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13 April, 01:12

The perfect square game is played as follows: player 1 says a positive integer, then

player 2 says a strictly smaller positive integer, and so on. The game ends when someone

says 1; that player wins if and only if the sum of all numbers said is a perfect square.

What is the sum of all n such that, if player 1 starts by saying n, player 1 has a winning

strategy?

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Answers (1)
  1. 13 April, 03:37
    0
    Answer: the number can be 9

    Step-by-step explanation:

    we want that

    n + n - 1 + n-2 + ... + 2 + 1 = X

    X must be a perfect square.

    And the last number, 1, must be said by player 1, so n must be an odd number.

    n + (n - 1) + (n - 2) + ... + (n - n + 1) = n*n - (1 + 2 + 3 + ... + n)

    and we know that:

    (1 + 2 + 3 + 4 + ... + n) = n * (n + 1) / 2

    So we have:

    x = n*n - n * (n + 1) / 2 = n*n - n*n/2 - n/2 = n*n/2 - n/2 = n * (n + 1) / 2

    Now we want to find X such that is a perfect square and n must be an odd integer.

    You can start giving different values for n until you reach a value of X that is a perfect square, for example, if you take n = 9, we have X = 36.

    and 36 = 6*6

    So if player 9, he will win always.
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