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#3401 Re: Puzzles and Games » Black and Red » 2006-02-27 05:53:35
Using the two equations Ganesh posted:
x, y, z are the three piles, xr is the number of red cards in x.
8xr + 5zr = 26
This is only true for integers when xr = 4, so xr = 4 and zr = 4. Thus, yr = 18
xr = 4, so xb = 12
yr = 18, so yr = 6
zr = 4, so zb = 8
x = 4+12 = 16
y = 18 + 6 = 24
z = 8 + 4 = 12
#3402 Re: Puzzles and Games » Progression Puzzle » 2006-02-27 05:33:51
I detect a hint of Fibonacci ...
You shouldn't Fib like that, MathIsFun...
That 12 really through me off Ganesh, but I think I finally got it:
1, 4, 1, 5
I'll leave it to others to find the next 4 numbers, it really is a great puzzle sequence.
#3403 Re: Puzzles and Games » Think of two cards » 2006-02-27 05:28:37
If there answer is n, s is the value of the spades, and h is the value of hearts:
s = floor(n/11)
h = n % 11 (where % is the remainder of n / 11)
#3404 Re: Help Me ! » Same Problem!!!!! » 2006-02-27 05:12:27
Oy, here we go again 
#3405 Re: Help Me ! » Desperate » 2006-02-27 05:07:59
Oh, and one thing I forgot to mention:
For example we could have the function f(x)=1/x
lim(1/x) when x goes to eternity is 0, RIGHT ?
Qn = f(1) + f(2) + f(3) + f(4) + f(5) + ... + f(n), when n goes to eternityNow if what you say is correct then Qn goes to 0 ????????
You picked a great example! As it turns out, even though that limit goes to 0, the sum diverges, that is, goes to infinity.
It is a great example to show that the converse of the 2nd thm I posted:
Is utterly false.
#3406 Re: This is Cool » u substitution » 2006-02-26 17:39:20
If you want to sub for cos x like that, then u = e^cos(x), so the domain of u is restricted by the range of e^cos(x). Thus, ln(u) is restricted by cos(x).
But with an example like this, du = -sin(x)e^cos(x)dx.
#3407 Re: Introductions » Why, hello. » 2006-02-26 16:01:30
We normally count sheep here, not horses. But I guess that doesn't really matter.
Anyways, welcome to the forums.
#3408 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2006-02-26 15:56:27
I was actually wondering that for quite a while, MathIsFun. Wouldn't it be more organized if Ganesh did one question per topic? I mean, we got an entire section of the forum for it, why not?
#3409 Re: Help Me ! » Desperate » 2006-02-26 15:48:38
Sorry to all other guys - you were correct with your conclusions but I didnt areed with the way how you got there.
Whether you agree or not, both of our methods are correct. There are theorems that say:
When n > m. Actually, that one I believe I can prove.
There is also a theorem that says if:
Then the sum of the series is divergent. Since infinity does not equal 0, both mine and Ansette's conclusions are not only correct, but well reasoned.
#3410 Re: Help Me ! » Desperate » 2006-02-26 02:15:15
affirmation, are you talking about an arbitrary, but non infinite, n? In your opening post, you wrote:
where we could find the sum of all members.
Which I thought you meant to be as n goes to infinity.
#3411 Re: Help Me ! » Quadratics Again.. » 2006-02-26 02:11:09
We have y = ax^2 + bx + c.
Plugging in (1, -2), we get:
-2 = a + b + c
And plugging in (2, -14) we get:
-14 = 4a + 2b + c
Let's try as best we can to "solve" one of these. The first one, when multipled by -2, becomes:
4 = -2a - 2b - 2c
Adding this to the second one, we get:
-10 = 2a - c
And this seems to be the only restriction. So let's choose an a and c that works in the above equation, and then solve for b.
-10 = 2(-7) - (-4) = -10
So a = -7 and c = -4. Putting these back into the very first equation:
-2 = -7 + b - 4
b = 9
a = -7 b = 9 c = -4. But do these work?
-2 = -7(1)^2 + 9(1) - 4 which is true, so it goes through (1,-2)
-14 = -7(2)^2 + 9(2) - 4 which is true, so it goes through (2, -14)
But keep in mind, there are infinitely many solutions. This is just one of them.
