Two sailors have equal rights in a pile of coconuts, and plan to divide them in the morning.
During the night, one sailor, fearing he will not receive his just share, secretly divided them into two piles, finding the division exact if he gave their pet monkey the extra coconut; he hid one pile for his own future use, and left the second pile for the next morning's sharing.
After he went to sleep, the other sailor awoke with the same idea, and treated the remaining pile in the same way as the first sailor, hiding his "half" after again giving the odd coconut to their pet monkey.
The next morning, the two sailors awoke and discovered that the pile of coconuts was exactly divisible into two piles...with nothing for the disappointed monkey!
How many coconuts were in the original pile? (Note: Is the answer unique?)
New Task: Repeat the problem for n sailors, for n = 3, 4, 5, ..... Is there a pattern in the resulting answers?
Source: Adapted from G. Merriman's To Discover Mathematics, 1942, p. 368
Hint: Take a guess, try it...and adjust....
Or, can you work backwards somehow?
Solution Commentary: I refuse to spoil this classic problem for you...but will note that for the case of n = 5 sailors, the smallest possible answer is 3121 coconuts.