I have a word problem about squirrels that I can not solve!?

Question

Here it is: 3 squirrels gather a pile of nuts and decide to store the pile outside their house overnight. During the night, 1 of the squirrels wakes up hungry and decides to steal 1/3 of the pile but he has to eat one in order to be able to take exactly 1/3. He leaves 2/3 of the nuts behind and goes back to bed. Exactly the same thing happens to the 2nd squirrel, who wakes up, eats 1 of the nuts, steals exactly 1/3 of the pile that's left, and goes back to bed. Then precisely the same thing happens to the third squirrel. She eats 1 nut, hides away 1/3 of the remaining nuts, and goes back to bed. In the morning, the 3 squirrels awake and divvy up the pile which is not empty. It divides exactly into three.
What is the amount of nuts that was in the original pile?

Wow! Thank you all for your help! You are all incredibly smart!!

Answer 1

The primary answer is 25. The general answer is 25 + 81i, for integer values of i (1, 2, 3, ...).

[Edit: James Chan's equation for b should be

b = 2/3 {2/3 [2/3 (a - 1) - 1] - 1}

For b=6, you get a=25, but for b=12, a is not an integer.
End edit]

This is a Diophantine problem (google Diophantine), a simple variation of the famous "coconut problem" (google that too) involving seven shipwrecked sailors. Doing three squirrels is easier than seven sailors. Anyway, Diophantus was an ancient Greek who worked on problems requiring integer solutions.

A long time ago I solved the coconut problem by myself, although I had to do some research to discover the technique. I've forgotten what I did, but we'll try it again on the squirrel problem.

Let n be the original number of nuts. After the first squirrel left, the pile contains

2/3 (n-1) = 2/3 n - 2/3

After the second squirrel leaves, the pile contains

2/3 (2/3 n - 2/3 - 1) = (2/3)^2 n - (2/3)^2 - 2/3

After the third squirrel leaves, the pile contains

2/3 [(2/3)^2 n - (2/3)^2 - 2/3 - 1]
= (2/3)^3 n - (2/3)^3 - (2/3)^2 - 2/3
= (2/3)^3 n - [(2/3)^3 + (2/3)^2 + 2/3]

Now, in the last expression, the part inside the brackets is a geometric series. When n=7 (the coconut problem), you need to use the formula for the sum of a series, but here, we can just use the common denominator of 3^3 = 27.

The last expression can be rewritten as

8/27 n - 1/27 (8 + 12 + 18) = 8/27 n - 38/27

Okay so far. This last expression, 8/27 n - 38/27, is the number of nuts left in the morning, and it is a number exactly divisible by 3, so we can write

8/27 n - 38/27 = 3m

where both m and n are integers. Multiplying both sides by 27, we have

8n - 38 = 81m
8n - 81m = 38

The tricky part is to get two integers, m and n, that satisfy the above equation. From here on, I'm not so sure of myself. 8n is an even number, and 38 is an even number, so m must be even. Let m = 2k, so

8n - 162k = 38
4n - 81k = 19

Here, 4n is even and 19 is odd, so k must be odd. Let k = 2j+1. Then

4n - 81(2j+1) = 19
4n - 162j - 81 = 19
4n - 162j = 100
2n - 81j = 50

2n and 50 are both even, so j must be even. Let j = 2i. Then

2n - 162i = 50
n - 81i = 25
n = 81i + 25

The smallest number that satisfies this is i=0. In that case, n=25. (Primary answer.) Other values for n include 81+25 = 106, 106 + 81 = 187, etc.

Answer 2

according to the question, the final number of nuts is divisible by 3 and also 3/2 (it is 2/3 of the previous number is the question) , so it must be the multiple of 6 , for example 6,12,18.....
we have theses, let a be the initial number of nuts
a - 1 is divisible by 3
2/3(a - 1) - 1 is divisible by 3
2/3[2/3(a - 1) - 1] is divisible by 3
2/3 { 2/3 [ 2/3 (a - 1 ) - 1 ] }is divisible by 3
b = 2 / 3 { 2/3 [2/3 (a - 1) -1} is a multiple of 6, so with each multiple of six, we will find an "a" , that is the result :
for example with b = 6, we have a = 25, plug a into the statements above, we have the result is 25 nuts

Answer 3

I suspect there is more than one answer, but assuming the smallest number that would work...

I would start at the end and work backward. the ending pile has to be a multiple of 3, and since it was 2/3 of the previous set, it also has to be a multiple of 2...