Clan x86
General Forums => Academic / School => Math and Other Problems => Topic started by: Ender on October 24, 2006, 07:36:10 pm
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How many ordered pairs of real numbers (s, t) with s and t both between 0 and 1 are there such that both 3s + 7t and 5s + t are integers?
I will post a hint on Thursday.
Edit: edited for clarity.
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What's this have to do with a flying gorilla? I'm disapointed, Ender.
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Most of these types of problems are tied to some sort of (uselessly entertaining) word problem.
I think the answer is infinitely many, by the way.
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Joe, that is for you to figure out. Go spend the rest of your life doing so.
[invisible_to_joe]It's meaningless.[/invisible_to_joe]
Sidoh, there are actually a finite number of solutions. This is a REALLY tough problem =), and it requires some creativity.
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(http://latex.sidoh.org/?render=\mathbb{Z}) has an infinite amount of negative components.
Do you mean ordered pair of numbers (s, t) in the set (http://latex.sidoh.org/?render=\mathbb{R}-\mathbb{Z})? :P
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0 < s, t < 1
It's all in the first quadrant.
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0 < s, t < 1
It's all in the first quadrant.
(http://sidoh.dark-wire.net/upload/files/435DEUNB3L-d9e7edae7c7caee2.jpg)
Is this what you were intending? It seems to me that there are an infinite amount of (real) solutions in both the first and fourth quadrants. :P
Aside from that, I don't see how the way numbers fall into quadrants relate to your choice of sets for this problem. :P
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Forget what I said about the first quadrant, it was meant to be a hint, not a clarification. It is a hint though.
And the question is correct. There are not an infinite number of ordered pairs (s, t).
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Forget what I said about the first quadrant, it was meant to be a hint, not a clarification. It is a hint though.
And the question is correct. There are not an infinite number of ordered pairs (s, t).
You said: (http://latex.sidoh.org/?render64=W3MsdF18c1xpbntcbWF0aGJie1J9fSxzID4gMCx0XGlue1xtYXRoYmJ7Un19LHQ8MSwzcys3dFxpbntcbWF0aGJie1p9fSw1cyt0XGlue1xtYXRoYmJ7Wn19).
Do you realize that (http://latex.sidoh.org/?render64=XG1hdGhiYntafVxzdWJzZXRuZXFcbWF0aGJie1J9)? ((http://latex.sidoh.org/?render=\mathbb{Z}) is a proper subset of (http://latex.sidoh.org/?render=\mathbb{R}))
Because of this, there are an infinite number of solutions such that (http://latex.sidoh.org/?render=s\in{\mathbb{Z}},r\in{\mathbb{Z}},s>0,t<1).
The numerical solutions to (http://latex.sidoh.org/?render64=M3MrN3Q=) and (http://latex.sidoh.org/?render64=NXMrdA==) given your restrictions are always members of (http://latex.sidoh.org/?render=\mathbb{Z}) if both s and t are members of (http://latex.sidoh.org/?render=\mathbb{Z}) themselves...
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s and t are never members of Z. s and t are both between 0 and 1, on an open-ended interval.
(Too lazy to use latex but will in the future.)
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s and t are never members of Z. s and t are both between 0 and 1, on an open-ended interval.
(Too lazy to use latex but will in the future.)
Not at all according to your own specifications. All you stated is:
s > 0 and t < 1. (http://latex.sidoh.org/?render=\mathbb{Z}) contains negative numbers as well...
At this point, I'm thoroughly convinced that: a) You don't understand what I'm saying, b) You wrote the question incorrectly or c) I'm right. :P
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At this point, I'm thoroughly convinced that: a) You don't understand what I'm saying, b) You wrote the question incorrectly or c) I'm right. :P
hehe =)
While some people may be offended by that, it gives me a good laugh. Thank you.
In the statement 0 < s, t < 1 the comma is a separation of variables, not a separation of conditions. As you must have already realized, what it means, or should mean, is that 0 < s < 1 and 0 < t < 1.
I agree it's ambiguous. I did, however, quote it from a reliable source, and don't consider myself an expert in math notation, so I'm not going to take any sides. I resign to ignorance and being bliss.
Anyways, forget the semantics, focus on the problem =) Or something else, just never spend so much time on the semantics of a problem.
I have a bad feeling there's going to be a prolonged discussion on this topic.
