Problem 40 - Page 3

ThePerfectHacker · #21 $Old$ 11-14-2007, 07:57 PM

By the way the numbers to do not have to be natural numbers. I specifically said that.

angel.white · #22 $Old$ 11-14-2007, 08:28 PM

Okay, in that case I have a start, maybe someone else can pick up what I don't know.

I would say we can find every instance by the following method:
A, B = random from the set {1,2,3,...,2007}
c,d = real numbers (this includes irrational and negative)

so:
$cd=A \Longrightarrow c=\frac{A}{d}$

$c+d=B$

Substitute:
$\frac{A}{d}+d=B$

Multiply by d:
$A+d^{2}=dB$

Subtract dB
$d^{2}-dB+A=0$

By Quadratic Formula:
$d=\frac{B\pm\sqrt{B^{2}-4A}}{2}$

-----

Therefore there will be a real solution so long as $B^{2} -4A \geq 0$

So I don't know how to find the probability that this will be the case, but I expect if you can find the probability of this, happening, then you can subtract that from100%, and that will be your answer.

angel.white · #23 $Old$ 11-14-2007, 09:59 PM

Okay, I think I have found a way to calculate it, but I would need computer software to make it work

so we need to know the probability that $B^{2}-4A>0$ is true. Lets modify the equation to make it more workable.

$B^{2}-4A>0$

Add 4A
$B^{2}>4A$

Square root (B must be positive because it is taken from a set of positive integers)
$B>2\sqrt{A}$

So now we need to find the probability that this is true for A and B are elements of {1,2,3,...,2007}

So if we assume, A we can figure out the probability for this to be true for B, based on some specific A. In this case, $2007-floor(2\sqrt{A})$ will give us the number of possibilities that B can be which will satisfy the equations. Then divide that by 2007 (the total number of possibilities that B can be, including those that do and those that do not satisfy the equations), and that will give you the percentage that the equations can be satisfied. Meaning cd=A and c+d=B is true (if A is true).

So for any A, the probability that it has real factors whose sum is B is
$\frac{2007-floor(2\sqrt{A})}{2007}$

And so because any given A has a 1 in 2007 chance of being true, then we can multiply that probability against the probability that B will be true for that A, and then count the total probabilities for each A.
$\frac{1}{2007}\cdot\frac{2007-floor(2\sqrt{A})}{2007}$

$\frac{2007-floor(2\sqrt{A})}{4028049}$

Now we just need to count them all. I'm not very familiar with summation notation, but I think this is correct

$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}$

As i cycles through every value, it will calculate the odds for an A of that value having a B that will satisfy cd=A and c+d=B, then add it to the running total, and when it is done there should be a percentage, between .96 and 1

Okay, so I don't actually know how to do that, (well, I know what it means, but I don't care to sit here for 2007 iterations and calculate the odds) but I think that it is the answer.

Does anyone know an online site which can do this type of equation? If not, perhaps someone can create a quick program to do it (I used to know how to, but it's been too long

)

Edit: Looks like if you have a graphing calculator, you can do it Summation (Sigma) Notation Using the Graphing Calculator I, unfortunately, do not have a graphing calculator, but if any of you would plug it in for me, I'd +rep you ^_^ (that's worth like an entire point)

ThePerfectHacker · #24 $Old$ 11-14-2007, 10:02 PM

You are doing good. But I have a way without software. There is a formula.

angel.white · #25 $Old$ 11-14-2007, 10:07 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

You are doing good. But I have a way without software. There is a formula.

I don't know how to do it without software, I'm still in Discrete Math 1, and this is the first summation notation equation I have written on my own, I had to look up how they work just to make sure I did that one right.

Perhaps you would be willing to educate me about the formula?

angel.white · #26 $Old$ 11-15-2007, 12:23 AM

Okay, I found a site that will calculate these:
Sigma
$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}=.970475036426$
So about 97.05%
^_^

Is that the correct answer?

ThePerfectHacker · #27 $Old$ 11-15-2007, 08:53 AM

Quote:

Originally Posted by angel.white $View Post$

Is that the correct answer?

I have no idea I did not solve the problem yet numerically.

CaptainBlack · #28 $Old$ 11-16-2007, 12:25 AM

Quote:

Originally Posted by angel.white $View Post$

Okay, I found a site that will calculate these:
Sigma
$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}=.970475036426$
So about 97.05%
^_^

Is that the correct answer?

That is the correct value for the sum you give.

RonL

angel.white · #29 $Old$ 11-16-2007, 11:26 AM

Quote:

Originally Posted by CaptainBlack $View Post$

That is the correct value for the sum you give.

RonL

Do you know whether the sum I've given the correct answer to the question?

ThePerfectHacker · #30 $Old$ 11-18-2007, 07:32 PM

Here is solution to #1. We want $x+y=B, xy=A$ that means $x^2+y^2+2xy=B^2\implies x^2+y^2=B^2-2A$ . But then $(x-y)^2 = x^2 + y^2 - 2xy = B^2 - 2A - 2A = B^2 - 4A \implies x-y = \sqrt\pm {B^2-4A}$ . And the system of equations $x+y = B \mbox{ and }x-y = \pm \sqrt{B^2-4A}$ always has a solution provided $B^2-4A \geq 0 \implies B^2 \geq 4A$ .

So we need to count the number of ways we can get $B^2\geq 4A$ for all pairs $(A,B)$ . Note if $B\geq 90$ . Then the maximum value $4A$ is $4(2007) = 8028$ which is always true for $B\geq 90$ . We we just need to count all pairs $(A,B)$ where $B\leq 79$ .

If $B=1$ then there is no $A$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 0.

If $B=2$ then there is just $A=1$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 1.

If $B=3$ then there is just $A=1,2$ in the pair $(A,B)$ so that $B^2 \geq 4A$ . So the count is 2.

If $B=4$ then we can pick $A=1,2,3,4$ . So the count is 4.

If $B=5$ then we can pick $A=1,2,3,4,5,6$ . So the count is 6.

The question is whether we can find a pattern. Yes! It is based on looking at even and odd cases. Say $B$ is even so $B=2c$ then $B^2 \geq 4A\implies 4c^2\geq 4A\implies c^2 \geq A$ in that case $A=1,2,3,...,c^2$ . So the count is $c^2$ .

If $B$ is odd, so, $B=2c+1$ then $(2c+1)^2 \geq 4A \implies 4c^2+4c+1\geq 4A \implies c(c+1)+.25\geq A$ . Which means $A=1,2,3,...,c(c+1)$ . So the count is $c(c+1)$ .

Now if we write out the numbers as we did for $B=1,2,3,4,5$ out further we will get:
$0,1,2,4,6,9,12,16,...$
Where (alternatively) $0=0\cdot 1,2=1\cdot 1, 6=2\cdot 3,12=3\cdot 4,...$
And (again alternatively) $1=1^2,4=2^2,9=3^2,16=4^2,...$ .
This list continous until $B=78,79$ on $78$ the value of $c=78/2 = 39$ . And for $79$ is will be $39(40)$ .

If we split this sum into even terms in the sequence and odd terms in the sequence we get:
$\sum_{n=1}^{39}n^2 + \sum_{n=1}^{39}n(n+1) = \sum_{n=1}^{39}2n^2+n = \frac{39(40)(79)}{3}+\frac{39(40)}{2} = 41080+780=41860$ .
But this is for $B\leq 79$ (I just realized it should have been $B\leq 89$ but I am too lazy to change it now. You get the idea).

Now just find the number of pairs $(A,B)$ for $B\geq 80$ which is just $80(2007)$ and add to the answer.