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WASD
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PostPosted: Tue 20 Oct, 2009  Reply with quote

Mew151 wrote:
Hmmm...

Well, Infinity, subtracted by itself, should be zero.

Because even though it is unlimited, no further numbers exist.
So even though Infinity + 5 = Infinity, Infinity = Infinity, never something greater, because nothing greater exists. So just treat Infinity like a number.
So Infinity - Infinity = 0. grin
If it is true, Infinity - Infinity - Infinity = -Infinity, because 0 - Infinity = -Infinity.

Something subtracted by itself is always 1.
edit: oops my fault, meant divided.

In my opinion Infinity+5=Infinity+5 just like x+5=x+5, even tho nothing is bigger then infinity...




Last edited by WASD on Fri 30 Oct, 2009; edited 1 time in total
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Lord Antares
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PostPosted: Tue 20 Oct, 2009  Reply with quote

You mean divided, don't you?

I didn't know 5-5=1


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tosxyChor
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PostPosted: Tue 20 Oct, 2009  Reply with quote

Infinity is only a symbol, mainly used in limits of a sequence. Its meaning is "this sequence, while composed of finite numbers, is approaching infinity, which means the generic term of this sequence will become greater than any number", and by generic term i mean every term of the sequence from some point onward.
Example: 5,4,10,9,15,14,20.... is a sequence of numbers, and if it goes on like this (adding 6, then subtracting 1) it is approaching infinity. For example, it will be bigger than 1000 from the 401st number on, it will be greater than 1000000 from the 400001st number on, etc.



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Bombax
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PostPosted: Tue 20 Oct, 2009  Reply with quote

Tosxychor wrote:
Infinity is only a symbol, mainly used in limits of a sequence. Its meaning is "this sequence, while composed of finite numbers, is approaching infinity, which means the generic term of this sequence will become greater than any number", and by generic term i mean every term of the sequence from some point onward.
Example: 5,4,10,9,15,14,20.... is a sequence of numbers, and if it goes on like this (adding 6, then subtracting 1) it is approaching infinity. For example, it will be bigger than 1000 from the 401st number on, it will be greater than 1000000 from the 400001st number on, etc.
Precisely. In other words, the infinity symbol represents growth without limit.

Quote:
Something subtracted by itself is always 1.
In my opinion Infinity+5=Infinity+5 just like x+5=x+5, even tho nothing is bigger then infinity...
AAAARGH! Never, ever, ever think that mathematics are defined by opinion. Mathematics is the most pure field of science existing, and does not depend on opinion at all. Tosxychors gives a nice explanation what infinity "represents" (it's not a number) in his post above.

And a number x € C subtracted by itself is 0, not 1.


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tosxyChor
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PostPosted: Wed 28 Oct, 2009  Reply with quote

Ohwell let's try and spice thing up a little:
1 . How many positive divisors does the number 5*4*3*2 = 5! have?
2 . Paul takes the number 3, erases it and writes 9, then erases it and writes 81, and so on, each time writing the square of the previous number, another 2006 times. What will be the units' digit of the last number?
3 . In how many ways you can order the letters L A P I S so that the first and the last letters are wovels?
Have fun ^^



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Fallen Youth
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PostPosted: Thu 29 Oct, 2009  Reply with quote

2. would be 2006 squared right?

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Bombax
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PostPosted: Thu 29 Oct, 2009  Reply with quote

Nope. The last digit in the mentioned number is 1.

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tosxyChor
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PostPosted: Thu 29 Oct, 2009  Reply with quote

Bombax wrote:
Nope. The last digit in the mentioned number is 1.

Any solution is better accompanied by a demonstration. I'll put it myself for this one:
The series starts with 3, 9, 81, 6561, 43046721... But the next number's units digit is only influenced by the units digit of the last number, as any other calculus would work torwards the bigger digits. IE. 81*81= 1*1 + 80*1 + 1*80 + 80*80= 1 + 80 + 80 + 6400 = 6461. And since 1*1 = 1, the lastr digit will always be 1 from that moment on.

Good luck with the other problems ^^



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Bombax
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PostPosted: Thu 29 Oct, 2009  Reply with quote

III: The first letter we can choose in two ways. The second we can choose in three ways, the third in two ways, the fourth in one way and the fifth in one way. The answer is 2*3*2*1*1 = 12.

