An interesting problem, which has multiple solutions.
Suppose you are given $2n+1$ coins $c_1, c_2, \dots, c_{2n+1}$, of positive real weights $w_1, w_2, \dots, w_{2n+1}$.
These coins have the property that if you remove any coin, then the remaining $2n$ coins can be split into two groups of exactly $n$ coins each, such that the sum of weights of coins in one group, is same as the sum of weights of coins in the other.
Show that $w_i = w_j$ for any $i, j$.
i.e. all the coins must be of equal weight.
[Solution]
Suppose you are given $2n+1$ coins $c_1, c_2, \dots, c_{2n+1}$, of positive real weights $w_1, w_2, \dots, w_{2n+1}$.
These coins have the property that if you remove any coin, then the remaining $2n$ coins can be split into two groups of exactly $n$ coins each, such that the sum of weights of coins in one group, is same as the sum of weights of coins in the other.
Show that $w_i = w_j$ for any $i, j$.
i.e. all the coins must be of equal weight.
[Solution]
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