Mathematics desk | ||
---|---|---|
< February 27 | << Jan | February | Mar >> | March 1 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Say, I'm doing a subtraction problem I really don't want to spend time on. Like 1235232 and 499876. Or if I was multiplying, adding, or dividing them, how could I find the digit to the far left? Thanks! Mαc Δαvιs X ( How's my driving?) ❖ 02:14, 28 February 2007 (UTC)
What was your question again? Is it, how do I find the left most digit if I subtract 499876 from 1235232? 202.168.50.40 03:33, 28 February 2007 (UTC)
I looked everywhere in the internet and I can not find the answer. I am frustrated and I need to know as soon as possible. Please help me!!!—Preceding unsigned comment added by 201.202.137.206 ( talk • contribs)
Go to www.google.com then type in "miles kilometer". The very first result is
kilometers to miles conversion calculator - length @ metric ... kilometers to miles conversion calculator, km, mile - length @ metric conversions . org.
202.168.50.40 20:29, 28 February 2007 (UTC)
I have found a new way to utilize the Fibonacci sequence and I would like to have opinions and/or help in publishing this on Wikipedia.
I have reduced the Fibonacci sequence to one digit (ex. 13=1+3=4). I have then seen that ones we reach the 24th digit of the Fibonacci sequence, the 24 numbers repeat themselves (ex. 25th number (reduced) is 1, the 26th is 1, the 27th is 2... etc.
This is the full sequence: 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9.
From here I copied and added a second sequence below the first but starting in the middle of the first (13th number: 8).
I soon realized that the numbers between top and bottom sequence always add up to 9. Nines would be lined up together.
1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9. (It seems Wikipedia doesn't allow for correct alignment).
8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9.
If we can find the Fibonacci sequence in nature and also it's derived 'golden ratio', I wonder if this mathematical beauty is at the base of some other sequencing in nature... I actually believe it's only time before it will be found...
Also, I noticed a interrelation between the pairs. For example, the 7-2 pair is spaced by 5 pairs from the other 7-2 pair and 7-2=5 (between two 9 pairs). The 6-3 pair is spaced by 3 pairs and the 5-4 pair is spaced by 1. I have created an excel sheet in which I've vertically lined up the double sequence and assign a color for each number. It is very beautiful and harmonious. 68.163.230.21 16:10, 3 March 2007 (UTC)
1, 1, 2, 3, 5, 1, 6, 0,
6, 6, 5, 4, 2, 6, 1, 0
Notice that each column adds to 7 or 0 (mod 7). I wouldn't be surprised if this works for any prime modulo. Dugwiki 20:52, 1 March 2007 (UTC)
More on the pattern. Let's say you're looking at the series mod M for some natural number M, and that the n'th member of the series equals 0 mod M. Then you have and . In other words, once you hit 0, you'll always get two of the same number in a row, which if that number is M-1 will mirror the start of the series "1, 1". Dugwiki 20:13, 2 March 2007 (UTC)
HERE IS THE RIDDLE..
if 1 chick costs 5 cents, 1 hen costs 1 dollar and 1 cock costs 5 dollars..and you're asked to buy 100 units consists of these 3 kinds with exactly 100 dollars, which variation of them can agree with this condition??
I've been told this riddle and its answer, but i want to know if there is a possible way to conclude this answer mathematically ??!!
the answer is (80,1,19) consecutively.......thanx in advance
What an offensive question! deeptrivia ( talk) 21:54, 28 February 2007 (UTC)
Thank you Spoon, but can you show the details of the step of using extended Euclidean algorithm, pls? (which led to: -21 * 99 + 26 * 80 = 1)
Would anybody who knows how to archive the Ref-Desks be able to assist in gettting the oldest day on each desk down to Feb 21, so that RefDeskBot can resume normal operation? Thanks for your help (you may be able to collaborate here if others are doing the same). Mart inp23 22:30, 28 February 2007 (UTC)
I've posted this on both sci.math and alt.sci.math.combinatorics without getting a single reply. I suppose the question I want answered is "Does this problem ring any bells with anyone"? Before my newsgroup postings, I would have bet money on my having rediscovered something well-known, but now I'm starting to wonder. So ...
... whilst designing an experiment recently, I evolved the following problem:
Given the sequence S of natural numbers 1...N (N odd), is it possible to find a permutation p(S) such that the pairwise difference S-p(S) gives the integers -(N-1)/2, ...,-1,0,1, ...,(N-1)/2?
At the time, I was specifically interested in S = 1...7.
It wasn't immediately obvious to me whether a solution was possible, and if it was, whether or not it would be trivial or obvious with hindsight. I shared the problem with a colleague, and we both set about independently trying to find a solution - which we each did, in a few minutes. We used a different strategy and came up with a different solution: his strategy also worked for the simpler case S = 1,2,3, whereas mine did not.