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tosxyChor
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PostPosted: Thu 29 Oct, 2009  Reply with quote

Great! *gives bombax cookie cookiemonster Now for another 2 problems to keep 3 of them up:
2a) On a chessboard (8x8 squares) Paul puts tokens on each square following this rule: both rows and columns are numbered, form 1 to 8, and for each square, he lookt up the corresponding row and column number, then he sums them up, and puts that many tokens on that square. How many total tokens has Paul put on the chessboard?
3a) That one's a little more difficult... see if you can do it ^^
Samantha has 2006 equilateral triangles of the same size, and wants to put them all on a table, in a way they don't overlap, and each one is sharing 2 sides with other 2 triangles. Can she make it? Can still she do it if she has 2005 of them? Is it possible for any number of them >12 ?



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Last edited by tosxyChor on Wed 04 Nov, 2009; edited 1 time in total
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Phoenyx
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PostPosted: Thu 29 Oct, 2009  Reply with quote

2a) the chessboard has 8x8=64 squares. He puts 2 tokens in 1 corner, and 16 in the opposite. The alternate corners are 9 and 9. Similarly each pair of squares totals 18 tokens, and the board has 32 pairs, resulting in 18x32= 576 tokens total. This feels sloppy though, is there a better way to arrive at the answer?

3a) in order to achieve 2 and only 2 shared sides, the triangles need to loop in some manner. The closed loop would be 6 triangles in a hexagon, 12 triangles to encompass a 1 triangle space, 14 for 2, 16 for 3, etc. etc.. I would venture a guess 2005 would NOT work, but any even number 12 and above, as well as 6, would meet the req's... assuming of course the table is large enough and/or the triangles small enough smile


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tosxyChor
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PostPosted: Thu 29 Oct, 2009  Reply with quote

2a) *gives Phoenyx a cookie cookiemonster The counting can be separated into rows and coloumns: tokens total = (tokens form row counting) + (tokens form columns) = 8*( 1+2+...+8 ) + 8*( 1+2+...+8 )= 8*36*2=576 indeed. ^^
3a) Nice, but I still need to know that it's impossible for any odd number of triangles to be used successfully (it's easy: try visualizing some examples)

Now for some counting! (seems difficult, but it's not wink5 )
2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?



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Last edited by tosxyChor on Wed 04 Nov, 2009; edited 1 time in total
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Phoenyx
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PostPosted: Fri 30 Oct, 2009  Reply with quote

GAHHH! lach1

Quote:
3a) Nice, but I still need to know that it's impossible for any odd number of triangles to be used successfully (it's easy: try visualizing some examples)


Working on this for quite awhile now, I keep coming up with ways to explain why it needs 2 at a time, but can't seem to figure how to explain why it CANT be an odd number.

I have a feeling it is something to do with it being impossible for two odd numbers to sum to an odd number, but... whatsthat help! eek2 I give up


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tosxyChor
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PostPosted: Fri 30 Oct, 2009  Reply with quote

Phoenyx wrote:
I have a feeling it is something to do with it being impossible for two odd numbers to sum to an odd number, but... whatsthat help! eek2 I give up

Guess I'l do it wink5
First things first, a graph where all dots (representing the triangles) are connected to two and two only of the other elements is necessarily made of separate loops (convince yourself of this)
Now, in any single loop of triangles, look at them on the table: there are traingles with one side facing up and others with one side facing down. The only way a triangle can connect to a side-up to share a side is to be side-down, and vice versa. So it's necessary for every loop to have an even number of triangles, since if it was odd, at one point we'd necessarily have 2 connected triangles that are oriented in the same way (and that's impossible ;D)

Now for another one!
3b) A, B and C are playing with 2008 skittles. 'A' brings down three times the skittles brought down by 'B', and 'B' him/herself brings down double the number of C's skittles. (They are all playing the same match, so the skittles they have brought down as a total can't exeed 2008)
What is the maximum number of skittles A could bring down?

And come on guys, 1) is not that hard, there is a nice formula to calculate the number of divisors, look "divisor" up on Wikipeida wink5



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PostPosted: Wed 04 Nov, 2009  Reply with quote

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.

3b. Is it really that easy? a+b+c=2008. a=3b. b=2c a=3(2c)=6c. 6c+2c+c=2008. 9c=2008. c=223.111 a = 1338.666. Of course, you can't have a fractional number of Skittles, so 1338.


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