Dudde Posted March 1, 2003 Share Posted March 1, 2003 I need some help with some relatively easy number theory problems, I'm just not sure how to go about them... Link to comment Share on other sites More sharing options...
fafalone Posted March 1, 2003 Share Posted March 1, 2003 Go ahead and post them, I'm sure me or someone else can be of help Link to comment Share on other sites More sharing options...
Dudde Posted March 2, 2003 Author Share Posted March 2, 2003 well like this one: how many ordered pairs (x,y) of positive integers are solutions to 4/x + 2/y = 1? I know this is probably mental math to someone somewhere, and I know the answer is probably easy to find, I just can't do it! and it's driving me nuts... also, if you could post how you arrived at the solution so I could see where I'm going wrong I would be very grateful;) Link to comment Share on other sites More sharing options...
blike Posted March 2, 2003 Share Posted March 2, 2003 v Link to comment Share on other sites More sharing options...
fafalone Posted March 2, 2003 Share Posted March 2, 2003 I feel compelled to note the following chat log: blike again: how many ordered pairs (x,y) of positive integers are solutions to 4/x + 2/y = 1? blike again: 1 ordered pair blike again: of positive integers blike again: (8, 4) blike again: right? fafaIone: unlimited blike again: howso fafaIone: (4n/n-2, n) blike again: k Link to comment Share on other sites More sharing options...
blike Posted March 2, 2003 Share Posted March 2, 2003 in my defense: blike again (10:58:27 PM): how many ordered pairs (x,y) of positive integers are solutions to 4/x + 2/y = 1? blike again (10:58:30 PM): infinite right? after receiving no response from faf (he was probably frantically working it out), i assumed i was wrong, so i switched my answer which led to the above exchange. as my bio professor always says: "Stick with your first answer and you'll save yourself slipups" Link to comment Share on other sites More sharing options...
fafalone Posted March 2, 2003 Share Posted March 2, 2003 Well actually, :inf: - 4 n:neq:{0,2} n:neq:{:inf:,-:inf:} Link to comment Share on other sites More sharing options...
Dudde Posted March 2, 2003 Author Share Posted March 2, 2003 wow.... I know I shoulda been able to figure that out, guess you can't win them all;) number theory isn't even taught at my school, my teacher just has some sheets lying around and gave me some because I was bored with his class (it was only intermediate algebra) I'll have to figure some of the others out...if I get stumped again I'll be sure to check here first;) thanks! Link to comment Share on other sites More sharing options...
blike Posted March 2, 2003 Share Posted March 2, 2003 did you understand how he arrived at that conclusion? Link to comment Share on other sites More sharing options...
Dudde Posted March 2, 2003 Author Share Posted March 2, 2003 not at all:D I was just working on trying to figure it out;) care to elaborate? Link to comment Share on other sites More sharing options...
blike Posted March 2, 2003 Share Posted March 2, 2003 What faf did was put the equation in terms of x= by manipulating it algebra style, like so: 4/x + 2/y = 1 [multiply everything by x, and y, to get rid of the x and y terms in the denominators] 4y + 2x = 1xy [subract 2x to get x's on one side] 4y = 1xy - 2x [take the common factor on the right side (which is x)] 4y = x(y - 2) [divide by (y-2) to get x alone] 4y / (y-2) = x --- since we now know what X equals, we can write the ordered pair like this: (4y / (y-2), y) ----------------------------- Now that I look at it, however, that answer is not sufficient, because you're asking how many INTEGER ordered pairs. If it were simply ordered pairs, the solutions would be infinite, as the equation indicates. However, i'll have to ge tback to you because its INTEGER ordered pairs. its probably something simple too, and were all just overlooking it Link to comment Share on other sites More sharing options...
Dudde Posted March 2, 2003 Author Share Posted March 2, 2003 AH!! the first part was simple! I guess I've been away from a math equation too long or something....I should've known that:lame: but thanks:D I'll see if I can't work it out myself as well;) Link to comment Share on other sites More sharing options...
Guest e-Monk Posted March 3, 2003 Share Posted March 3, 2003 >> fafaIone: unlimited >> fafaIone: (4n/n-2, n) well, like blike said, the question asks about positive integers, so this is only true when: n-2|4n this begs the question when is such a thing true. by definition, if n-2|4n, then: 4n = k(n-2) for some (positive, in our case) integer k. but, if n>4, then 4n=k(n-2)>kn/2 so it follows that: k<8 so all we have to do is check for satisfying n's for all k's less than 8: k=1: 4n=n - 2 ----> no pos. int. solution K=2: 4n=2n - 4 ----> no pos. int. solution k=3: 4n=3n - 6 ----> no pos. int. solution k=4: 4n=4n - 8 ----> no pos. int. solution k=5: 4n=5n - 10 ------> n = 10 k=6: 4n=6n - 12 ------> n = 6 k=7: 4n=7n - 14 ----> no pos. int. solution But, since we assumed n>4 we have to check all n's up to 4: n=1: 4n/n-2 is negative n=2: 4n/n-2 is division by zero n=3: 4n/n-2 = 12 ; good n=4: 4n/n-2 = 8 ; good so we got 4 positive integer solutions: (5,10) (6,6) (12,3) (8,4) corresponding to n=3,4,6,10 and proved no other gives such a solution. So there are only four, not infinitly many, positive integer solutions. That's a very nice problem, by the way. Is there a chance you could post another one of those? Link to comment Share on other sites More sharing options...
blike Posted March 3, 2003 Share Posted March 3, 2003 That was sexy, e-monk thanks for clearing that up for us. Link to comment Share on other sites More sharing options...
Dudde Posted March 3, 2003 Author Share Posted March 3, 2003 wow...that was awsome... I'm not sure I have any more of those types of problems, but I certainly have more regular ol' number theory problems;) most of which I'm sure wouldn't be too hard..lemme see... how about: solve the differential equation dy/dx with the initial condition y(0)=C Link to comment Share on other sites More sharing options...
Dave Posted March 3, 2003 Share Posted March 3, 2003 Originally posted by Dudde solve the differential equation dy/dx with the initial condition y(0)=C hmm solve the differential equation dy/dx = what? it's not an equation otherwise Link to comment Share on other sites More sharing options...
Dudde Posted March 3, 2003 Author Share Posted March 3, 2003 you understand my hatred of this paper;) I'll post another one I know answerable later, but I'm kind of in school right now:D Link to comment Share on other sites More sharing options...
Dave Posted March 3, 2003 Share Posted March 3, 2003 Originally posted by Dudde you understand my hatred of this paper;) pretty strange paper putting "solve the differential equation dy/dx with initial conditions y = 0" doesn't make any sense whatsoever. mind you, edexcel did put an unsolvable problem on one of the a-level pure maths exam paper, which wasn't really all that clever of them, so nothing ceases to amaze me Link to comment Share on other sites More sharing options...
Dudde Posted March 4, 2003 Author Share Posted March 4, 2003 yeah, the weird thing is that I'm not even in a math class, I just do this stuff for fun, but because I can already do most of the stuff my old teacher would have to give me (I've already borrowed a text book) he gave me some stuff he knew I would have fun with^_^ like this one: for distinct real numbers x and y let P(x,y) be the larger of x and y and let p(x,y) be the smaller of x and y. If a<b<c<d<e, then evaluate P(P(a,p(b,c)), p(d,p(a,e))) I know this is easy, but the last couple times I tried it I was distracted and now my mind is stuck in this wrong answer mode for this... Link to comment Share on other sites More sharing options...
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