Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax/Solution/Addition Case

by : $95$

 * Beeswax

Solution
There is indeed no solution for the addition case, as claimed:


 * The detectives assumed that it was an addition sum, and utterly failed to solve it.:

If this were an addition, we have immediately from the units place:
 * $X = 0$

Now in the ten-millions place, if there is no carry from the millions place, we would have:
 * $A + P = P$

which gives a repeated $A = 0$.

Hence there is such a carry, and $A = 9$, giving a carry to the hundred-millions place.

The hundreds place and the hundred-millions place have the identical sum:
 * "$E + B = K$"

So there must be a carry from the tens place and no carry to the thousands place.

From the thousands place we have:
 * $S + S \equiv E \pmod {10}$

so $E$ is even.

We have from the millions place:
 * "$S + W = E$"

so this place must have received a carry from the hundred-thousands place, and $W + 1 = S$.

As we now have:
 * $S + W + 1 = 10 + E$

there must be a carry from the thousands place to the ten-thousands place.

So far we have: E 9 S E B S B S 0 B P W W K S E T Q + 1 1 1 ? 1 0 1 0 - <-- carries --- K P E P W E K K Q

Notice that we have:
 * $S + T = 10 + K$
 * $B + E + 1 = K$

which gives the inequality:
 * $4 = 1 + 2 + 1 \le K \le 7 + 8 - 10 = 5$

However, if $K = 5$, $B + E = 4$.

As $B, E \ne 0$ and $E$ is even, we would end up with:
 * $B = E = 2$

which is impossible.

Hence $K = 4, B = 1, E = 2$.

From the thousands place:
 * $S + S = 12$

so $S = 6$.

From the ten-thousands place:
 * $W = B + K + 1 = 6 = S$

which repeats a digit.

So we have shown that there are no solutions to the addition case without repeating digits.