Make $3$ piles of $120$ from $241$ coins Starting with $241$ coins with positive integer values, you can make $3$ piles so that each of them has value exactly $120$. Coins together have value of $360$. Prove this is true for all values of coins. 
I tried sorting the coins. Then picking three with largest values. Then somehow I need to fit all the remaining coins so that they fill the piles. How can this be done?
Minimum number of coins with value $1$ is $122$.
 A: HINT:
What is the minimum number of coins with value $1$? And what is therefore the maximum value of the sum of all coins with a value of 2 or more?
Answer
It is easy to see that there need to be at least $122$ coins with a value of $1$, and thus the total value of all the coins with a value of $2$ or more is at most $238$.
It is also easy to see that no individual coin can have a value of more than $120$
So, if there are at least two coins with a value between $60$ and $120$, then put each coin in a separate pile. Since the total value of the coins with a value of $2$ or more is at most $238$, and since the first two coins have a sum value of at least $120$, that means that the sum value of the rest of the coins with a value of $2$ or more is below $120$, and so they can all be put into a third pile, and now the three piles can be filled up with the coins of value $1$
If there is one coin with a value between $60$ and $120$, then put that one by itself, and now create a second pile of coins with a value of $2$ or more, adding at a time, until you either are out of those coins, or until that pile reaches a sum value of $60$ or more; as you are adding one coin at a time, it is impossible to go from a sum value from below $60$ to a sum value of above $120$, so if the sum value of the remaining coins with a value of $2$ or more is more than $60$, then you are guaranteed to get a second pile of such coins with a sum value of between $60$ and $120$. And, once you have that, you can once again put all the other coins with a value of $2$ or more into a third pile without going over $120$, by the same reasoning as before. And then you can fill up the three piles with the coins with a value of $1$
Finally, if there are no coins with a value of over $60$ at all, then by the above reasoning you can create two piles with coins with a value of $2$ or more, each with a sum value of between $60$ and $120$ (unless you can create only such pile, but then it is even more trivial to fill up the piles), and then once again you can put all the rest of those coins with a value of $2$ or more in a third pile, and then fill up the three piles with coins of value $1$
