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  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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#1526 Re: Help Me ! » Very interesting problems.. » 2005-12-26 10:27:07

1. Yes <> means, that it can't be.
2. No, I haven't computed it for n=9 yes so i'm not sure. But i'm sure for n<9.

#1527 Re: Help Me ! » Very interesting problems.. » 2005-12-26 09:42:48

Here's what I think. In the question we haven't N and n. We have only n. So we take all the numbers between 1 and n and put them on a sircular table.
Example:n=3 we should find is there an order of the numbers 1,2,3 on a sircular table so the rules to be observed.
Seerj must answer,too. Because we may solving a quite different quest then the original.

#1528 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-26 07:43:56

#k+38-bertrand's postulate. I think there isn't a simple proof. I wrote something with the chebishev's  theta function.

#1529 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-26 06:53:54

In mathsy's case b = sqrt((sqrt(5)-1)/2)q=0.7861513778...q

#1531 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-26 05:24:44

Let the sides be

,
and
. Then


Let
. Then



, p>0 so

Then the hypotenuse is

#1532 Re: Help Me ! » Very interesting problems.. » 2005-12-26 05:09:58

n<>8
Now I'm waiting the case n=9.

#1534 Re: Help Me ! » Very interesting problems.. » 2005-12-26 03:36:39

So my program did some work. This was the question I wanted to solve in the begining. n>7. And you don't have to check form 1 to 100. It's enough ot check from 1 to n.

#1535 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-23 18:05:48

And for k+47
I think I got it. But I'm not so sure.
The sum of all the ten numbers must be "random" number too, because if remainders by 10 are ramdon we can't say anyting about the remainder of the sum. I think that the probability is 0.1

#1536 Re: Help Me ! » Very interesting problems.. » 2005-12-23 13:03:07

But wait a minute...
If i've understood correctly
N cant be greater than n so some numbers may be repeated.

#1537 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:59:54

Seerj, I understood now. n may be greater than N.

#1538 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:56:45

For the ricky's comments. Ijust have seen them:
1. "The sum of all perfect squares is 2n(n+1)/2=n(n+1)"
I wanted to say that the sum of all the squares that are sum of 2 numbers in the table is exactly the doubled sum of the numbers, because each number takes part to two squares.
2. " The the remainder of the sum of every two consecutive numbers when it's divided by 4 must be 0 or 1."- I don't wanted to say that the two numbers are consecutive at all. I just wanted to say that the sum of two consecutive at the table numbers is square so it's remainder when it's divided by 4 is 0 or 1. This property of the succesive at the table numbers may reduce the computation very much.

#1539 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:45:00

...n<>7
when I tryed n=8 my vb compiler crushed because I had very much AND-s in one row. Tomorrow I can improve my program and I can write a program that calculates by my algoritm.

#1541 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:31:07

I can't wait until tomorrow. n<>5 I wrote fast vb6.0 program:
Private Sub Command1_Click()
Const n = 5
For i1 = 1 To 5
For i2 = 1 To 5
For i3 = 1 To 5
For i4 = 1 To 5
For i5 = 1 To 5
If i1 <> i2 And i1 <> i3 And i1 <> i4 And i1 <> i5 And i2 <> i3 And i2 <> i4 And i2 <> i5 And i3 <> i4 And i3 <> i5 And i4 <> i5 And Squ(i1 + i2) + Squ(i2 + i3) + Squ(i3 + i4) + Squ(i4 + i5) + Squ(i5 + i1) = 5 Then Form1.Text1.Text = i1 & i2 & i3 & i4 & i5
Next
Next
Next
Next
Next
End Sub

Function Squ(x)
If Sqr(x) = Math.Round(Sqr(x)) Then Squ = 1 Else Squ = 0
End Function

#1542 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:18:52

Great, Ricky, but you don't have to compute all the numbers between 1 and 100.
Between 1 and n is enough. I think that will accelerate the algoritm.

#1543 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:15:29

I've loved this sequence!
I'm starting with making programs tommorow morning, because it's too late now.
See you!
wink

#1544 Re: Help Me ! » Very interesting problems.. » 2005-12-23 12:10:03

Here's an algoritm that creates infinite series:
1.a1=1
Do
I is the smallest suqare that is greater than ai
if (I-ai) is unequal to all the numbers a1,a2, ... a3 then a(i+1) = (I-ai)
Next

That's how I created the upper sequenses.
I'll be glad if somebody test my algoritm. I think it will be one of the minimals.

#1545 Re: Help Me ! » Very interesting problems.. » 2005-12-23 11:46:01

And here's ss with 13 numbers:
1 24 12 4 5 11 14 2 7 9 27 22 3!

#1546 Re: Help Me ! » Very interesting problems.. » 2005-12-23 11:41:02

Here is ss with 7 numbers:
1 8 17 19 6 10 15

#1547 Re: Help Me ! » Very interesting problems.. » 2005-12-23 11:29:04

So somebody could write a program.
The number of combinations between n numbers in circular table is exactly (n-1)!/2
if n = 50 this makes exactly 304140932017133780436126081660647688443776415689605120000000000 combinations. If there isn't any mathematical way the hard computation will be very hard.

#1548 Re: Puzzles and Games » Secret message » 2005-12-23 09:41:45

The first letter is X, I guess.

#1549 Re: Dark Discussions at Cafe Infinity » jU. » 2005-12-23 09:31:05

For Polly:
And you REALLY won't understand beacuse it's gone...

#1550 Re: Dark Discussions at Cafe Infinity » Rubic's Cube solution » 2005-12-23 09:27:14

At last summer vacantion I've solved this game for a week. I haven't got any books about rubik cube. My strategy truely isn't the best, but it solves RK for about 5 minutes.

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