Henry Ernest Dudeney/Puzzles and Curious Problems/99 - The Conspirators' Code/Solution
Puzzles and Curious Problems by Henry Ernest Dudeney: $99$
- The Conspirators' Code
- Two conspirators had a secret code.
- Their letters sometimes contained little arithmetical sums related to some quite plausible discussion,
- and having an entirely innocent appearance.
- But in their code each of the ten digits represented a different letter of the alphabet.
- Thus, on one occasion, there was a little sum in simple addition which, when the letters were substituted for the figures, read as follows:
F L Y F O R + Y O U R ----------- L I F E
- It will be found an interesting puzzle to reconstruct the addition sum with the help of the clue that $I$ and $O$ stand for the figures $1$ and $0$ respectively.
Solution
5 9 8 5 0 7 + 8 0 4 7 ----------- 9 1 5 2
Proof
For each place value, the carry can only be $0, 1$ or $2$.
Let the carries from the units, tens and hundreds be $c_1, c_2$ and $c_3$ respectively.
We are given that $O = 0$ and $I = 1$.
So from the hundreds column we have:
- $F + F + 0 + c_2 \equiv 1 \pmod {10}$
Since the left hand side is an odd number:
- $c_2 = 1$
and thus $2 F \equiv 0 \pmod {10}$.
Since $F \ne 0$, we must have $F = 5$.
This gives $c_3 = 1$, and thus $Y + 1 = L$.
From $c_2 = 1$ we also have, in the tens column:
- $L + U + c_1 = 15$
so:
- $L = 15 - U - c_1 \ge 15 - 9 - 2 = 4$
We test each possible value of $L$.
Suppose $L = 4$.
Then $Y = 3$, $U = 9$ and $c_1 = 2$.
From the units column:
- $3 + R + R \ge 20$
- $R \ge 8.5$
but since the value of $9$ is taken by $U$, this is impossible.
Thus $L > 4$.
Since the value of $5$ is taken by $F$:
- $L \ne 5, 6$
Suppose $L = 7$.
Then $Y = 6$.
From the tens column:
- $7 + U + c_1 = 15$
Since $c_1$ can take on $0, 1, 2$:
- $U = 8, 7, 6$
but both $6$ and $7$ are taken.
Thus $U = 8$ and $c_1 = 0$.
From the units column:
- $6 + R + R = E < 10$
This gives $R = 1$ but $1$ is also taken.
Thus we cannot have $L = 7$.
Suppose $L = 8$.
Then $Y = 7$.
From the tens column:
- $8 + U + c_1 = 15$
Since $c_1$ can take on $0, 1, 2$:
- $U = 7, 6, 5$
but both $5$ and $7$ are taken.
Thus $U = 6$ and $c_1 = 1$.
From the units column:
- $7 + R + R = 10 + E$
The possible values of $R$ are $2, 3, 4, 9$.
But:
- $7 + 2 + 2 = 11$
- $7 + 3 + 3 = 13$
- $7 + 4 + 4 = 15$
- $7 + 9 + 9 = 25$
and all these possibities will use the same digit twice.
Thus $L \ne 8$.
Now we arrive at $L = 9$.
So we have $Y = 8$.
From the units column:
- $10 c_1 + E = 8 + R + R > 8 + 1 + 1 = 10$
so $c_1 \ge 1$.
From the tens column:
- $9 + U + c_1 = 15$
Since $c_1$ can take on $1$ and $2$:
- $U = 5, 4$
but $5$ is taken by $F$, so $U = 4$ and $c_1 = 2$.
Since $0$ is taken by $O$:
- $8 + R + R > 20$
This gives $R > 6$.
Since both $8$ and $9$ are taken, we must have $R = 7$ and thus $E = 2$.
Hence we arrive at the solution:
5 9 8 5 0 7 + 8 0 4 7 ----------- 9 1 5 2
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $99$. -- The Conspirators' Code
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $158$. The Conspirators' Code