Three sailors come across a pile of coconuts. The first sailor takes half of them plus half a coconut.
The second sailor takes half of what is left, plus half a coconut. The third sailor also takes half of
what remains, plus half a coconut. Left over is exactly one coconut, which they toss to a monkey.

How many coconuts were in the original pile?

*Credit: This puzzle, created by Martin Gardner in 1961, is a simplified version of a more difficult
puzzle by Ben Ames Williams published in 1926.*
**ANSWER**:

*15*.

**EXPLANATION**: Let

**S1**,

**S2**, and

**S3**
represent the number of coconuts taken by the first, second, and third sailor respectively and let

**X** represent the number of coconuts in the original pile.

S1 = X/2 + 1/2

S2 = (X - S1)/2 + 1/2

S2 = ((X - (X/2 + 1/2)) /2) + 1/2

S2 = X/2 - (X/4 + 1/4) + 1/2

S2 = 2X/4 - (X/4 + 1/4) + 2/4

S2 = 2X/4 - X/4 - 1/4 + 2/4

S2 = X/4 + 1/4

S3 = (X - S1 - S2)/2 + 1/2

S3 = (X - (X/2 + 1/2) - (X/4 + 1/4))/2 + 1/2

S3 = X/2 - (X/4 + 1/4) - (X/8 + 1/8) + 1/2

S3 = X/2 - X/4 - 1/4 - X/8 - 1/8 + 1/2

S3 = 4X/8 - 2X/8 - 2/8 - X/8 - 1/8 + 4/8

S3 = X/8 + 1/8

X - S1 - S2 - S3 = 1

X - (X/2 + 1/2) - (X/4 + 1/4) - (X/8 + 1/8) = 1

X - X/2 - 1/2 - X/4 - 1/4 - X/8 - 1/8 = 1

8X/8 - 4X/8 - 4/8 - 2X/8 - 2/8 - X/8 - 1/8 = 1

X/8 - 7/8 = 1

X - 7 = 8

X = 15

Do you have a

suggestion for this puzzle (e.g. something that should
be mentioned/clarified in the question or solution, bug, typo, etc.)?