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tosxyChor
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PostPosted: Wed 04 Nov, 2009  Reply with quote

Phi_guy wrote:
1. We have 1-5, all possible combinations of 2, 3, 4, and 5 of those numbers. There's only one possible combination for all 5. For 4, we have 5 combinations. For 3, 10; for 2, I got 10 as well. But there's some overlap; 2-5 can also be expressed as combinations of 1*(2-5), so we can ignore those. For similar reasons, we can drop 5 from the group of 3's, 2 from the 4's, and 1 from the group of 5. So 5+1+5+10+10-1-4-5-2 = 19.
I really didn't understand the explanation, however the answer is wrong. I'll give an hint: start writing 5!=120 as 2^3 * 3 * 5;

Phi_guy wrote:
3b. Is it really that easy?
Yes, it was. ^^ Cookie earned cookiemonster and new problem!!
3c) (I'll leave the 3rd problems easy) In a stadium there must be a precise order to enter through the 5 gates: first 1 person goes in the 1st gate, then 2 in the 2nd, then 3 in the 3rd, 4 in the 4th, and 5 in the 5th, and then the cycle starts over. Which gate will the 2007-th person pass through?



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Phi_guy
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PostPosted: Wed 04 Nov, 2009  Reply with quote

1. Let's see, we have 2, 3, 5, 2*3, 2*5, 2*3*5, 2^2, 2^2*3, 2^2*5, 2^2*3*5, 2^3, 2^3*3, 2^3*5, 2^3*3*5, and 3*5, and 1. That makes 16.

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tosxyChor
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PostPosted: Wed 04 Nov, 2009  Reply with quote

cookiemonster Counting is good for little numbers, but now go for 10! = 10 * 9 * ..... * 2 * 1. I want a formula! lachgroen


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WASD
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PostPosted: Sun 08 Nov, 2009  Reply with quote

Anyone tried doing math/homework in a dream? tounge2

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tosxyChor
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PostPosted: Sun 08 Nov, 2009  Reply with quote

It was in a past challenge: do all 4 operations with numbers bigger than 5 digits or so (writerscube was in it, but I can't really find it right now).
Some other also tried to go by the powers of 2, and didn't manage to get to his usual big numbers.
I'm guessing math is a good way to get the left hemisphere (rational thinking) back in gear before starting the real LD, so you can have more control and stability ;D



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Phi_guy
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PostPosted: Mon 09 Nov, 2009  Reply with quote

So if I give you a formula, do I get a cookie recipe?

Let's see. 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 * 3^2 * 2 * 5=
2^8 * 3^4 * 5^2 * 7
I want to say it's just 9*5*3*2 .

Here's a math-related puzzle. You'll have to think very creatively to solve it. I lol'd when I found the secret.

Draw that without lifting your pencil, going over a line twice, or manipulating the paper (i.e. folding it).


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tosxyChor
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PostPosted: Mon 09 Nov, 2009  Reply with quote

Phi_guy wrote:
So if I give you a formula, do I get a cookie recipe?

Let's see. 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 * 3^2 * 2 * 5=
2^8 * 3^4 * 5^2 * 7
I want to say it's just 9*5*3*2 .

Well I'll just give you 2 cookies, beacuse you didn't explain the formula. cookiemonster cookiemonster That makes three cookies for a single problem, you should be more than happy lach1 I'll demonstrate it in your place.

We have a number of prime divisors, (namely 2, 3, 5 and 7), each wth his own exponent, and to form a divisor, we can combine any available power of those divisors freely. We can take 2^3, 3 and 5 to form 8*3*5=120, take only 2^7 and form 128, and so on.

To form all possible divisors, we just have to multiply all possibilities for each prime divisor, and if that divisor has n as its maximum exponent, that means we have n+1 choices for it. (IE for 5^2, we can ignore it and take 5^0, take a single 5, or take both and get 5^2=25, three total choices). Multiplying all choices (that are completely independent) we get (8+1)*(4+1)*(2+1)*(1+1) = 9*5*3*2 = 270 total divisors. (Notice how it is completely irrelevant from the kind of prime divisors we have)

Quote:
Draw that without lifting your pencil, going over a line twice, or manipulating the paper (i.e. folding it).
It is indeed mathematically impossible to do it the standard way, because there are 4 points from which an odd number of lines is leaving. (And the maximum number to make it possible is 2). So, we could draw extra lines here and there to help us, perhaps over another paper so we don't ruin the drawing. Is it so?
But if you didn't say not to fold the paper, I would have immediately thought of it first



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Phi_guy
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PostPosted: Mon 09 Nov, 2009  Reply with quote

Actually, I was just going for "it's impossible."

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WASD
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PostPosted: Mon 09 Nov, 2009  Reply with quote

Tosxychor wrote:
cookiemonster Counting is good for little numbers, but now go for 10! = 10 * 9 * ..... * 2 * 1. I want a formula! lachgroen

I found this one...

uppsidedown L = gamma
gamma(n+1)=n!

read more on http://en.wikipedia.org/wiki/Gamma_function


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tosxyChor
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PostPosted: Tue 10 Nov, 2009  Reply with quote

Phi_guy wrote:
Actually, I was just going for "it's impossible."
I definitely think too hard into stuff. I seriously LOL'd when I saw your answer lach2 and what, no cookie? smile

WASD wrote:
gamma(n+1)=n!
Quite right, but we were talking more about a formula to calculate the munber of divisors of any given number

Ok so, 2 problems are still up, feel free to add a 3rd one to the mix, or I will post one tonight.



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Phi_guy
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PostPosted: Tue 10 Nov, 2009  Reply with quote

2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?

Not going to work on this until I'm actually awake, but I'd say it's somewhere around 5,250. Close?

And fine, here's your cookie. cookiemonster


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tosxyChor
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PostPosted: Tue 10 Nov, 2009  Reply with quote

Moar liek 4000+.
I'll give a little hint: 12= 3*4, and divisibility by 3 and 4 are easily studied. wink

EDIT: While we're at it, http://en.wikipedia.org/wiki/Mathematical_joke



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tosxyChor
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PostPosted: Sat 14 Nov, 2009  Reply with quote

Man, I totally forgot to add a 3rd problem back. lach1
1b) Geometry! ^^
If the circle has diameter 2, how much's the coloured area?



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tosxyChor
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PostPosted: Thu 17 Dec, 2009  Reply with quote

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WASD
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PostPosted: Fri 18 Dec, 2009  Reply with quote

Socks-ychor wrote:
Man, I totally forgot to add a 3rd problem back. lach1
1b) Geometry! ^^
If the circle has diameter 2, how much's the coloured area?

I figured that one out with a little help from my math teacher, http://i50.tinypic.com/1jvlte.jpg
So the area is... http://www.wolframalpha.com/input/?i=hexagon+area
s is the radius, so of the the diameter is 2, s=1
Area=2.59808 cm^2 (or some other unit)


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