MathProblemSolver101

1/3 * 120 = 40

1/4 * 120 = 30

1/6 * 120 = 20

1-(1/3 + 1/4 + 1/6) = 1 - (9/12)=1/4

1/4 * 120 = 30

2^2 mod 10 = 4

2^6 mod 10 = 4

2^10 mod 10 = 4

any 2^(4k+2) mod 10 = 4, k≥0, can you see why?

(118-2)/4 + 1 = 30

30 * 4 mod 10 = 120 mod 10 = 0 mod 10.

answer is 0

(20*80+8*90+2*94)/30=83.6

so i was thinking stars and bars, but the people arent distinguishable and the bracelets are ....

So casework is the only option.

Complementary casework has less cases, so lets jump in

well.. we know that there can be 4^4 = 256 total cases

3 bracelets: choose 1 person to receive the 3 and one more out of the 3 remaining to get teh rest

and since distinguishable we have 3! * 1 * 3 = 18

4 bracelets: choose 1 person out of 4: 4 * 4! = 96

so 256-(18+96).... im not so sure ...

73_ _: 8*8

_ 73 _: 7*8

_ _ 73: 7*8

7373: 1

No 7 or 3: 7*8*8*8

8*8+7*8+7*8+1+7*8*8*8

I am not evaluating that, heavens.

sus numbers

"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. :)