Clever Trader

1)Initially the trader pays at 1st checkpoint. His sacks left with 29 each. As he was smart, he transferred 2 coconuts from the 1st sack to the other two, keeping others at 30. 2)This will continue up to 10 checkpoints. After that his 1 sack will contain 0 coconuts & other two contain 30 each. 3)Now he has to pay only from 2 sacks (as the third one is empty now). He gives 2 coconuts, left with 29 in each two. Transferred one coconut to the second one, keeping 30 in one. This will continue upto next 15 checkpoints. After that, he will left with 30 coconuts in 1 sack and other 2 sacks are empty. Now for the last 5 checkpoints, he will give one coconut. 4)And finally left with 25 coconuts.

In order to give away the least possible coconuts you must have the minimum possible number of bags with coconuts in.

You must start with 90 coconuts (30 in each bag). In the first 10 checkpoints you will have given away 30 coconuts. So you can then have two bags with 30 each in.

For the next 15 checkpoints, you can have 2 bags - so lose 2 coconuts at each checkpoint. Then all your remaining coconuts can go into one bag.

With the final five checkpoints (25-30) you will have only one bag - losing 5 coconuts from the 30 you had at checkpoint 25.

This leaves you with 25 coconuts at the end.

First 10 checkpoint he will give 10 from each sack, then in each sack he will be left with 20, 20X3 sack=60, so he will make 2 sack now. Again in next 15 checkpoint he will give 15 from each sack, then in each sack he will be left with 15 in each sack. 15X2 sack=30, so he will now make 1 sack. In next 5 checkpoint he will give 5 so he will be left with 25 coconuts.

He loses three nuts per post, After crossing the tenth post throw 1 sack and put all the 60 coconuts in two sacks, now he loses 2 nuts per post, after crossing 25th post throw another sack and place all left over 30 nuts in one sack and pass throw remaining 5 posts giving one nut at each post thus at most 25 nuts could be taken over finally

The plan is just to minimize the coconut loss by making as many baskets empty as possible. After 10 checkpoints the count in each basket is 20. So, the trader can divide the 20 coconuts in one basket among the other 2 baskets keeping 10 in each. Thus now the trader has 2 baskets having 30 coco. and 1 basket is empty. So, now the trader will lose only 2 coconuts instead of 3 at each checkpoint. After crossing another 15 checkpoints ,he will have 15 coconuts in each of the 2 baskets. So he will take 15 from one and put it in the other basket. So, now 1 basket has 30 coco. and the other 2 are empty. Now, the loss will be minimum with only 1 coconut losing at each checkpoint. So through the next 5 checkpoints he will lose 5 coconuts. So the maximum coconuts he can have= 30-5=25 (ANS)

The idea for this problem is to reduce the number of sacks the trader is carrying since he has to give a cocunut from each sack he is carrying. If the trader doesn't do anything, after 30 checkpoints, he will end up with 0 coconuts.

For checkpoints 1 - 10: The trader has to give 3 coconuts away, 1 from each bag. However, after each time he can take 2 coconuts from 1 bag to top up the other 2 bags back to 30. This means 1 bag gets reduced by 3 coconuts each checkpoint and after 10 checkpoints, that bag is empty.

*For checkpoints 11 - 25: * The trader has to give 2 coconuts away, 1 from each of the remaining bags. Once again, he can top up 1 bag by taking a coconut from the other bag. This means 1 bag gets reduced by 2 each checkpoint and after 15 checkpoints, the 2nd bag will be empty.

For checkpoints 26 - 30: * The trader only has to give 1 coconut away. Since he has been refilling the bag each time, that bag still holds 30 coconuts and he will end up giving a total of 5 away. This means by the end of the trip, he will have * 25 coconuts left over.

For the 1st 10 checkpoints he will give one coconut from eah sack, thus 20 coconuts will remain in each sack, he can now transfer those 60 coconuts into two sacks containing 30 coconuts each and leave the third sack, then he will follow the same procedure till 25th checkpoint,after which there will remain 30 coconuts,which he can now transfer to a single sack and after 5 more checkpoints,he will be left with 25 coconuts!! - subham

At the beginning he had 30 coconuts in each bag (we'll call them A, B, C)(A:30 means there are 30 coconuts in bag A). For the first 10 checkpionts, he'll give 30 coconuts, and will be left with A:20, B:20, C:20. Before entering the 11th checkpoint he'll put all of his coconuts from C into A and B, so that he's left with A:30, B:30, C:0.Because there are no coconuts in the C, he'll now give out 2 coconuts per checkpoint. After 15 checkpoints (after checkpoint No 25) he'll be left with A:15, B:15, C:0. Now he'll move all the coconuts into A. That will leave him with A:30, B:0, C:0. Now all he has to do is to go to the 30th checkpoint leaving only 1 coconut on each checkpoint. That will leave him with 25 coconuts in the end.

