Henry Ernest Dudeney/Puzzles and Curious Problems/99 - The Conspirators' Code/Solution

by : $99$

 * The Conspirators' Code

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