orl wrote: There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that if any coin with weight $ w_k$, $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists. All n is $ n\equiv 1 (\mod 2)$ and $ n > 5$ The first we will prove that n must be an odd number . Exist a set such that $ \sum_{i = 1}^n w_i$ which is smallest. From condition we have : $ S_n\equiv w_i,\forall i = 1,..,n$ where $ S_n$ is sum of all $ w_i$ Therefore all element of set if parity . If all of them is even then the set $ \{\frac {w_i}{2}\}$ also satisfy condition ,this is contradiction from the chose $ S_n$ is minimum . It gives that all elements of set must be odd number . Now suppose n is even then when we remove from the set a $ w_i$ then exactly one of two set after split has an odd numbers of element ,it gives that sum of all element is an odd number , but the other set has sum of all element is a even . Therefore thay can't equal . So n must be an odd number . The case $ n = 3,5$ it is easy to prove there is no set satisfy . We will prove for all odd number great than 5 by induction . Let : $ f = 1*3*....*(2n - 1)$ ( n is odd number) where * can take $ + , -$ The case n=7 is obvious by some carculation . Suppose it true for $ 2n + 1$ we will prove for $ 2n + 3$ Easy to check that when we make a transform : chose two odd consecutive and change them sign then we have $ |f' - f| = 2$ If we move a term from the set ${ \{1,...,2n - 1}$ then use the result $ |f' - f| = 2$ we can chose sign for $ 2n + 1,2n + 3$ for which $ f = 0$ (not contain the object which had been moved ) . Take all terms which have sign + in a set ,the others in a set . Two sets have equal sum. If we move $ 2n + 3$ or $ 2n + 1$ ,in two case it is quite easy to find the ways by consider two case $ n\equiv 1 (\mod 4)$ and $ n\equiv 3(\mod 4)$ So all all n odd and $ n > 5$ satisfy condition .

i think there no N to satisfy; (**)it is clear that all Wi shoud have same parity! consider W1<W2...<Wn,then if this set satisfy the condition then set of Wi-k or k.Wi do well,hence set of Wi-W1 can do that. let Wi-W1=2^Mi.Xi ,(Xi is odd). let MIN(Mi)=p ,then set of Ri=(Wi-W1)/2^p can satisfy the condition,but we know that one of the Ri is even(R1=(W1-W1)/2^p=0) and other Rj are odd BUT stand on (**)they shoud have same parity!unless Wi-W1=0,but it is appose to our data!! is it true?

honav wrote: i think there no N to satisfy; (**)it is clear that all Wi shoud have same parity! consider W1<W2...<Wn,then if this set satisfy the condition then set of Wi-k or k.Wi do well,hence set of Wi-W1 can do that. let Wi-W1=2^Mi.Xi ,(Xi is odd). let MIN(Mi)=p ,then set of Ri=(Wi-W1)/2^p can satisfy the condition,but we know that one of the Ri is even(R1=(W1-W1)/2^p=0) and other Rj are odd BUT stand on (**)they shoud have same parity!unless Wi-W1=0,but it is appose to our data!! is it true? Try to first fully read the conditions! orl wrote: The number of coins in the two groups can be different. Then carefully read the proposed solutions (as you are new to this forum, the number of stars of the poster is most often a good indication on the good quality of his work - shurely TTsphn is one of the most authoritative users, so chances he is wrong - and that noone pointed yet that till now - are very slim). What you wrote is only true partially. Indeed, the set $kW_i$ (homothety) works, indifferently of the splitting into groups; however, the set $W_i - k$ (translation) works only if the splitting is always made into groups of same cardinality. Indeed, with this extra constraint, the result is all weights must be equal (and it may even be extended to the case when the weights are real numbers, not necessarily integer!).

TTsphn wrote: Let : $ f = 1*3*....*(2n - 1)$ ( n is odd number) where * can take $ + , -$ Easy to check that when we make a transform : chose two odd consecutive and change them sign then we have $ |f' - f| = 2$ I think chose two odd consecutive and change them sign then we have $ |f' - f| = 4$. Since $(2n+3)+(2n+7)=(2n+1)+(2n+5)+4$, suppose it true for odd number $n$, we can prove for $n+2$.

