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

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