These are from Smullyan’s “The Riddle of Scheherazade”. They are of mathematical nature, and serve for appetizers to a nice collection of logic riddles. They go like this:

"Well," replied Scheherazade, "one night a thief stole into Abdul's shop—"

"He should be drawn and quartered!" interrupted the king.

"True, Your Majesty," replied Scheherazade, "but, to get on with my story, the thief joyfully came across a pile of diamonds.  His first thought was to take them all, but then his conscience bothered him, and he decided to content himself with only half."

"Hmm!" said the king.

"And so he took half of the diamonds and started to leave the shop."

"Ho!" said the king.

"But then he thought: `I'll take one more,' which he did."

"Hoo!" said the king.

"And so he left the shop, having stolen half the diamonds plus one."

"Then what happened?" asked the king.

"Strangely enough, a few minutes later, a second thief entered the shop and took half of the remaining diamonds plus one.  Then a third thief entered the shop and took half the remaining diamonds and one more. Then a fourth thief entered and took half the remainder and one more.  Then a fifth thief entered, but took no diamonds since they were all gone."

"So what is the problem?" asked the king.

"The problem," she said, "is how many diamonds were in the pile to start with?"

"Now, how should I know?"

"It's not difficult to figure out," she replied.

How many diamonds were in the pile?

I will present two ways to go about the solution: a “forward” and a “backward” way. The latter seems to give a much faster solution without less computations, while the former goes in a very straightforward fashion, making use of the basic techniques of algebra.</p>

### Forward method

Like in all problem-solving, we start by labeling the unknown quantity with a variable, say $$x$$, which indicates the number of diamonds in the original pile—this is what we are looking for. The first thief then took $$\frac{x}{2} +1$$ diamonds, and thus after he leaves the store, there will be $$x - \frac{x}{2} - 1$$ remaining diamonds in the pile.

Let us construct a table with three columns where we indicate, on each row, the amount of diamonds that each successive thief encountered, how much they took, and how much remained after their pillage. For the sake of neatness, I will reduce some of the fractions in the columns of diamonds before and diamonds after.

\begin{equation} \begin{array}{|c|||c|c|c|} \hline \text{Thief} & \text{before} & \text{takes} & \text{after} \\ \hline \text{first} & x & \frac{x}{2} + 1 & x - \frac{x+2}{2} \\ \hline \text{second} & \frac{x-2}{2} & \frac{x-2}{4} + 1 & \frac{x-2}{2} - \frac{x+2}{4} \\ \hline \text{third} & \frac{x-6}{4} & \frac{x-6}{8} + 1 & \frac{x-6}{4} - \frac{x+2}{8} \\ \hline \text{fourth} & \frac{x-14}{8} & \frac{x-14}{16} + 1 & \frac{x-14}{8} - \frac{x+2}{16} \\ \hline \end{array} \end{equation}

The value of the last entry of the table must be zero by hypothesis (that is how many diamonds did the fifth thief find!). We can then solve that equation to find the solution:

\begin{equation} \frac{x-14}{8} = \frac{x+2}{16}. \end{equation}

This gives the amount of diamonds in the original pile. What is the value of $$x$$?

### Backwards method

The backwards method starts by realizing that the fourth thief finds only two diamonds in the pile (since “one half of the diamonds plus one” is all the diamonds present in the store).  Algebraically, this reads $$x_4 = x_4/2 +1$$, where $$x_4$$ is the amount of diamonds found by the fourth thief.

How many diamonds did the third thief find in the store? (let’s label that value $$x_3$$) We know that, after taking half of them plus one, there remained only 2.  The only possibility is thus $$x_3 = 6$$.  Again, algebraically this is the solution to the equation $$x_3 - (x_3/2 + 1) = 2.$$

Continue in this fashion, finding the values of $$x_2$$ and $$x_1$$.  The latter is the value we are looking for.

### Miscellaneous

Which method suits you better?

Smullyan doesn’t stop here.  He presents us with two more versions of the same riddle.  Can you find solutions to those too?

According to a second version, each of the first four thieves took half of what he found plus two, instead of plus one, and again the fifth thief found none.  According to this version, how many diamonds were originally in the pile?

The third version is the same as the second, except that the fifth thief found one diamond.  If this version is correct, then how many diamonds did the first thief find?