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?

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.

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure

step-by-step explanation: