A crew needs to divide oranges [duplicate]

Question

A boat sinks in the ocean and the 5 survivors each jump into a different lifeboat. They agree to meet on the deserted island pointed out in the distance. The next morning, the first castaway walking along the beach finds a huge pile of oranges. He decides to divide them into 5 equal parts. After the division, there is one left, which he throws into the sea. He then leaves to explore the island with his share of the oranges.

The second castaway arrives at the pile of oranges, and also thinking he is the first to arrive, he decides to divide it into 5 equal parts. After the division, there is one left, which he throws into the sea, then he leaves to explore the island with his share.

same thing with the 3d,4th and 5th.

a) What is the minimum number of oranges that was on the beach initially?

I'm really strugling to find a path to the response wich is 3121, I'm trying to use mods with multiple equations but chinese rest theorem does not apply here.

Answer 1

I think I have seen a similar problem elsewhere involving a monkey and coconuts, but the final answer was quite different as the food was divided a total of six times, not five. I think the solution method works though so here is my version of that solution:

Imagine, if you can, that we start with a pile of $-4$ oranges.

Now, since the orange thrown into the sea can happen at the start or the end, do it now, so we have a pile of $-5$ oranges.

Take your $\frac{1}{5}$ share which is $-1$ oranges and put the remaining $-4$ back and the original problem is unchanged.

So, if we are allowed a negative answer, $-4$ works.

Now, since the oranges are divided into 5 piles a total of 5 times, we can add $5^5=3125$ to any solution to get another valid solution.

This means the smallest valid solution is $-4+3125=3121$.

Answer 2

This is an old problem, which I remember in a form that uses coconuts and a monkey.

To solve the problem with positive oranges, let $A$ be the number of oranges taken by explorer 1, $B$ be the number taken by explorer 2, and so on to $E$ for explorer 5. Then we have a Diponantine syatem of equations:

$4A=5B+1$

$4B=5C+1$

$4C=5D+1$

$4D=5E+1$

Now define successors $A'=A+1,B'=B+1,$ etc, and render the equations in terms of these successors:

$4A'=5B'$

$4B'=5C'$

$4C'=5D'$

$4D'=5E'$

So $A'$ must be a multiple of $5^4=625$, which sets the minimum positive value as $A=A'-1$ as $624$, and then the minimum size of the initial pile must be $5A+1=3121$.

Answer 3

Here is my solution:

Let T be the total number of oranges that the first person finds, so:

a) T = 1 + 5k1 (k1 being the number of oranges in each of the 5 piles after throwing away one orange) Since person 1 takes one of the 5 piles for himself we're left with:

b) 4k1 = 1 + 5k2, with the same logic as above, we have

c) 4k2 = 1 + 5k3

d) 4k3 = 1 + 5k4

e) 4k4 = 1 + 5k5

by adding 4 to both sides of the equations we can simplify each equation and make the sums a bit easier:

a) T + 4 = 5(1 + k1)

b) 4(k1 + 1) = 5(1 + k2)

c) 4(k2 + 1) = 5(1 + k3)

d) 4(k3 + 1) = 5(1 + k4)

e) 4(k4 + 1) = 5(1 + k5)

now starting with equation e and solving for k4 + 1, and then substituting the answer in equation d and so on, you reach the final equation, as follows: T + 4 = (5^5(1+k5)/4^4) now since we know that T is the total number of oranges that the first person had found, we know that this number has to be a positive integer, Since 5^5/4^4 are relative primes 1+K5 has to be a multiple of 4^4, for example, 4^4n. now substituting 1+k5 for 4^4n we have: T = 5^5n - 4 => T = 3125n - 4 and the smallest amount for n being at n = 1, therefore: n=1, T = 3121