Three sailors land on an island where they find a pile of whole coconuts and a monkey. They agree to sleep overnight and divide up the pile of coconuts in the morning. During the night, one sailor wakes up, gives one coconut to the monkey, takes the number of whole coconuts that is EXACTLY one-third of what is left, and goes back to sleep. Then the second sailor awakens and does the same. Later, the third sailor wakes up and does the same. In the morning, there are fewer than 10 coconuts left. Each sailor takes a number of whole coconuts that is exactly one-third of what is in the pile. HOW MANY COCONUTS WERE IN THE ORIGINAL PILE?
Here is how I did it:
Let x be the number of coconuts to start with. We are given that it is a whole number.
After Sailor 1 gives one to the monkey and takes a third. Then the number of coconuts remaining is (x-1) - (x-1) (1/3) = 2(x-1)/3.
Sailor 2 gives 1 coconut to the monkey, so there are 2(x-1)/3 - 1 = (2x-5)/3. He takes a third and leaves [(2x-5)/3](2/3) = (4x-10)/9.
Sailor 3 gives 1 to the monkey. so now (4x-10)/9 -1 = (4x-19)/9 are left. He takes a third, leaving behind [(4x-19)/9](2/3) = (8x-38)/27. This is the number of coconuts in the pile that the sailors found next morning. Let (8x-38)/27 = y, say, which gives 8x = 27y + 38.
Now what do we know about this number y? We are given that it is less than 10 AND it is a whole number divisible by 3 (because each sailor took a third and nothing was left). Therefore possible values for this number are 0, 3, 6 and 9. But we are also given that the initial pile had a whole number of coconuts. Hence, the problem reduces to finding that y from which 8x = 27y + 38, gives an x as a whole number.
You will see that the only y that gives x as a whole number is y = 6!!! Then 8x = 200, so x = 25, as obtained by Philllip B. Any comments?
Dattaprabhakar (Dr. G.)