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Neather2
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Welcome back! I'm posting more maths games for train your brains. I'll give also 1 points for each puzzle resolved.

1)

Two old friends, with the passion of mathematical puzzles, meet by chance on the road after many years. The first says:
- How many children do you have?
- Three. One male and two females.
- And how old are they?
- Considering their age as integers, their product is 36 and the sum is equal to the number of home here in front.
The friend thinks a bit and then says:
- You don't have given me enough information!
And the second replies:
- It's true: The biggest female child has 2 big eyes!
(Resolved by TripleM)

2)

What is the sum of the first 2450 natural integers numbers? (Resolved by TripleM)

3)

I wrote a two-digit number (not ending with zero). Then I delete the first digit from left. Finally, I multiplied the number remained (one digit) for 9. Surprise: I found the number from which i was start!
What was that number?
(Resolved by TripleM)

4)

According to what order are the following numbers?
5-2-9-4-7-3-1
(Resolved by TripleM)

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Conspiracy, my new hold, has just been released.

[Last edited by Neather2 at 02-06-2009 09:02 AM]
02-06-2009 at 06:45 AM
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eb0ny
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quote:
Neather2 wrote:
2)
What is the sum of the first 2450 natural integers numbers?
33636
sum([ int(x) for x in ''.join( [ str(y) for y in xrange(1, 2451) ] ) ])


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[Last edited by eb0ny at 02-06-2009 06:57 AM]
02-06-2009 at 06:55 AM
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Neather2
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No, it isn't.

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Conspiracy, my new hold, has just been released.
02-06-2009 at 07:10 AM
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TripleM
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No, its
Click here to view the secret text

02-06-2009 at 08:45 AM
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TripleM
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1 and 3:

Click here to view the secret text


02-06-2009 at 08:48 AM
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TripleM
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And finally, I presume the answer to 4 is:
Click here to view the secret text

02-06-2009 at 08:51 AM
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Neather2
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Correct! take this points!

I spent a lot of mods points here, lol. I must start doing harder puzzles or you will take all them.

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Nothing to say. I just play the game. And you, sir, should play it too.

Conspiracy, my new hold, has just been released.

[Last edited by Neather2 at 02-06-2009 09:04 AM]
02-06-2009 at 09:00 AM
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Neather2
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Here's another one:

triangoli2.JPG

Don't read italian writes here. Just explain me how that is possible.

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Conspiracy, my new hold, has just been released.
02-06-2009 at 09:12 AM
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eb0ny
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quote:
Neather2 wrote:
2)

What is the sum of the first 2450 natural integers numbers?
Sums of natural integers and their numbers are different.

Sum of integers from 10 to 15 is 10+11+12+13+14+15=75
Sum of integers' numbers from 10 to 15 is 1+0+1+1+1+2+1+3+1+4+1+5=21

I thought you were asking for the latter. The answer I gave you is correct, it's just for the wrong problem.

EDIT: Regarding triangles.

Lines that make up the red triangle are 3 and 8 squares long. It's hypotenuse is sqrt(73) or approx. 8.554. The sine of the smallest angle is 3/sqrt(73) = 0.351123, which gives us angle's size - 20.556 degrees.

If we do the same thing for the green triangle, the respectable sine is 0.371391, and angle's size is 21.801 degrees.

These angles mismatch. This means, that the figures are not triangles in the first place: the "hypotenuse" in the top figure is shaped inwards, while the bottom one it is shaped outwards, accomodating the extra square.

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[Last edited by eb0ny at 02-06-2009 10:01 AM]
02-06-2009 at 09:49 AM
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Neather2
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Yeah. Anyone is allowed to post here maths games too, of course.

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Conspiracy, my new hold, has just been released.
02-06-2009 at 12:40 PM
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Neather2
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More Games

6) Find a number such that moving the first digit after the last make the half of this number.
For example, it has 315, moving the three after the five you get 153 that is close but not the half of 315.

7) Find four numbers such that the fourth power of one of them is equal to the sum of the fourth powers of the other three.


____________________________
Nothing to say. I just play the game. And you, sir, should play it too.

Conspiracy, my new hold, has just been released.

[Last edited by Neather2 at 02-06-2009 12:53 PM]
02-06-2009 at 12:50 PM
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brian_s
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I have one:

Eight points are placed on the surface of a sphere with a radius of 1. The shortest distance between any two points is greater than 1.2. How can the points be arranged?
02-06-2009 at 09:21 PM
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The Stew Boy
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quote:
Neather2 wrote:
Here's another one:

triangoli2.JPG

Don't read italian writes here. Just explain me how that is possible.

Click here to view the secret text

02-06-2009 at 09:30 PM
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lopsidation
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Inscribe a cube in the sphere, then place the points at the eight vertices of the cube. Each edge of the cube is 2/sqrt(3), which is around 1.15.

Here's another one: Find 4 polynomial functions (f(x), g(x), h(x), and j(x)) that all meet at 0, don't meet anywhere else, and:
When x<0, f(x) > g(x) > h(x) > j(x)
When x>0, h(x) > g(x) > j(x) > f(x)

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02-06-2009 at 09:47 PM
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stigant
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quote:
Inscribe a cube in the sphere, then place the points at the eight vertices of the cube. Each edge of the cube is 2/sqrt(3), which is around 1.15.


quote:
The shortest distance between any two points is greater than 1.2.



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02-06-2009 at 09:58 PM
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lopsidation
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Oh. *slaps self* I need to stop making stupid mistakes. Well then, it seems impossible, unless by "shortest distance" you mean going across the surface of the sphere. Then the cube method works.

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02-06-2009 at 11:06 PM
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TripleM
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quote:
lopsidation wrote:
Oh. *slaps self* I need to stop making stupid mistakes. Well then, it seems impossible, unless by "shortest distance" you mean going across the surface of the sphere. Then the cube method works.


It's definitely possible :) (with the normal definition of distance)

[Last edited by TripleM at 02-06-2009 11:08 PM]
02-06-2009 at 11:08 PM
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Kwakstur
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If this world were governed by a slightly different set of physical laws where sp3d4 orbital hybridization exists, I would've learned this two months ago. Dang.

What if you started with the 8 points of a cube, but took one set of 4 coplanar points and rotated it by 45 degrees?

Edit: Nope, the points on the same plane are still closer to eachother than 1.2.

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[Last edited by Kwakstur at 02-06-2009 11:55 PM]
02-06-2009 at 11:52 PM
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lopsidation
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Okay, how about this: inscribe a tetrahedron in a sphere, and put four points at each of its vertices. Then put four other points on the sphere near the the middle of each face of the tetrahedron, so those four other points form another tetrahedron that's upside-down. This looks like it should actually work.

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02-07-2009 at 01:35 AM
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TripleM
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I think Kwakstur has the right idea. The solution indeed has a square with a square rotated by 45 degrees opposite (with the right lengths, that is).
02-07-2009 at 01:57 AM
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Jutt
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I can get extremely close by taking the following points:

(3/5,0,4/5)
(-3/5,0,4/5)
(117/125,0,-44/125)
(-117/125,0,-44/125)
(0,3/5,-4/5)
(0,-3/5,-4/5)
(0,117/125,44/125)
(0,-117/125,44/125)

The smallest occuring distance between two points is about 1.1987 and the next smallest exactly 1.2.
I think that a solution can be found by nudging the points slightly into the right direction.

On second thought, going from Kwakstur's idea and my near solution, I think an inscribed square antiprism would indeed do the trick.

EDIT: Wikipedia confirms

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[Last edited by Jutt at 02-07-2009 02:41 AM : added link]
02-07-2009 at 02:37 AM
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