$4$ piles of stones in to $5$ piles

Question

I wanted to ask for verification of "proof" of the following. We are given $4$ piles of stones and we form $5$ piles of stones from those. The question asks to prove that there are at least $2$ stones in a smaller pile

My attempt: Since we are forming more piles , this suggests that there is a pile with less amount of stones , since otherwise we would have more stones than what we started with. Thus there is a pile with at least $1$ stone . But this stone came from another pile , so there is a decrease there too. It's possible when we formed the new piles , we tried to add and compensate for the decreases pile , but in any case one of the piles has less stones , since if they had the same or more stones there would be more stones which is not true. The proof is quite formal in the book , that's why I asked here. Thanks

Second attempt: Say the pile containing not less than any of the other 3 piles has f stones . Now there must a pile in the 5 piles that has less than f stones .since if there is not there is at least 5f stones which is contradictory.Now exclude this pile containing less than f stones from the system and we have 4 piles now . Now this new 4 piles must have in total less than the original set since excluded pile has at least 1 stone. Now there must be a pile that has less than one of the previous piles , sinc if each one has at least the same number of stones , then those 4 piles would have more stones than they should have. Combining those two piles , they are the ones q was looking for.(i hope)

Answer 1

Your solution has a gap in the beginning. Suppose the four piles are 1,5,5,5, and we put them into five piles 2,2,4,4,4, which pile has "less stone"?

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The solution is as follows.

When the stones are in the 4 piles, write a fraction on each of them: For each stone in a pile of size $n$, write $\frac1n$. Note that all the written numbers sum to 4. Now form 5 piles. Write a new fraction on each of the stone the same way. Note that all the newly written numbers now sum to 5. A bunch of numbers previously summed to 4, now they summed to 5. Can you see why two of them must have grown bigger?

Usually, doing a "microscopic" analysis like you did is very hard and easy to screw up. The best way is to do it "macroscopically".

Answer 2

Suppose you have $4$ piles of stones with $n$ stones in each and now you form $5$ piles from them. Then, the newer of the $5$ piles cannot have all $\ge n$ stones because then we would have a total $5n$ stones which is $>4n$. So, now one of the piles must have $1, 2, ... n-1$ stones. Clearly, this pile has $\ge2$ and $\le n$ stones if we exclude the $n=1$ case. If this pile has $n=1$ stone then, some other pile must have a number of stones $\le n$ (since otherwise, you would have $>4n$ stones) and that smaller pile has $<n$ and $>1$ stones (so atleast two condition is satidfied).