I then investigated the problem systematically using Mathcad, and determined somewhat to my surprise that there were in fact 24 possible solutions. Clearly, every solution must have one member of S that remains fixed, to give the value 0. My colleague had fixed the starting value 1, whereas I had fixed the central value 4. It turns out that for S = 1...7, there are solutions for every choice of fixed value. Some choices are more "difficult" than others. Fixing the value 1 or 7 gives six possible solutions each. Fixing the value 2, 4, or 6 gives four possible solutions each. Fixing the value 3 or 5 gives two possible solutions each.
The solutions divide into three classes: those consisting, in addition to the fixed point, of one six-cycle, two three-cycles and three two-cycles. From my dim memory of group theory, this sounds like Sylow's theorem is lurking somewhere.
The simplest case S=1,2,3 shows that there are not always solutions for every choise of fixed value.
It seems certain that the general version of this problem has been studied, possibly indirectly.
Is it necessarily the case that the number of solutions increases hugely as N increases, or are there "interesting" values of N for which solutions are uncommonly sparse, or require a restricted choice of fixed value - or even are non-existent?
Spargeus 22:39, 28 February 2007 (UTC)
That's a very pretty solution (actually, it's two solutions, depending which half we "arbitrarily" assign the middle number to)! I'd love to know how you came up with it. I particularly like the fact that it's a constructive proof, not just an existence proof. As you say, looking for (N-1)/2 2-cycles simplifies the problem a bit - it's the logical place to start. Now we know that there is always at least one pair of solutions, I'm wondering if the method gives any clues as to how we might go about identifying "higher-order" solutions.
I'm sure there must be a Sylow thing going on - any solution must consist of a set of equal-order cycles - so if the order is O, there are C such cycles and OC = N-1. So O and C must always divide (N-1). I can't prove that, it just seems obvious, and my investigation of N=7 provides a little evidence. But even if it is the case, it's only a necessary condition, not a sufficient one.
O=2 is always such a divisor, so it's neat that there is always a corresponding solution. C=2 is also always such a divisor, so it would be really neat if there was also always a corresponding solution. I don't see any promising line of enquiry to extend your method from 2-cycles to [(N-1)/2]-cycles, though.
Spargeus 12:11, 4 March 2007 (UTC)
Mathematics desk | ||
---|---|---|
< February 27 | << Jan | February | Mar >> | March 1 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Say, I'm doing a subtraction problem I really don't want to spend time on. Like 1235232 and 499876. Or if I was multiplying, adding, or dividing them, how could I find the digit to the far left? Thanks! Mαc Δαvιs X ( How's my driving?) ❖ 02:14, 28 February 2007 (UTC)
What was your question again? Is it, how do I find the left most digit if I subtract 499876 from 1235232? 202.168.50.40 03:33, 28 February 2007 (UTC)
I looked everywhere in the internet and I can not find the answer. I am frustrated and I need to know as soon as possible. Please help me!!!—Preceding unsigned comment added by 201.202.137.206 ( talk • contribs)
Go to www.google.com then type in "miles kilometer". The very first result is
kilometers to miles conversion calculator - length @ metric ... kilometers to miles conversion calculator, km, mile - length @ metric conversions . org.
202.168.50.40 20:29, 28 February 2007 (UTC)
I have found a new way to utilize the Fibonacci sequence and I would like to have opinions and/or help in publishing this on Wikipedia.
I have reduced the Fibonacci sequence to one digit (ex. 13=1+3=4). I have then seen that ones we reach the 24th digit of the Fibonacci sequence, the 24 numbers repeat themselves (ex. 25th number (reduced) is 1, the 26th is 1, the 27th is 2... etc.
This is the full sequence: 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9.
From here I copied and added a second sequence below the first but starting in the middle of the first (13th number: 8).
I soon realized that the numbers between top and bottom sequence always add up to 9. Nines would be lined up together.
1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9. (It seems Wikipedia doesn't allow for correct alignment).
8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9.
If we can find the Fibonacci sequence in nature and also it's derived 'golden ratio', I wonder if this mathematical beauty is at the base of some other sequencing in nature... I actually believe it's only time before it will be found...
