You have 100 coins on the table, all tails up. One move is turning any 93 coins over.How many moves will it take to get all the coins on heads?
It's question that I found in mathematical "olimpics" for high school and I found it interesting.
In the brackets there was a hint that I should use method of counting two ways(or double counting).
On
The answer is: 16.
Consider any solution. Each turn we are flipping 93 coins, and overall we are flipping each coin an odd number of times. This there are 100 coins, the total number of coin flips is even. Since 93 is odd, it follows that the number of moves must be even.
Since the number of moves is even, instead of flipping 93 coins each time, we can flip the the remaining 7 coins each time - the end result will be the same, since there is an even number of moves.
In 14 moves we can touch at most 98 coins, so since the number of moves is even, we need at least 16 moves. In the first 12 moves, we can flip the first 84 coins. We flip the other 16 coins in 4 moves as follows:
$$ 1111111000000000 \\ 1111110100000000 \\ 1111110010000000 \\ 0000000001111111 $$
As each move changes the parity of the number of heads, it is clear that we need an even number of moves. Consider the effect of two consecutive moves: The two sets of $93$ turned coins must largely overlap: They must have at least $86$ coins in common. Hence their combined effect is to turn an even number $\le 14$ of coins. Hence $14$ moves can at most turn $98$ coins, i.e., we need at least $16$ moves.
On the other hand, we can turn any even number $2m\le14$ of coins with two moves: Partition these in two sets of $m$ coins, pick $93-m$ coins from the rest (possible because $2m+93-m\le 100$) and turn these and the first/suecond subset in the first/second move. It follows that the lower bound of $16$ moves is achievable.