Creampuff

Dont worry he's there! if you click the down arrow on the leaderboard hes on the top i think its a glitch

sorry for me it says that the link doesnt work

Nevermind 

ohh ok

so ..... the answer would be 2 3/4 please correct me if im wrong

Let X be total number of membersX = r m (1)X = (m + 1) *(r -2) = r m + r – 2m – (2)(2) – (1): r – 2m – 4 = 0  r = 2m (3)(3) to (2): X = (m+1)(2m+4-2)=2(m+1) (4)Since X < 100, (m+1)2 = X/2 < 50Largest possible integer m is 7. Answer: X = 98
 

For a triangle exists, the side lengths of the triangle must be such that the sum of the two shorter sides must be greater than the third side.
This also is equivalent to any two sides must have a sum greater than the third side.
So
7+10 > x^2, => x^2 < 17 => x < sqrt(17)     (maximum)
7+x^2 > 10, => x^2 >3 => x > sqrt(3)
Therefore
sqrt(3) < x < sqrt(17)
If x must be an integer,
2< x < 4, or x = {2,3,4}

wait its right thank you!!