Challenging Math Problems (Or not)

[carneyf1d]

yea i did something wrong. i dont think you can split the series up and multiply them. but i know that the original 1/2 * (-1)^n *n*(n+1) is right.

and you don't want me to post problems. they won't be easy.

[lllazar]

yea i did something wrong. i dont think you can split the series up and multiply them. but i know that the original 1/2 * (-1)^n *n*(n+1) is right.

and you don't want me to post problems. they won't be easy.

Plz?

And what grade are u in...

BTW guys, if you even care, im keeping records...

[carneyf1d]

fine. given the very first problem in this post with the 5 terms in a set and average.

give me the number of unique solutions for the set of 5 numbers. This means that each number must be different than every other in the set.
To make this problem easier, assume that the highest number in the set can only be 9
Hint** there are over 100 possible combinations**
But i want to know the exact number of possible combinations

and i'm a 3rd year at GT.

[lllazar]

fine. given the very first problem in this post with the 5 terms in a set and average.

give me the number of unique solutions for the set of 5 numbers. This means that each number must be different than every other in the set.
To make this problem easier, assume that the highest number in the set can only be 9
Hint** there are over 100 possible combinations**
But i want to know the exact number of possible combinations

and i'm a 3rd year at GT.

Ahh, GT is my dream school

[winneratlife]

fine. given the very first problem in this post with the 5 terms in a set and average.

give me the number of unique solutions for the set of 5 numbers. This means that each number must be different than every other in the set.
To make this problem easier, assume that the highest number in the set can only be 9
Hint** there are over 100 possible combinations**
But i want to know the exact number of possible combinations

and i'm a 3rd year at GT.

Tbh...this problem is much easier when there is no restriction on the highest number (I'm assuming whole numbers, no 0)

Basically, you line up 35 ones, then decide how many ways you can line up dividers, which is 34C4, then account for double counting (divide by 4!) and non-uniques

Another way is to tree/brute force it...but that could get ugly real quick

But now I'm stuck the 9 restriction...

So I guess I divide it into four zones of 8,9,9,8...but then I still have say....slot 1 on zone 1 and slot 9 of zone 2 make a number greater than 9

Am I on the right track?

[carneyf1d]

the number 9 came from the lowest possible maximum for the set to exist. The highest possible number of combinations occurs first when the set maximum is 25. I wrote a program that can solve it by brute forcing it...but brute forcing it in under 5 seconds.

if you want, you can tell me the maximum number of combinations.

[andrewwski]

Ok from now on if i post a geometry question, ill post a picture with it, ill have the pic up in a few minutes.

Guys how do i paste a pic to a post directly? I hate having to link you guys to photobucket...

http://s658.photobucket.com/albums/uu30 ... oblem1.png

There you go, the smaller circle is O and the bigger is Q, find the ratio of O to Q

The ratio of the radii is one to nine.

Meh, this looks fun!

So, find:



Because this is bashable, you must explain how you did it. NO brute force allowed!

I think i got it....you factor each pair like this:
becomes

As i did this for each "pair", i found that the sequence decreases by 4 each time like this:

(98-97)(98+97) = 195
(96+95)(96-95) = 191

So then i went to the 51st number in the sequence, like this:

When i did 199-(51*4), i got -1. So i knew the sequence ended with 3, since -1 + 4 = 3.

The answer is 199+195+191+187...+3, or 5050.

Right. When you reduce it to (100+99), you end up with the series 100+99+98+97..., or simply , which is equal to 5050.

Could also make the series equal to ((n)+(n-1))(n-(n-1)), which equals (2n+1)(1)=(2n+1) and then use a step size of 2 (only evaluate for n=2, 4, 6...).

[winneratlife]

the number 9 came from the lowest possible maximum for the set to exist. The highest possible number of combinations occurs first when the set maximum is 25. I wrote a program that can solve it by brute forcing it...but brute forcing it in under 5 seconds.

if you want, you can tell me the maximum number of combinations.

Ye...but I just found out I'm a failure...

CUZ: There is only 1 combination: (5,6,7,8,9)

This is the maximum for any 5 distinct numbers with a max of 9, so it is the only way to achieve the maximum

[binary010101]

Find the formula for the volume of the right-angle intersection of two right circular cylinders without using calculus.

http://en.wikipedia.org/wiki/File:Bicyl ... _solid.gif

EDIT: It was in my AP Calc textbook, and I still can't figure it out...

Bump.

fine. given the very first problem in this post with the 5 terms in a set and average.

give me the number of unique solutions for the set of 5 numbers. This means that each number must be different than every other in the set.
To make this problem easier, assume that the highest number in the set can only be 9
Hint** there are over 100 possible combinations**
But i want to know the exact number of possible combinations

and i'm a 3rd year at GT.

Are we assuming that the values for this set are all integers for the purposes of solving this problem?

[carneyf1d]

well if order does not count in the set yes, there is only 1 combination. 9,8,7,6,5.
however, if it does, there are 120 solutions.
sorry should've been more clear about combinations vs permutations.

if anybody would like to solve for the maximum number of permutations assuming the set can have a range of all possible numbers, be my guest.