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# The Sailors and the Coconuts

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?

Some generalizations!!! Intended for worthy graduate students?  Not meant for David!

(1) Let s be the number of sailors, m be the number given to tje monkey and r = p / q (a rational number, p > 0, q > 0, are whole numbers) is the fraction taken by each sailor.  In the end each sailor gets the same rational fraction of (and a whole number) of coconuts, with nothing left.

(2) Same as (1), but the i_th sailor gives m_sub_i to the monkey and takes r_sub_i, a known rational fraction.  At the end, however, what is left is a multiple of s.

(3) Same as (2) above, but the i_th sailor gives m_sub_i to the monkey with probability p_sub_i and takes r_sub_i, a known rational fraction with a probability u_sub_i.  What is the distribution of the number of coconuts left at the end?

Enough is enough!

Dattaprabhakar (Dr. G.)

### 2 Answers by Expert Tutors

Dattaprabhakar G. | Expert Tutor for Stat and Math at all levelsExpert Tutor for Stat and Math at all le...
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David:

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.)

Two small errors:  1. Not Phillip B., but Phillip R.  Sorry.  2. Just the above bold underscored line the word.... from ....  should be ...for....

Excuse me for the errors.

Dr. G.
Phillip R. | Top Notch Math and Science Tutoring from Brown Univ GradTop Notch Math and Science Tutoring from...
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25 coconuts were in the original pile.

The first sailor gave 1 to the monkey and took 1/3 of 24 = 8 which left 16.

The second sailor gave 1 to the monkey and took 1/3 of 15 = 5 which left 10.

The third sailor gave 1 to the monkey and took 1/3 of 9 = 3 which left 6

Then all three sailors took 2 more each.

In summary,
1st sailor got 10
2nd sailor got 7
3rd sailor got 5
monkey got 3

Total = 25