The statement is not true. Consider 5 coins index 0 to 4. Starting from 0. In this way only 0 2 and 4 are flipped.

Well actually it is true since, following your example,we would flip coins in this order: 0-2-0-4-4,so in fact, the coin number 2 is the only one that will remain face down in the end. Anyways,thanks for your help, and any other help would be appreciated as well

L.Lawliet03 wrote: Well actually it is true since, following your example,we would flip coins in this order: 0-2-0-4-4,so in fact, the coin number 2 is the only one that will remain face down in the end. Anyways,thanks for your help, and any other help would be appreciated as well Ah! you are right. I was thinking about the complete residue family, which is not correct in this problem. It is in fact quite straightforward to find the flip pattern with recurrence relation: sol = RSolve[{a[n + 1] == a[n] + n + 1, a[1] == 0}, a, n][[1]] =>{a -> Function[{n}, 1/2 (-2 + n + n^2)]} (*Example*) With[{n = 3}, Table[Mod[a[i] /. sol, 2*n + 1], {i, 2*n + 1}]] => {0, 2, 5, 2, 0, 6, 6} (*Example*) With[{n = 10}, Table[Mod[a[i] /. sol, 2*n + 1], {i, 2*n + 1}]] =>{0, 2, 5, 9, 14, 20, 6, 14, 2, 12, 2, 14, 6, 20, 14, 9, 5, 2, 0, 20, 20} Tally[%] => {{0, 2}, {2, 4}, {5, 2}, {9, 2}, {14, 4}, {20, 4}, {6, 2}, {12, 1}} (*only 13th coin is flipped once*) for $ 0 \le k_1 < k_2 \le 2n+1$, $a[ k_1] - a[k_2] \equiv 0 \pmod{2 n+ 1} \equiv \frac{1}{2} \left( -2 + k_1+{k_1}^2 \right) - \frac{1}{2} \left( -2 + k_2+{k_2}^2 \right) \equiv \frac{\left( k_1+k_2 +1 \right) \left( k_1- k_2 \right)}{2}$ when $k_1+k_2 +1$ is $2n+1$. By parity analysis you know $2$ always divides $\left( k_1+k_2 +1 \right) \left( k_1- k_2 \right)$. Also there are $n$ pairs of $k_i$ and $k_j$ in $2n$ numbers so that $k_1+k_2 = 2n$ and there is one $k_{l}$ left unpaired. Thus, all coins are divided into three classes: 1. coins flipped zero times, of which the index is not $a[k]$ 2 coins flipped even times 3. one coin flipped once, indexed $a[k_l] \pmod{2n+1}$ $\blacksquare$

You basically have to check that the sequence of remainders a1 , a2 , a3.....,a(2n+1) defined by a(j)= 1+2+3+....+j(mod 2n+1) has the property that all but one term of the sequence appears an even no. Of times in the sequence.this can be checked by pairing up ai with a(2n-i) for i = 1,2...,(n-1) and pairing up a(2n) and a(2n+1). Now it is easy to see that only the remainder a(n) will appear an odd no. of times in the sequence.(possibly once or once+an even no. Of times among the rest of the terms(due to pairing) Also any other residue not equal to a(n) will appear an even no. Of times due to pairing.So basically,you even know the position of the single heads down coin remaining.

By pairing up I mean that the terms paired up are equal to each other

Thanks for your solutions, they are really helpful.Cheers!