Answer:
This is a classic game theory problem known as "Nim". To determine who will win, we can use binary notation and calculate the "nim-sum" of the two piles. The nim-sum is simply the binary XOR (exclusive OR) of the two numbers, which means that we add the two binary numbers without carrying over:
100011 (35 in binary)
XOR 11001 (25 in binary)
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110010 (nim-sum in binary)
The nim-sum is 110010 in binary, which is equal to 50 in decimal. This means that the game is equivalent to starting with a pile of 50 stones.
In a game of Nim, the player who starts with an even number of stones will lose if both players play perfectly, and the player who starts with an odd number of stones will win. Since 50 is even, the second player will win if they play perfectly.
To see why, suppose the second player removes k stones in their first move, leaving the first player with a pile of 50 - k stones. The key is that the first player can always mirror the second player's move by removing 50 - k stones from the other pile. This will leave both piles with the same size, and the second player will be forced to make the same move again. This process will repeat until one of the piles is empty, and the other player will win.
Therefore, the second player will win if they play perfectly, and they can guarantee this outcome by following the above strategy.
Explanation: