# Math Help - five men, a monkey, and a pile of coconuts

1. ## five men, a monkey, and a pile of coconuts

On a desert island, five men and a monkey gather coconuts all day, then sleep.
The first man awakens and decides to take his share. He divides the coconuts into five equal shares with one left over. He gives the extra one the, hides his share and goes back to sleep. Later the second man awakens and takes his fifth from the remaining pike; he too finds one extra one and gives it to the monkey. Each of the remaining three men does likewise in turn. Find the minimum number of coconuts originally present.

I tried starting off with $\frac{x}{5} \cdot \frac{5}{1}$

Would I do this but adding 1 each time?

2. would this work?

$(\frac{5}{1}(\frac{5}{1}(\frac{5}{1}(\frac{5}{1}(\ frac{x-1}{5})-1)-1)-1)-1)$

dividing it by 1/5 each time and minusing 1?

3. Originally Posted by Mukilab
On a desert island, five men and a monkey gather coconuts all day, then sleep.
The first man awakens and decides to take his share. He divides the coconuts into five equal shares with one left over. He gives the extra one the, hides his share and goes back to sleep. Later the second man awakens and takes his fifth from the remaining pike; he too finds one extra one and gives it to the monkey. Each of the remaining three men does likewise in turn. Find the minimum number of coconuts originally present.

I tried starting off with $\frac{x}{5} \cdot \frac{5}{1}$

Would I do this but adding 1 each time?
The Ultimate Puzzle Site - Brain-Teasers

4. Originally Posted by mr fantastic

I had the wrong numerator :/

thanks

5. er, at first I thought this was right

It's a different question

I tried using the same technique on my question and it didn't work

6. Using the fact that the difference between the amount of coconuts after each step must be an integer, and the difference is multiplied by 4/5 after each step, the first difference must be divisible by 5^4 so that the last difference is an integer. The smallest such difference is 625, which gives you an initial coconut count of 3121.

7. Originally Posted by icemanfan
Using the fact that the difference between the amount of coconuts after each step must be an integer, and the difference is multiplied by 4/5 after each step, the first difference must be divisible by 5^4 so that the last difference is an integer. The smallest such difference is 625, which gives you an initial coconut count of 3121.
that is what I got but it doesn't work if you do the calculation.

Also, your notes are very hard to follow

8. five men, a monkey, and a pile of coconuts
wait ...

I thought it was Churchill, Stalin, Roosevelt, and a monkey in a bar?

10. Hello, Mukilab!

This is a classic problem: "The Monkey and the Coconuts".
It has a long (but very elementary) solution.

On a desert island, five men gather coconuts all day, then sleep.

The first man awakens and decides to take his share.
He divides the coconuts into five equal shares with one left over.
He gives the extra one to the monkey, hides his share,
puts the remaining coconuts in a pile, and goes back to sleep.

Later the second man awakens and takes his fifth from the remaining pile.
He too finds one extra one and gives it to the monkey.

Each of the remaining three men does likewise in turn.

Find the minimum number of coconuts originally present.
The original problem concluded with:
. . In the morning, the five men innocently divided the remaining coconuts.
. . This time the division came out even; no coconut for the monkey,.

Let $N$ = the original number of coconuts.

The first man divided them into 5 equal piles with one left over: . $N \:=\:5A + 1 \;\;[1]$
He hid his $A$ coconuts and put the $4A$ coconuts in a pile.

The second man divided them into five piles with one left over: . $4A \:=\:5B + 1\;\;[2]$
He hid his $B$ coconuts and put the $4B$ coconuts in a pile.

The third man divided them into five piles with one left over: . $4B \:=\:5C + 1\;\;[3]$
He hid his $C$ coconuts and put the $4C$ coconuts in a pile.

The fourth man divided them into five piles with one left over: . $4C \:=\:5D + 1\;\;[4]$
He hid his $D$ coconuts and put the $4D$ coconuts in a pile.

The fifth man divided them into five piles with one left over: . $4D \:=\:5E + 1\;\;[5]$
He hid his $E$ coconuts and put the $4E$ coconuts in a pile.

In the morning, they divided them into five equal piles: . $4E \:=\:5F\;\;[6]$

From [6]: . $E \:=\:\frac{5F}{4}$

Since $E$ is an integer, $F$ is a multiple of 4: . $F \:=\:4p$

Then we have: . $E \:=\:\frac{5(4p)}{4} \:=\:5p$

Substitute into [5]: . $4D \:=\:5(5p)+1 \:=\:25p + 1$

We have: . $D \:=\:\frac{25p+1}{4} \:=\:6p + \frac{p+1}{4}$

Since $D$ is an integer, $p+1$ must be a multiple of 4: . $p+1\:=\:4q \quad\Rightarrow\quad p \:=\:4q-1$

Then we have: . $D \:=\:6(4q-1) + \frac{(4q-1)+1}{4} \quad\Rightarrow\quad D\;=\;25q-6$

Substitute into [4]: . $4C \:=\:5(25q-6) + 1 \:=\:125q - 29$

Then: . $C \:=\:\frac{125q-29}{4} \:=\:31q - 7 + \frac{q-1}{4}$

Since $C$ is an integer, $q-1$ must be a multiple of 4: . $q-1 \:=\:4r \quad\Rightarrow\quad q \:=\:4r+1$

Then we have: . $C\:=\:31(4r+1)-6 + \frac{(4r+1)-1}{4} \quad\Rightarrow\quad C \:=\:125r + 24$

Substitute into [3]: . $4B \:=\:5(125r + 24) + 1 \:=\:625r + 121$

Then: . $B \:=\:\frac{625r + 121}{4} \:=\:156r + 30 + \frac{r+1}{4}$

Since $B$ is an integer, $r+1$ must be a multiple of 4: . $r+1\:=\:4s \quad\Rightarrow\quad r \:=\:4s-1$

Then we have: . $B \:=\:156(4s-1) + 30 + \frac{(4s-1)+1}{4} \quad\Rightarrow\quad B \:=\:625s - 126$

Substitute into [2]: . $4A \:=\:5(625s - 126) + 1 \:=\:3125s - 629$

Then: . $A \:=\:\frac{2125s-629}{4} \:=\:781s - 157 + \frac{s-1}{4}$

Since $A$ is an integer, $s-1$ is a multiple of 4: . $s-1 \:=\:4t \quad\Rightarrow\quad s \:=\:4t+1$

Then we have: . $A \:=\:781(4t+1) - 157 + \frac{(4t+1)-1}{4} \quad\Rightarrow\quad A \:=\:3125 + 624$

Substitute into [1]: . $N \:=\:5(2125t+ 624) + 1$

. . . . and we have: . $N \:=\:15625t + 3121$

The least $N$ occurs when $t = 0$

. . Therefore: . $N \:=\:3121$