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26 
Posts: 548 Joined: 06 Sep 2009 Last Visit: 17 Sep 2013
Location: Sweden  


Posted: Tue 20 Oct, 2009 


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.
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



 
 
   





Posts: 1074 Joined: 29 Jan 2009 Last Visit: 26 Jan 2013
Location: Croatia  


Posted: Tue 20 Oct, 2009 


You mean divided, don't you?
I didn't know 55=1



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Location: Present Moment  


Posted: Tue 20 Oct, 2009 


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.
Current LD goal(s): Healing



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
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Posted: Tue 20 Oct, 2009 


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.



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Location: Present Moment  


Posted: Wed 28 Oct, 2009 


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 ^^
Current LD goal(s): Healing



 
 
   




24 
Posts: 346 Joined: 29 May 2009 Last Visit: 22 Nov 2011
Location: California  


Posted: Thu 29 Oct, 2009 


2. would be 2006 squared right?



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
LD count: Thrice a week
 


Posted: Thu 29 Oct, 2009 


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



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Location: Present Moment  


Posted: Thu 29 Oct, 2009 


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 ^^
Current LD goal(s): Healing



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
LD count: Thrice a week
 


Posted: Thu 29 Oct, 2009 


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.



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Posted: Thu 29 Oct, 2009 


Great! *gives bombax cookie 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 ?
Current LD goal(s): Healing
Last edited by tosxyChor on Wed 04 Nov, 2009; edited 1 time in total



 
 
   





Posts: 103 Joined: 04 Oct 2009 Last Visit: 09 Dec 2018
 


Posted: Thu 29 Oct, 2009 


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



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Posted: Thu 29 Oct, 2009 


2a) *gives Phoenyx a cookie 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 )
2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?
Current LD goal(s): Healing
Last edited by tosxyChor on Wed 04 Nov, 2009; edited 1 time in total



 
 
   





Posts: 103 Joined: 04 Oct 2009 Last Visit: 09 Dec 2018
 


Posted: Fri 30 Oct, 2009 


GAHHH!
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... I give up



 
 
   




31

Posts: 2415 Joined: 11 Jul 2009 Last Visit: 30 Dec 2013
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Posted: Fri 30 Oct, 2009 


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... I give up 
Guess I'l do it
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 sideup to share a side is to be sidedown, 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
Current LD goal(s): Healing



 
 
   




27 
Posts: 493 Joined: 03 Feb 2007 Last Visit: 16 Dec 2009
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Posted: Wed 04 Nov, 2009 


1. We have 15, 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; 25 can also be expressed as combinations of 1*(25), 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+101452 = 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.



 
 
   
 
 