We will prove more general result: For any odd $n \geqslant 7$, there exit a partition $\left\{ {1,3,...2n - 1} \right\} = A \cup B$, $\sum\limits_{x \in A} x = \sum\limits_{x \in B} x $, and $\left\{ {1,3,...2n - 1} \right\} = C \cup D$, $\sum\limits_{x \in C} x =2+ \sum\limits_{x \in D} x $. A)For $n = 7$: if cut $1$, then $13 + 11 = 3 + 5 + 7 + 9$; and $\left( {5 + 9 + 11} \right) - \left( {3 + 7 + 13} \right) = 2$. if cut $3$, then $13 + 9 + 1 = 5 + 7 + 11$; and $\left( {11 + 13} \right) - \left( {1 + 5 + 7 + 9} \right) = 2$; if cut $5$, then $13 + 9 = 1 + 3 + 7 + 11$; and $\left( {3 + 7 + 13} \right) - \left( {1 + 9 + 11} \right) = 2$; if cut $7$, then $13 + 3 + 5 = 1 + 9 + 11$; and $\left( {9 + 13} \right) - \left( {1 + 3 + 5 + 11} \right) = 2$; if cut $9$, then $13 + 7 = 1 + 3 + 5 + 11$; and $\left( {13 + 7 + 1} \right) - \left( {3 + 5 + 11} \right) = 2$; if cut $11$, then $13 + 5 + 1 = 3 + 7 + 9$; and $\left( {1 + 3 + 7 + 9} \right) - \left( {13 + 5} \right) = 2$; if cut $13$, then $11 + 7 = 1 + 3 + 5 + 9$; and $\left( {11 + 7 + 1} \right) - \left( {3 + 5 + 9} \right) = 2$. B)If the result is true for a odd $n \geqslant 7$, then for the $n + 2$, we plus $2n + 1,2n + 3$. If cut a $1\leqslant {w_i} \leqslant 2n - 1$, then there exit a partition $\left\{ {1,3,...2n - 1} \right\} = A \cup B$, $\sum\limits_{x \in A} x = \sum\limits_{x \in B} x $, and $\left\{ {1,3,...2n - 1} \right\} = C \cup D$, $\sum\limits_{x \in C} x =2+ \sum\limits_{x \in D} x $. Let $A' = D \cup \left\{ {2n + 3} \right\}$, $B' = C \cup \left\{ {2n + 1} \right\}$, then $\sum\limits_{x \in A'} x = \sum\limits_{x \in B'} x $. Let $C' = A \cup \left\{ {2n + 3} \right\}$, $D' = B \cup \left\{ {2n + 1} \right\}$, then $\sum\limits_{x \in C'} x - \sum\limits_{x \in D'} x = 2$. Next we suppose cut $2n + 1$ or $2n + 3$. C)For $n = 4k - 1$, if cut $2n + 3$, we can make $2k$ pairs, and the sum of each pair is $2n + 2$, take $k$’s pairs to be $A$, and $\left\{ {1,2n + 1} \right\} \subseteq A$, the rest to $B$.Then $\sum\limits_{x \in A} x = \sum\limits_{x \in B} x $; Let $C = B \cup \left\{ 1 \right\}$, $D = A\backslash \left\{ 1 \right\}$, we have $\sum\limits_{x \in C} x - \sum\limits_{x \in D} x = 2$. If cut $2n + 1$, let $C' = A \cup \left\{ {2n + 3} \right\}\backslash \left\{ {2n + 1} \right\}, D' = B$, then $\sum\limits_{x \in C'} x - \sum\limits_{x \in D'} x = 2$；$A' = D \cup \left\{ {2n + 3} \right\}\backslash \left\{ {2n + 1} \right\},B' = C$, so $\sum\limits_{x \in A'} x = \sum\limits_{x \in B'} x $. D)For $n = 4k + 1$, if cut $2n + 3$, without $n,n + 2$, we can make $2k$ pairs, and the sum of each pair is $2n + 2$, take $k$’s pairs(include $\left\{ {1,2n + 1} \right\}$) and $\left\{ {n + 2} \right\}$ to $C$, rest to $D$, then $\sum\limits_{x \in C} x - \sum\limits_{x \in D} x = 2$. Let $C\backslash \left\{ 1 \right\} = A$, and $D \cup \left\{ 1 \right\} = B$, then $\sum\limits_{x \in A} x = \sum\limits_{x \in B} x $. If cut $2n + 1$, without $1,3,2n - 5,2n - 3,2n - 1,2n + 3$, we can make $2k - 2$ pairs, and the sum of each pair is $2n - 2$, take $k - 1$’s pairs$ \cup \left\{ {2n + 3,2n - 5,1} \right\}$ to $A$, rest to $B$, then $\sum\limits_{x \in A} x = \sum\limits_{x \in B} x $. Let $C = B \cup \left\{ 1 \right\}$, and $D = A\backslash \left\{ 1 \right\}$, then $\sum\limits_{x \in C} x - \sum\limits_{x \in D} x = 2$. So we done.