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just never spend so much time on the semantics of a problem.
The problem doesn't make any sense in the way that I (correctly, I may add :P) interpreted it.
In any case, though, these sorts of problems are widely viewed as a waste of time. You may get the occasional answer, but unless you're posting something that's seemingly more profound, then don't expect to get many replies. It's not that people here aren't capable of solving these sorts of questions. :p
I am unaware of any formal mathematical notation where a comma suggests what you're saying this did. If you're going to post riddles from elsewhere, at least clean up the garbage. :P
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This problem is not a waste of time. It just takes a very creative step to simplify it. It exemplifies creativity in mathematics.
Here's the hint that I promised I would give on Thursday. Consider this problem in two dimensions. Let x = 3s + 7t and y = 5s + t. If x and y are both to be integers, then (x, y) will be a lattice point on the graph (a lattice point is the intersection of x = c and y = k where c and k are both integers, or, to put it otherwise, where the gridlines incrementing by one in both dimensions intersect). There are an infinite number of lattice points on the xy plane, but only a finite number in which 3s + 7t and 5s + t are both integers AND where s and t are both between 0 and 1. Find the bounds of the containing shape and then count the lattice points.
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This problem is not a waste of time. It just takes a very creative step to simplify it. It exemplifies creativity in mathematics.
I didn't say it was. Re-read what I said.
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I never accused you of saying it was ;)
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Your reply sure made it sound like you were. :P
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Your reply sure made it sound like you were. :P
just never spend so much time on the semantics of a problem.
The problem doesn't make any sense in the way that I (correctly, I may add :P) interpreted it.
In any case, though, these sorts of problems are widely viewed as a waste of time. You may get the occasional answer, but unless you're posting something that's seemingly more profound, then don't expect to get many replies. It's not that people here aren't capable of solving these sorts of questions. :p
I am unaware of any formal mathematical notation where a comma suggests what you're saying this did. If you're going to post riddles from elsewhere, at least clean up the garbage. :P
0 < s, t < 1
It's all in the first quadrant.
Is this what you were intending? It seems to me that there are an infinite amount of (real) solutions in both the first and fourth quadrants. :P
Aside from that, I don't see how the way numbers fall into quadrants relate to your choice of sets for this problem. :P
(http://latex.sidoh.org/?render=\mathbb{Z}) has an infinite amount of negative components.
Do you mean ordered pair of numbers (s, t) in the set (http://latex.sidoh.org/?render=\mathbb{R}-\mathbb{Z})? :P
COUNT THE TONGUE SMILEYS!!! JUST COUNT THEM!!!! THEY ARE STARTING TO BURN MY EYES I SEE THEM SO OFTEN!
Edit: Zeus told me 4 answers if 0,0 isn't a possibilty.
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Solution:
The shape from which we will count the lattice points has four vertices. Each vertice is a solution of the ordered pair (3s + 7t, 5s + t) with (s,t) as (0, 0), (0, 1), (1, 0), (1, 1), since these represent the bounds of s and t: 0 < s < 1 and 0 < t < 1. There are 31 lattice points, and thus 31 solutions.
Draw it out and it will make more sense.
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Solution:
The shape from which we will count the lattice points has four vertices. Each vertice is a solution of (3s + 7t, 5s + t) with (s,t) as (0, 0), (0, 1), (1, 0), (1, 1), since these represent the bounds of s and t: 0 < s < 1 and 0 < t < 1. There are 31 lattice points, and thus 31 solutions.
Draw it out and it will make more sense.
so that's actually a greater/lesser than or equal to sign?
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so that's actually a greater/lesser than or equal to sign?
() indicates an open interval (all points between the components, but not including the components). [] indicates a closed interval, including all points between the components as well as the components.
IE, a domain of (0,10) would indicate 0 < x < 10. A domain of [0,10] would indicate 0 <= x <= 10.
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() indicates an open interval (all points between the components, but not including the components). [] indicates a closed interval, including all points between the components as well as the components.
IE, a domain of (0,10) would indicate 0 < x < 10. A domain of [0,10] would indicate 0 <= x <= 10.
Interval notation is lame; I know it, but I just prefer the symbols.
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Interval notation is lame; I know it, but I just prefer the symbols.
Lazy shorthand becomes perfered after you work for an extended period of time on the subject.
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Yes, notation can be daunting but it makes math much easier to communicate.