OK. Here's a good one?

Question

4 sailors spend a few days picking coconunts on a island to make a bit of money. On their last night, the 1'st sailor starts thinking, "I'm not sure I trust these guys. I'm going to take my 1/4 now." So he divides the pile of coconuts into 4 equal piles and has one coconut remaining which he throws into the bushes. He takes his pile and goes to his boat. During the rest of the night his 3 partners have the same thought and do the same thing. Each time dividing the remaining pile into 4 equal piles and having one left over which is thrown into the bushes.

Next morning the pile is much smaller, but nobody says a word as they're all equally 'guilty'. So they just divide the remaining pile into 4 piles and this time the pile divides exactly into 4 equal piles with no remainder.

Question: What is the *smallest* number of coconuts that could have been in the pile to begin with? And how do you find all of the numbers of coconuts that would work in this puzzle?

Doug

Nobody has it so far.
1109-1 = 1108
1108/4 = 277 so 4 pikes of 277 then
277*3 = 831 Then
831-1 = 830 and
830/5 = 207.5 Oops........

Answer 1

It is really quite simple, here it goes:

Let x be the number of coconuts that are in the pile to begin with

- The first sailor takes x the number of coconuts in the pile to begin with and he divides into 4 equal piles and discards 1 of the coconuts, this means that (x - 1) is perfectly divisible by 4 where we have subtracted 1 to indicate the coconut that was discarded, hence (x - 1)/4 is an exact whole number, to be more specific it is a positive integer (that is an element of the family of positive integers Z+ = {1,2,3,4,.....})

- So the amount that is remaining on the beach when the second sailor comes along and decides to do the same thing is = (x - 1)/4. And the second sailor also divides this amount of coconuts into four equal piles with one coconut remaining which he discards. So, [(x - 1)/4 - 1]/4 is a positive integer ( that is an element of Z+ = {1,2,3,4,....}), since the amount (minus the left over coconut) (x - 1)/4 - 1 is perfectly divisible by 4.

- Now the third sailor comes along and he sees the left over coconuts from above (that being: [(x - 1)/4 - 1]/4 ), and again he does the same thing hence, [[(x - 1)/4 - 1]/4 - 1]/4 is a positive integer, since the left over coconuts subtract 1 is perfectly divisible by 4 (because it was divided into 4 equal piles)

- The fourth sailor does the same thing as the 3 previous sailors and as a result we have [[[(x - 1)/4 - 1]/4 - 1]/4 - 1]/4, which is an element of the positive integers, since he took the pile that he saw discarded 1 coconut (hence we subtract 1 again) and divides into four equal piles.

- Now, the next morning the sailors divide the remaining pile into 4 equal piles but now the pile divides exactly into 4 piles with no remainder, hence we do not need to discard the coconut as we did before, we simply have the amount of coconuts remaining from above and divide that by 4.
Hence [[[[(x - 1)/4 - 1]/4 - 1]/4 - 1]/4]/4 = y, which is an element of the positive integers (Z+ = {1,2,3,4,.....}). Hence y is an element of the positive integers and represents the number of coconuts that are shared amongst the four sailors at the final dividing of the coconuts. so y = {1,2,3,4,....}, which means that each sailor at the last sharing will each receive 1 coconut or 2 coconuts etc...., depending on what x (the number of coconuts to begin with) is.

- So to determine what is the smallest number of coconuts that could have been in the pile to begin with, we just substitute into the equation above y = 1, since y is the y = 1 is the smallest vale that y can assume.

- First lets simplify the above equation
( [[[[(x - 1)/4 - 1]/4 - 1]/4 - 1]/4]/4 = y ) to get x as a function of y:

- x = 4(4(4( 16y + 1) + 1) + 1) + 1
= 4(4(64y + 5) + 1) + 1
= 4(256y + 21) + 1
= 1024y + 85

- Hence, we have x = 1024y + 85, where y is an element of the positive integers ( y = Z+ = {1,2,3,4,.....} ).
- So substituting y = 1 gives x = 1109 (the smallest number of coconuts in the pile to begin with)
- To find all of the numbers of coconuts that could have been in the pile simply substitute all possible values of y into the equation
hence we get: x = {1109, 2133, 3157, 4181,...., etc...}.

- I hope this does it!!!!!

Answer 2

I think u ve to use chinese remainder theorem (CRT)to solve this.
(let = mean congruent)
x=1 mod 4
3/4*(x-1) =1 mod 4
so on... and use CRT to solve