I haven't looked at any of the solutions here, but I would like to solve this problem. Can someone please clarify the following: (1) can the weights be negative integers or 0? (2) when we split the remaining coins into 2 groups, can one of the groups be empty?

Common sense says that weights are never negative, and most likely not zero either. The solutions in this thread use that weights are positive. (When weights can be negative, I find exactly one additional $n$ that works.) If the weights are all positive, naturally neither group can be empty (except the trivial case $n=1$). Turns out that with weights can be negative, that still doesn't matter.

The solutions written above are not understandable; they are written in broken english and not clearly explained. Can someone pls put the above solutions into a readable form, or better explain the solution. Thank you.

Lemma 1. The number of weights $n$ must be an odd number. Proof. Assume that $S_n = \sum_{i=1}^n w_i$ is minimal. If any coin is removed the remaining sets sum must be a multiple of $2$. Thus, \[w_2+w_3+\cdots+w_n \equiv 0 \mod 2\]\[w_1+w_3+\cdots+w_n \equiv 0 \mod 2\]\[\vdots\]\[w_1+w_2+\cdots+w_{n-1} \equiv 0 \mod 2.\]Summing these up we get $(n-1)(w_1+w_2+\cdots+w_n) \equiv 0 \mod 2$. Thus, we have two cases: $n \equiv 0 \mod 2$ and $\sum_{i=1}^n w_i \equiv 0 \mod 2$ and $n \equiv 1 \mod 2$ and $\sum_{i=1}^n w_i \equiv 1,0 \mod 2$. In the first case, $w_i \equiv 0 \mod 2, \forall i = 1,\ldots,n$. This is impossible as the new set $\left \{\dfrac{w_i}{2}\right \}$ also works but is smaller contradicting the minimality of $S_n$. Thus, $n$ is odd. Lemma 2. The number of weights $n>5$. Proof. For $n = 3$, no such set exists since we would be left with two numbers $a,b$, which can't be equal since they are distinct. For $n=5$, we will have $4$ numbers left which must be broken into two groups that are equal. Let the five weights be $w_1 < w_2 < w_3 < w_4 < w_5$. Suppose we remove $w_1$. Then we must have $w_2+w_5 = w_3+w_4$ and/or $w_5 = w_2+w_3+w_4$. The second case is not possible as in this case we would have $w_5 = w_2+w_3+w_4 > w_1+w_3+w_4$, which means the coins cannot be split when we remove $w_2$. Thus we must have $w_2+w_5 = w_3+w_4$. This shows that $w_1+w_5 < w_3+w_4$. But then no splitting is possible when we remove $w_2$ since $w_5 < w_3+w_4$ tells us that the group with $w_5$ must contain exactly $2$ coins. The second coin cannot be $w_1$, and as $w_5 > w_4 > w_1$ the second coin is $> w_1$ we have no other choice. Lemma 3. Any odd n > 5 works. Proof. Let $f = 1 \ast 3 \ast \cdots \ast (2n-1)$ for some odd number n and the $\ast$ symbol can take either $+,-$ where the weights of the coins are $1,3,\ldots,2n-1$. Now suppose that we can satisfy the condition of the problem for some $2n + 1$ and then we need to show that $2n + 3$ is possible. Then define a transformation on $f$ call it $f'$ where $|f'-f| = 2$ which we form by taking two odd consecutive integers and changing their sign to form $f'$. We then move a term from the set $\{1,\ldots,2n-1\}$ then by using the result $|f'-f|$ we change the sign on $2n+1$ or $2n+3$ using $|f'-f| = 2$ to get $f = 0$. Thus two sets have equal sum if we move $2n+3$ or $2n+1$ which holds for $n \quad 1 \pmod4$ or $n \equiv 3 \pmod{4}$.

