MathProblemSolver101

"Other way around" is the same answer. Because here, you can consider absolute value since x and y are nonzero.

Thanks for clearing the problem up... this turned out to be correct and all the steps are fairly logical. also I realized that diagonal was impossible..because then you cant construct a line of different shpae

Unfortunately i dont have it TeXed out.

I can try

You can have something like

{triangle} {triangle} {triangle} 

{square} {square} {square}

{triangle} {triangle} {square} 

Thank you melody and GA for your thoughtful answers! I made some progress on 2, considering combinatorics, but got an answer that I'm sure is incorrect or overcounted for: \(\binom{3}{2} \cdot 3! \cdot 3!\). This answer feels wrong, but I'm not sure why - anyone want to take a stab?

x_1 + x_2 = -a

x_1 x_2 = 1

x_1 + x_2 = -1

x_1 x_2 = -a

-a=-1

1=-a

a=-1, 1

The question is vague. It can be any odd multiple of 15.

Instead of casework, try complementary.

P(odd) happens when it is odd * odd = (1/2)^3 = 1/8 (all odd)

1-1/8 = 7/8

The link provided below has a correct answer, but the answerer didn't explain why it only works if it is divisible by 3.

First off, if you have a fraction n/2, then you can have any n in order to repeat

If you have a fraction n/3, then we need a multiple of 3

if we have a fraction n/5, then you can have any n

so we need a multiple of 3 in order to repeat

thus, the answer is 474/3 (if its divisible) or ceil(474/3) (if it's not divisible.)

it's divisible, so the answer is 474/3=158

There is a reason for the pattern - can you see why? Do you see anything intesting if we generalize \(x\) for not just 2, but for all integral \(x\)? Try finding a general formula. Also notice - does your 1, 0, etc. pattern have anything to do with \(x=2\)? Maybe there is, and maybe not. Try to find out. :)