26. There are four piles of stones: One with 6 stones, two with 8 , and one with 9. Five players numbered $1,2,3,4$ and 5 take turns, in the order of their numbers, choosing one of the piles and dividing it into two smaller piles. The loser is the player who cannot do this. State the number of the player who loses.
26. Ans: 3 .
Observe that we begin with 4 piles of stones and end up with $6+8+8+9=31$ piles of (one) stones. At the end of each turn, there is exactly one more pile of stones than the beginning of the tum. Thus, there can be exactly $31-4=27$ legal turns. Hence, Player 3 is the first player who cannot make a move.