I need help solving this problem

Question

3 monkeys lived on an island and hunted for food together. They put all of the coconuts they found in a large pile and went to sleep.

During the night, the 1st monkey woke up hungry. He ate 1 coconut, then he divided the rest of the coconuts into 3 equal piles. He pushed 2 piles back together, then took the 3rd pile off into the forest to eat some more before returning to bed.

After a short time, the 2nd monkey woke up hungry and went to the pile of coconuts. He ate 1 coconut and then he divided the rest of the coconuts into 3 equal piles. He pushed 2 piles back together, then took the 3rd pile off to the beach to eat some more before returning to bed.

Soon, the 3rd monkey woke up, went to the pile of coconuts and ate 1 coconut because he was hungry. He divided the rest of the coconuts into 3 equal piles. He pushed 2 of the piles together then took the 3rd pile back to his favorite tree to eat some more before returning to bed.

In the morning, the 3 monkeys went to the pile of coconuts and were surprised to only see 6 coconuts left. “I know there were more than this yesterday,” said the 1st monkey. The other 2 monkeys nodded in agreement. “I only took a couple last night,” the 1st monkey said. “I only took a couple last night,” the 2nd monkey echoed. “Me too!” said the 3rd monkey.

How many coconuts were in the pile when the monkeys went to sleep?
How many coconuts did the 1st monkey take?

The 2nd monkey?
The 3rd monkey?
Why couldn’t the 3 monkeys be fair and nice to each other even if they wanted to?

Answer 1

This is a fun problem to work on in groups as there are several approaches that can lead to the equation used in solving such problems. (Search Diophantine equations to learn more.)

 

And not to ruin all your fun, how about I just get you started and let you try to finish the problem on your own. If you are still struggling to finish, post an update and I'll come back and give you another clue.

 

Start with x = # of coconuts they started with and y = # of coconuts they had left in the morning (which for this problem is 6). 

 

We'll use m₁ for the 1st monkey and m₂ for the 2nd monkey.

 

 

MONKEY 1

 

So m₁ = # coconuts in the 1st monkeys pile after he eats 1 and divides the rest into 3 equal piles. 

   As x = original number of coconuts, when he eats 1 he is left with x - 1 coconuts. 

   Then he divides the x - 1 coconuts into 3 equal piles or into thirds. Thus m₁ = 1/3(x - 1) coconuts.

 

   So the 1st monkey takes 1/3(x - 1) coconuts, therefore leaving 2/3(x -1) for the 2nd monkey. 

 

Are we good so far?

 

MONKEY 2

 

So m₂ = # coconuts in the 2nd monkeys pile after he eats 1 of what is left after the 1st monkey, then divides the rest into 3 equal piles. Remember the 2nd monkey is starting with 2/3(x -1) coconuts.

 

  Starting with 2/3(x -1), he eats 1 coconut and is left with ________ - 1 coconuts.

  Then he divides the _________ - 1 coconuts into 3 equal piles or into ______ . Thus m₂ = 1/3(__________) coconuts.

 

  So the 2nd monkey takes ___________coconuts, therefore leaving ____________ for the 3rd monkey.

 

 

Monkey 3 and Original # of Coconuts

 

Now do the same for monkey 3. What is the equation for the 3rd monkey? Now that you have the equation for the 3rd monkey, how can you solve for the original number of coconuts? 

 

And what if there were 5 monkeys or 10 monkeys? Would they be any more fair to each other?! :)

Answer 2

Hi Tyshea,

 

Lets follow through step by step to find this answer

 

let x = number of coconuts in the pile at the start

 

 

then the first monkey took 1  

 

so you have x-1 

 

now he splits the pike in 3 and takes 1/3 of the pile so that means he left 2/3

 

 

so we multiply by 2/3 and get 2x/3 - 2/3  (this is the number left after the first monkey

 

now the second monkey 

 

he takes one coconumt

 

2x/3 - 2/3 -1  = 2x/3 - 5/3

 

now he splits the pile in 3 and takes  1/3 so he leaves 2/3

 

2/3(2x/3-5/3)  = 4x/9 -10/9

 

now the third monkey

 

he takes one coconut

 

4x/9 - 10/9 -1 = 4x/9 - 19/9

 

now he splits the pile in 3 and takes 1/3 so he leaves 2/3

 

2/3(4x/9 - 19/9)  = 8x/27 - 38/27

 

and this is the final number and we know this equals 6  so now we can solve for x

 

8x/27 - 38/27 = 6

 

Multiply both sides by 27

 

8x - 38 = 162

 

8x = 200

 

x = 25

 

the first monkey eat one and then took 1/3 of 24  so in total he took 9 (if you count the one he eat

 

At that point there were 16 left

 

the second monkey eat one then took 1/3 of 15 so in total he took 6 (if you count the one he eat

 

at that point there were 10 left

 

the third monkey eat one then took 1/3 of 9  so in total he took 4 (if you count the one he eat

 

The monkeys could not be fair at the beginning because the total number of coconuts was not divisible by 3 and also they were monkeys and loved monkey business.

 

Hope this helps