Ruby Pile Split

If your pile were greater than 10, then the other two piles together would be less than 20, and at least one of those piles would be less than 10. So the scoundrel would have chosen your pile instead, and by contradiction your pile cannot be greater than 10. So the maximum number of rubies you can keep is 10 (3 piles of 10 each).

Your maximum is $\boxed{\text{10 rubies}}$ which you can guarantee by forming three equal piles of ten rubies.

Why can't you do better than this? Because if your piles are unequal in size the scoundrel will leave you with the smallest one!

And of course with unequal piles the smallest pile must have less than ten rubies. Because if it had ten or more rubies then the three piles together would contain more than thirty rubies, which is impossible.

Case 1 : Suppose the piles have

10+a , 10+b and 10-a-b rubies,

Then obviously scoundrel will choose 1st two piles and you will be left with less than 10 rubies.

Case 2: piles have 10 , 10+a and 10-a rubies, Scoundrel will again choose first two piles and again you will get less than 10 rubies.

Case 3: piles have 10 -a, 10-b and 10+a+b rubies such that b > a. Then scoundrel chooses 10-a and 10+a+b piles and you will again have less than 10 coins

CASE 4: piles have 10 , 10 , 10 rubies. Here,scoundrel can have only 20 rubies and you will get 10 rubies.

Thus , you will get the maximum number of rubies in case 4 and that too 10 rubies

Let's say that you split them into the three piles. The first pile, let's call it Pile A, would have x rubies. The second pile, Pile B, would have y rubies. And the final pile, Pile C, would have whatever is left: in this case, 30-(x+y) rubies.

Because the scoundrel would always pick the biggest two piles, you need to maximise the minimum value of the three piles. However, the highest the smallest pile could be is 10

Why is that? Well, let's assume that the smallest pile - pile A, for example - is 11 rubies. Due to this, Pile B would have y rubies, and Pile C would have 19-y rubies. However, no matter what value of Y you use, either one or both of the other piles would be smaller than Pile A, therefore disproving that the smallest pile is 11 rubies.

However, if we assume that pile A has 10 rubies, then we can give Pile B 10 as well, leaving 10 for Pile C. As all three piles are the same, the scoundrel could pick any two of them, leaving the remaining one for you - and regardless of what he picked, the leftover one would have 10 rubies.

Hmm..and we assume that the scoundrel have the opportunity to choose his piles consecutively? The question is not clear on that matter...and if not, the answer is not 10 but 14 (15+14+1)...:)

We want to maximize the number of rubies we get, so if the scoundrel will choose the piles with the most rubies then we will get 30- the largest 2 piles. The largest 2 piles will have an average more than the smallest pile because they are both greater than the smallest pile. If we have piles with 10, 10 and 10 rubies, then we will not have any largest pile, so the average of the 2 "largest piles" will equal the smallest pile. If there are piles with more than or less than 10, then we will get 30-2* (the largest 2 piles)= 30-4 (the average of the largest 2 piles). Since the average of the largest 2 piles is more than the smallest pile, 30-4 (the average of the largest 2 piles)<30-2 (the smallest pile). In a pile with 10, 10 and 10 rubies, 30-2 (the average of the 2 largest piles)=30-2 (the smallest pile), because the average of the 2 largest piles is equal to the smallest pile. Since these 2 quantities are equal, this implies that (your quantity of rubies)=30-2 (the average of the largest 2 piles for piles of only 10), and (your quantity of rubies)<30-2*(the average of the largest 2 piles for piles of different quantities). This implies that 30-(The average of the largest 2 piles for piles of only 10)>30-2(the average of the largest 2 piles for piles of different quantities), which implies your quantity of rubies for piles of only 10>your quantity of rubies for piles of different quantities, which implies you will maximize your quantity of rubies if you have piles of only 10. So you will have a maximum amount of 30-10-10= 10 .

Any division other than 10, 10, 10 would result in two piles being greater than 20 and your pile being less than 10. So only an even 10, 10, 10 split results in your maximum of 10.

If the scoundrel is clever and discerns a difference in pile sizes, you will be left w. the smallest pile. To guarantee that you are left w. the largest pile stack 3 piles of 10 rubies, each being equal to the largest, and let the scoundrel choose any 2 piles. You are left to choose the largest pile possible.C.H

The scoundrel will get 2 piles which of them has 10 rubies because 30÷3=10 so you will keep the one of them because he will take 2 and you will have 10 rubies .

A slightly more interesting problem might be: A king has challenged you to a contest against a scoundrel.

You have to divide 30 rubies into 3 piles. (You can choose how many rubies go in each pile.) The scoundrel and you will then each choose a pile for the scoundrel to keep.

Assuming the scoundrel is greedy and wants as many rubies as possible, what is the maximum number of rubies that you can keep?

The greedy scoundrel wants to keep maximum roubie contained piles. If I keep more than ten roubies in a pile the greedy one will capture mine because of the rest two at least one will contain roubies less than ten. So I must keep 10 roubies to get maximum achievement.

Having a pile of any number other than 10 would make piles of different sizes. The scoundrel will obviously choose the biggest and next to it in size ( bigger the pile , more the rubies) leaving you with the smallest pile and less number of rubies. So in order to maximize the number of rubies left when two piles are to be given away to the scoundrel, the best strategy is to have them equally distributed in three piles, that is 10 in this case

It's an easy question. Divide it by 3.

You have to spilt the piles into 3 equal piles of 10 rubies each because if there is variation in the piles, the scoundrel will take the 2 largest ones and give you the smallest pile, and since there is variation the smallest pile has to have less than 10 rubies so the most you can get is 10 .

I actually thought that this was a trick question, so I was thinking about it for a really long time...

30/3=10 and if he takes at least 2 then 1 like would be lower than 10

Let there be there piles having x, y, z rubies.

Let x>= y >= z

Now the greedy scoundrel will Pick up piles having x and y rubies . So we have z rubies left with us. Also x + y + z = 30 We need to find maximum value of z

x, y >=z, so x + y + z >= 3z

=> 30>= 3z

=> z <= 10

Thus maximum value of z is 10

Thus is possible when you divide rubies equally

We need to simply maximize the $min(a,b,c)$ which could be obtain if $a=b=c=10$

Just divide by 3

so stupid.. too obvious..