#3412 Re: Help Me ! » Desperate » 2006-02-25 08:42:02
I think that f(1), f(1) + f(2), f(1) + f(2) +f (3) etc or f(1), f(2), f(3)/sperately/ should be proved to be part of arithmetical or geometrical progression where we could find the sum of all members.
#3413 Re: This is Cool » who woulda thunk it? » 2006-02-25 06:12:30
I'm not really suprised at 1, but I really like the way mikau took the limit. Never would have thought of doing that.
#3414 Re: Jokes » Mathematical proof 2=1 » 2006-02-24 03:56:31
a = b, so a - b = 0. And we all know what happens when you divide by 0, don't we?
#3415 Re: Puzzles and Games » Hidden countries » 2006-02-24 03:46:13
10/10, but I made a program to scan all the sentences and compare them with names of countries 
Don't you just love alternate solutions?
#3416 Re: Puzzles and Games » Number of spheres within a cube. » 2006-02-23 13:53:25
1" diameter or radius?
#3417 Re: This is Cool » Quantum computer gives answer when not running? » 2006-02-23 13:51:19
Is that a joke? Computers don't produce errors, the people who programmed (and designed) them do.
I think what they mean is that a bit won't be misread or a piece of hardware is less likely to fail. Something along those lines.
#3418 Re: Help Me ! » differential equation xy"+y'-mxy=0 » 2006-02-23 13:40:10
In differential equations, you want to find an equation, not a value. Thus, you are solving for y, not x.
It's been a long time for me as well (ok... 2 months... but still...), I know you can solve any 2nd degree equation (that is solvable) using matricies. But I don't believe there is any simpiler form for this equation.
How much about diff eq do you know? Is this a course in it or are you just studying yourself? Have you used matricies to solve differential equations in the past?
#3419 Re: Puzzles and Games » fun with logic » 2006-02-22 17:20:49
Just like stacking cups, you put one glass inside another.
It's a cheat, but that's what I'm best at. 
#3420 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2006-02-22 17:16:31
#3421 Re: This is Cool » Number theory #1 » 2006-02-22 12:18:00
Alright, I'm bout ready to give up. It seems like I'm missing some fact that being 0 mod gives, but I can't seem to find it. Being divisible by a+b doesn't seem to be enough.
#3422 Re: Puzzles and Games » fun with logic » 2006-02-22 09:33:21
How do you measure the distance?
Assuming that you measure distance by the center of the glass, and you are only measuring in a 2d plane...
Take one glass on the table, and put the other three classes inside of it (stacked). Then the distance of each glass is.... 0.
#3423 Re: Help Me ! » Discreet proof help needed » 2006-02-22 06:03:40
1. Since we want to show a remainder when dividing by 5, we must first get something that is divisble by 5. So lets consider the 5 cases:
n = 5k
n = 5k + 1
n = 5k + 2
n = 5k + 3
n = 5k + 4
Note this covers all possible values for n. Now lets go through them one by one:
n = 5k. Then n^2 = 25k^2, which has a remainder of 0 when divided by 5.
n = 5k + 1. Then n^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2) + 1, which has a remainder of 1 when divided by 5.
Go through the others.
#3424 Re: Help Me ! » Need help with a injective (one-to-one) proof » 2006-02-22 05:39:50
Ah, sorry, I got mixed up with terms.
The image of f is a subset of B, such that b∈Im f iff there exists an a∈A such that f(a) = b.
Assume |A| < |Im f|. Then either there exists an a0 and a1∈A, such that a0=a1 and f(a0) ≠ f(a1) or there exists a b∈Im f such that there does not exist an a∈A such that f(a) = b. These are the only two ways there can be more elements in Im f. In either case, they lead to contradictions, and so |A| ≥ |Im f|.
Now assume |A| > |Im f|. Then there must exist a0 and a1∈A such that a0 ≠ a1 and f(a0) = f(a1). But f must be injective. Contradiction, so |A| = |Im f|.
Did that make more sense?
#3425 Re: Help Me ! » Need help with a injective (one-to-one) proof » 2006-02-21 14:22:11
For every element in A, you know there has to be one element in B. Thus, it has to be |A| ≤ |B|. If it was |A| > |B|, then there would be at least one element in A for which there was no element in B.
Edit: Heh, I don't quite know why this was in the proof for A. It doesn't belong there.