Henry Ernest Dudeney/Puzzles and Curious Problems/97 - Letter Multiplication/Solution

by : $97$

 * Letter Multiplication

Solution
4 9 7 3 x        8 --- 3 9 7 8 4

Proof
Note that:


 * $M < T$, because $MEATS = SEAM \times T < 10000 \times T = T0000$.


 * $M < S$, because $MEATS = SEAM \times T < SEAM \times 10 = SEAM0$, and $M \ne S$.

Also $S \ne T$.

By observing the lower triangular part of the modulo $10$ multiplication table below:

$\begin{array}{c|cccccccc} T \backslash M & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 2 & 2 \\ 3 & 3 & \color {red} 6 \\ 4 & 4 & \color {red} 8 & 2 \\ 5 & 5 & 0 & 5 & 0 \\ 6 & 6 & 2 & \color {red} 8 & 4 & 0 \\ 7 & 7 & \color {red} 4 & 1 & \color {red} 8 & 5 & 2 \\ 8 & 8 & \color {red} 6 & \color {red} 4 & 2 & 0 & 8 & 6 \\ 9 & 9 & \color {red} 8 & \color {red} 7 & \color {red} 6 & 5 & 4 & 3 & 2 & \\ \end{array}$

we see that only the red numbers satisfy the conditions for $S$.

Now we consider an additional condition:
 * $M \ge \floor {\dfrac {S \times T} {10}}$

i.e. the tens digit of $S \times T$ cannot exceed $M$.

Otherwise $\floor {\dfrac {S \times T} {10}} \ge M + 1$,

and $SEAM \times T > S000 \times T = 10000 \times \dfrac {S \times T} {10} \ge 10000 \paren {M + 1} > MEATS$.

Hence only the red numbers in:

$\begin{array}{c|ccc} T \backslash M & 2 & 3 & 4\\ \hline 3 & \color {red} 6 \\ 4 & 8 & 2 \\ 5 & 0 & 5 & 0 \\ 6 & 2 & 8 & 4 \\ 7 & \color {red} 4 & 1 & 8 \\ 8 & 6 & \color {red} 4 & 2 \\ 9 & 8 & 7 & 6 \\ \end{array}$

remain.

Using only the values of $M, T, S$, we can determine the value of $SEAM$ by the formula:
 * $SEAM = \dfrac {S00M0 - M00TS} {10 - T}$

since:

Now we inspect the $3$ remaining cases using the formula.

$M = 2, T = 3, S = 6$ gives:
 * $6EA2 = \dfrac {60020 - 20036} {10 - 3} = 5712$

$M = 2, T = 7, S = 4$ gives:
 * $4EA2 = \dfrac {40020 - 20074} {10 - 7} = 6648 \tfrac 2 3$

$M = 3, T = 8, S = 4$ gives:
 * $4EA3 = \dfrac {40030 - 30084} {10 - 8} = 4973$

Only $4973$ matches the required form of $SEAM$.

Hence the solution given is the unique solution.