To save the maximum number of coconuts, the trader must not have the number of coconuts in the sacks equal after each checkpoint. Rather, he has to empty each sack as soon as possible so that later he has to give less than $3$ coconuts at consequent checkpoints as $1$ or $2$ of the sacks will be empty. This can be achieved by making sure that after giving coconuts at each checkpoint, the coconuts from a particular sack be used to replenish the other sacks so as to empty the particular sack quickly. We can show the coconut distribution at checkpoints from sacks $A,B,C$ as follows---

After $1$ st checkpoint, $A=29,B=29,C=29 \implies A=30,B=30,C=27$

After $2$ nd checkpoint, $A=29,B=29,C=26 \implies A=30,B=30,C=24$

If we continue like this upto after $10$ th checkpoint, we will see that sack $C$ can be fully emptied and coconuts are now only in $A,B$ as $A=30,B=30,C=0$ . So, from now on, at each checkpoint the trader will give $2$ coconuts, one from each coconut filled sack.

After $11$ th checkpoint, $A=29,B=29,C=0 \implies A=30,B=28,C=0$

After $12$ th checkpoint, $A=29,B=27,C=0 \implies A=30,B=26,C=0$

If we continue like this upto after $25$ th checkpoint, we will see that now sack $B$ and $C$ are fully emptied and coconuts are now left in only A as $A=30,B=0,C=0$ . So, from now on, at each checkpoint the trader will give only $1$ coconut from sack $A$ which is filled with coconuts.

After $26$ th checkpoint, $A=29,B=0,C=0$

After $27$ th checkpoint, $A=28,B=0,C=0$

$.....$

After $30$ th checkpoint, $A=25,B=0,C=0$

Thus, we can see that this is the most coconut saving method and this leaves us with $25$ coconuts in sack $A$ and no coconuts in sacks $B$ and $C$ .

So, total number of coconuts that are left in the end $=\boxed{25}$

Having given 1 coconut from each sack on very 1st checkpoint, the trader will fill 2 sacks by taking 2 coconut from the 3rd sack and filling one-one coconut in each of the first two sack. Repeating this pattern for first 10 checkpoints the trader will be left with 2 sacks only. From eleventh checkpoint the trader will give away 1 coconut from each of the 2 sacks and again take 1 coconut from 2nd sack and put it in 1st sack to make it full. Repeating this pattern again for next 15 checkpoints the trader will be left with only 1st sack from 16th checkpoint and will give 1 each on remaining 5 checkpoints keeping 25 with him.

The way is to at every checkpoint he gives 1 coconut from each sack and then fill the two sacks full. The way to do this is give one coconut from each sack and then from the first sack take 2 coconuts out and put one each in other two sacks. In this way in the first 10 checkpoints the first sack would be empty (taking 3 coconuts at each checkpoint) and he would have lost 30 coconuts out of 90. The other two sacks contain 30 coconuts each. Now from these two sacks give one coconut from each to next checkpoint and take one coconut from this 2nd sack and put it in the third one. in this way in next 15 checkpoints sack 2 will be empty and he would have passed 25 checkpoints. Also he is left with just 30 coconuts and 5 checkpoints. Give one coconut at each checkpoint and remaining coconuts are 25!!

He has 90 Coconuts in three sacs.

To first 10 check post he will give 3x10= 30 coconuts

Remaining 60 coconuts he will shift in two sacs. To next 15 post he will give 2x15= 30 coconuts

Remaining 30 Coconuts he will shift in one sac. To remaining post he will give 1x5= 05 coconuts

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Total Number of coconuts he give to 30 check post will be = 65 coconuts

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Coconuts he retained after 30 check post will be 90 minus 65 = 25 Coconuts

ow? As he has to give one coconut for every sack he carries, he will try to get rid of the sacks as early as possible. To get rid of the first sack: He should be able to fill coconuts from one sack to other two sacks, lets say he is able to do it after x checkpoints, to get x we can say

(space left in first sack)x + (space left in second sack)x = (remaining coconuts in third sack)30 -x; x = 10;

So after 1st checkpoints, he will be left with 2 sacks containing 30 coconuts each and he will throw the third sack. To get rid of the second sack: He should be able to fill coconuts from second sack to first sack, lets say he is able to do it after another y checkpoints, to get y we can say

(space left in first sack)y = (remaining coconuts in second sack)30 -y; y =15

Thus after 25th check point he will be left with one sack containing 30 coconuts, at this point he will throw away the second sack as well. Now we can easily say that after 30th checkpoint he will be left with 30 -5 = 25 coconuts.