The answer is all odd $n\ge 7$. Without loss of generality, assume that $w_1< w_2 <\dots < w_n$ and that $\gcd(w_1,w_2,\dots,w_n)=1.$ Now, note that if $S=w_1+w_2+\dots+w_n$ then $S-w_1,S-w_2,\dots,S-w_n$ are all even. Therefore, $w_k$ have a fixed parity for $1\le k\le n.$ Since they are not all even, they are all odd. This implies that $S$ is odd. Since $S$ is odd, $n$ must be odd. This is not possible for $n=3$ because removing $w_1$ would imply $w_2=w_3$. For $n=5$, removing $w_5$, we see that $A_5=(w_2+w_3,w_1+w_4)$ or $B_5=(w_1+w_2+w_3,w_4)$ are the only possible ways to split them into two groups, for size reasons. Similarly, removing $w_4$, we see that $A_4=(w_2+w_3,w_1+w_5)$ or $B_4=(w_1+w_2+w_3,w_5)$ are the only possible splitting ways. If we use $A_5$ and $A_4$ or $B_5$ and $B_4$ then we get $w_4=w_5$, absurd. If we use $A_4$ and $B_5$ then $2w_1+w_5=w_4$, absurd. Thus, we must use $A_5$ and $B_4$. Now, if when we remove $w_3$ we use $w_1+w_2+w_4=w_5$ then from $B_4$ we know $w_3=w_4$. Therefore, $w_1+w_5=w_2+w_4$ so $2w_1+w_3=w_4.$ However, $w_4-w_3=w_2-w_1$ so $2w_1=w_2-w_1$ so $3w_1=w_2.$ If when we remove $w_2$, we use $w_1+w_5=w_3+w_4$ then this contradicts $w_1+w_5=w_2+w_4.$ Thus, $w_1+w_3+w_4=w_5.$ This implies that $w_2=w_4$ by $B_4$, so that is also impossible. Therefore, $n=5$ is impossible. For $n\ge 7$ we claim that $1,3,5,\dots,2n-1$ will work. All we need to prove, in fact, is that there is some way to place $+$'s or $-$'s between the numbers $1,3,5,\dots,2n-1$ and before $1$ so that the result can be any number $1,3,5$, to $2n-1$. For $n=7$, we use \begin{align*} 1-3-5-7-9+11+13 &= 1\\ 1+3-5-7+9-11+13 &= 3\\ 1+3+5+7-9+11-13 &= 5\\ 1-3-5+7+9+11-13 &= 7\\ 1+3+5-7+9+11-13 &= 9\\ 1-3+5-7-9+11+13 &= 11\\ 1+3+5-7+9-11+13 &= 13 \end{align*}Now, suppose it is possible with $(1,3,\dots,2n-1)$ Then we'll prove it's possible with $(1,3,\dots,2n+3).$ If we are trying to make some number $1\le k\le 2n-1$, then take any way to make $k$ with $(1,3,\dots,2n-1)$. Then, we subtract $2n+1$ and add $2n+3$ to make $k+2.$ Now, add a negative sign before the $1$ to make $k$. If the sign before $1$ is already negative, then we add $2n+1$, subtract $2n+3$, and then make the sign before $1$ positive. If we are trying to make $2n+1$, then you take the way to make $2n-1$, add $2n+3$ then subtract $2n-1$ and then leave it alone. Now, to make $2n+3$, if the $1$ is negative, then do what we did for $2n+1$ but change that negative before the $1$ to positive. Otherwise, we can always find two consectuvie numbers $2i-1,2i+1$ such that $2i-1$ is positive and $2i+1$ is negative. Now, if we flip those two signs then we can make $2n+3$ with $(1,3,\dots,2n-1)$. However, using what we did for $1\le k\le 2n-1$ we can also make $2n+3$ using $(1,3,\dots,2n+3)$ as desired.