Five sailors and the coconuts
Five sailors gather a pile of coconuts and go to sleep. During the night, each sailor in turn wakes up, divides the pile into five equal parts with one coconut left over, gives the extra coconut to a monkey, hides one part for himself, and puts the other four parts back together as the pile.
In the morning the remaining pile divides into five equal shares with nothing left over.
What is the smallest possible number of coconuts in the original pile?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Each night step takes a pile to . Try adding 4 imaginary coconuts to the pile.
With 4 extra coconuts, each step is just “multiply by ”.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Four imaginary coconuts make every step a clean multiplication by : . Five such steps require the shifted pile to be divisible by .
Step 1: the shift. If a sailor finds coconuts, he leaves . Then So with , each sailor's step takes to .
Step 2: five sailors. After five steps the shifted pile is . For every intermediate pile to be a whole number, must be divisible by . The morning pile is then , and with it is , which divides by 5. So the smallest pile is
Check. The pile goes , and each of these minus 1 (before a sailor's turn) divides by 5. In the morning, .
The original version. In the 1926 story the monkey also gets one coconut in the morning; the same trick then needs , and the answer is .