Also, I noticed a interrelation between the pairs. For example, the 7-2 pair is spaced by 5 pairs from the other 7-2 pair and 7-2=5 (between two 9 pairs). The 6-3 pair is spaced by 3 pairs and the 5-4 pair is spaced by 1. I have created an excel sheet in which I've vertically lined up the double sequence and assign a color for each number. It is very beautiful and harmonious. 68.163.230.21 16:10, 3 March 2007 (UTC)
1, 1, 2, 3, 5, 1, 6, 0,
6, 6, 5, 4, 2, 6, 1, 0
Notice that each column adds to 7 or 0 (mod 7). I wouldn't be surprised if this works for any prime modulo. Dugwiki 20:52, 1 March 2007 (UTC)
More on the pattern. Let's say you're looking at the series mod M for some natural number M, and that the n'th member of the series equals 0 mod M. Then you have and . In other words, once you hit 0, you'll always get two of the same number in a row, which if that number is M-1 will mirror the start of the series "1, 1". Dugwiki 20:13, 2 March 2007 (UTC)
HERE IS THE RIDDLE..
if 1 chick costs 5 cents, 1 hen costs 1 dollar and 1 cock costs 5 dollars..and you're asked to buy 100 units consists of these 3 kinds with exactly 100 dollars, which variation of them can agree with this condition??
I've been told this riddle and its answer, but i want to know if there is a possible way to conclude this answer mathematically ??!!
the answer is (80,1,19) consecutively.......thanx in advance
What an offensive question! deeptrivia ( talk) 21:54, 28 February 2007 (UTC)
Thank you Spoon, but can you show the details of the step of using extended Euclidean algorithm, pls? (which led to: -21 * 99 + 26 * 80 = 1)
Would anybody who knows how to archive the Ref-Desks be able to assist in gettting the oldest day on each desk down to Feb 21, so that RefDeskBot can resume normal operation? Thanks for your help (you may be able to collaborate here if others are doing the same). Mart inp23 22:30, 28 February 2007 (UTC)
I've posted this on both sci.math and alt.sci.math.combinatorics without getting a single reply. I suppose the question I want answered is "Does this problem ring any bells with anyone"? Before my newsgroup postings, I would have bet money on my having rediscovered something well-known, but now I'm starting to wonder. So ...
... whilst designing an experiment recently, I evolved the following problem:
Given the sequence S of natural numbers 1...N (N odd), is it possible to find a permutation p(S) such that the pairwise difference S-p(S) gives the integers -(N-1)/2, ...,-1,0,1, ...,(N-1)/2?
At the time, I was specifically interested in S = 1...7.
It wasn't immediately obvious to me whether a solution was possible, and if it was, whether or not it would be trivial or obvious with hindsight. I shared the problem with a colleague, and we both set about independently trying to find a solution - which we each did, in a few minutes. We used a different strategy and came up with a different solution: his strategy also worked for the simpler case S = 1,2,3, whereas mine did not.
I then investigated the problem systematically using Mathcad, and determined somewhat to my surprise that there were in fact 24 possible solutions. Clearly, every solution must have one member of S that remains fixed, to give the value 0. My colleague had fixed the starting value 1, whereas I had fixed the central value 4. It turns out that for S = 1...7, there are solutions for every choice of fixed value. Some choices are more "difficult" than others. Fixing the value 1 or 7 gives six possible solutions each. Fixing the value 2, 4, or 6 gives four possible solutions each. Fixing the value 3 or 5 gives two possible solutions each.
The solutions divide into three classes: those consisting, in addition to the fixed point, of one six-cycle, two three-cycles and three two-cycles. From my dim memory of group theory, this sounds like Sylow's theorem is lurking somewhere.
The simplest case S=1,2,3 shows that there are not always solutions for every choise of fixed value.
It seems certain that the general version of this problem has been studied, possibly indirectly.
Is it necessarily the case that the number of solutions increases hugely as N increases, or are there "interesting" values of N for which solutions are uncommonly sparse, or require a restricted choice of fixed value - or even are non-existent?
Spargeus 22:39, 28 February 2007 (UTC)
That's a very pretty solution (actually, it's two solutions, depending which half we "arbitrarily" assign the middle number to)! I'd love to know how you came up with it. I particularly like the fact that it's a constructive proof, not just an existence proof. As you say, looking for (N-1)/2 2-cycles simplifies the problem a bit - it's the logical place to start. Now we know that there is always at least one pair of solutions, I'm wondering if the method gives any clues as to how we might go about identifying "higher-order" solutions.
I'm sure there must be a Sylow thing going on - any solution must consist of a set of equal-order cycles - so if the order is O, there are C such cycles and OC = N-1. So O and C must always divide (N-1). I can't prove that, it just seems obvious, and my investigation of N=7 provides a little evidence. But even if it is the case, it's only a necessary condition, not a sufficient one.
O=2 is always such a divisor, so it's neat that there is always a corresponding solution. C=2 is also always such a divisor, so it would be really neat if there was also always a corresponding solution. I don't see any promising line of enquiry to extend your method from 2-cycles to [(N-1)/2]-cycles, though.
Spargeus 12:11, 4 March 2007 (UTC)