, which constrains $C$ and $J$
\(\ds \leadsto \ \ \) \(\ds H \times J\) \(=\) \(\ds 6 \times 4 = 24\) \(\ds = K J\) We about know everything now anyway

$\blacksquare$


Sources

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