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Title: MATHS STUFF Post by NICOLA on Jan 1st, 2005, 2:46pm f I buy a certain 4 items priced at: $1.20 $1.25 $1.50 $3.16 - To get the total of these figures, it does not matter if the prices are added together as one would expect or if the prices are multiplied. The total bill will be the same: $7.11. What mathematical principle is being displayed in this problem? PLEASE HELP AS SOON AS POSSIBLE, I NEED UR BRAINSSS |
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Title: Re: MATHS STUFF Post by Sir Col on Jan 1st, 2005, 3:19pm [e]Edited because I just re-read my post and I felt that I was somewhat rude with my comments. Sorry, Nicola![/e] Isn't this is about the third time we've seen this question? Anyway, I guess this goes part of the way to answering your question... The example displays one of infinitely many solutions to the equation: abcd=a+b+c+d. Writing abcd-a=b+c+d, a(bcd-1)=b+c+d, so we get a=(b+c+d)/(bcd-1). In other words, pick ANY values you like for b,c,d and it will give you the necessary value for a to make it work (as long as bcd[ne]1). For example, b=1, c=2, d=3, so a=(b+c+d)/(bcd-1)=6/5=1.2 That is, 1.2+1+2+3=1.2x1x2x3=7.2. |
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Title: Re: MATHS STUFF Post by Nicola on Jan 1st, 2005, 3:41pm You are a complete loser lmao |
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Title: Re: MATHS STUFF Post by THUDandBLUNDER on Jan 1st, 2005, 3:48pm Quote:
b = 1.11 c = 1.17 d = 2.13 Exact change please! on 01/01/05 at 15:41:22, Nicola wrote:
You should do your own homework! |
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Title: Re: MATHS STUFF Post by Sir Col on Jan 1st, 2005, 4:18pm ;D on 01/01/05 at 15:48:59, THUDandBLUNDER wrote:
Good point! I suppose that I should give the original question poser some credit in that the set of numbers were chosen in such a way that bcd-1 divided b+c+d. I'm almost tempted to investigate a set of terminating decimal forms for four variables. The two variable version, ab=a+b is fairly trivial: b=a/(a-1), or writing a=(m+n)/n, b=(m+n)/m. Has anyone else explored it with more than two variables? |
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Title: Re: MATHS STUFF Post by Sir Col on Jan 1st, 2005, 4:32pm on 01/01/05 at 15:41:22, Nicola wrote:
I cannot be a complete loser: either I am a loser or not; it is a tautology. Although in your defence, I suppose the word, complete, could be part of an ellipsis for my state of preparation for losing: before I was not quite there, whereas now I am fully qualified to be a loser. In which case your whole statement is a pleonasm: nothing we didn't already know has been said. Did you really want to say, LMAO? That would suggest you or I said something amusing, and as we've just established that you said nothing, it must have been me. Why, thank you very much, Nicola! :-* |
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Title: Re: MATHS STUFF Post by THUDandBLUNDER on Jan 2nd, 2005, 12:23am on 01/01/05 at 16:18:39, Sir Col wrote:
Well, that is not (cough) completely correct. If a = 4, b = 1.333333333333333333333333333333333... We need to find integers n and m such that nm = 100(n + m) ;) |
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Title: Re: MATHS STUFF Post by Sir Col on Jan 2nd, 2005, 3:17am Actually I wasn't completely wrong either. :P That is why I qualified it with writing a=(m+n)/n and b=(m+n)/m. As long as m and n are of the form 2j5k, we have terminating decimal types. For example, m=2350=8, n=2-252=6.25, a=1.78125, b=2.28; and sure enough, 1.78125+2.28=1.78125*2.28=4.06125. |
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Title: Re: MATHS STUFF Post by THUDandBLUNDER on Jan 2nd, 2005, 6:48am Quote:
But, sure enough, 100ab is not an integer. Even 10000ab is not an integer! nm = 100(n + m) So n = 100m/(m - 100) Hence n,m = 100 + 2j5k [smiley=leqslant.gif] 10100 and a,b = 1 + (2j5k/100) [smiley=leqslant.gif] 101 In fact, we can express the solutions parametrically as a = 1 + v b = 1 + 1/v where 1/100 [smiley=leqslant.gif] v = 2j5k/100 [smiley=leqslant.gif] 100 Then ab = a + b = (1 + v)2/v There ought to be corresponding expressons for 3 and more variables. With N variables we need for i = 1 to N 100xi [in] [bbz]+ and 100[smiley=prod.gif]xi [in] [bbz]+ Given the first constraint, the chance of satisfying the second constraint decreases as N increases. |
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Title: Re: MATHS STUFF Post by Sir Col on Jan 2nd, 2005, 7:49am Oops! Sorry, I didn't appreciate that we were talking money (2 d.p.), I just thought we were trying to avoid recurring decimals. :-[ |
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Title: Re: MATHS STUFF Post by THUDandBLUNDER on Jan 2nd, 2005, 8:00am on 01/02/05 at 07:49:51, Sir Col wrote:
I was never talking anything else. Have you [completely] lost it? ;D |
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Title: Re: MATHS STUFF Post by JocK on Jan 3rd, 2005, 2:01am on 01/01/05 at 14:46:56, NICOLA wrote:
I wouldn't refer to it as a number theoretical 'principle', but the following observation can be made: For positive integer values of the parameter N, the equation [prod]k=1..K Xk/N = [sum]k=1..K Xk/N has at least one integer solution in X1, X2, .. , XK-2, XK-1, XK: N, N, .. , N, N+1, N(KN-N+1) For N=1 and K=2, 3 or 4 this is the only solution: 2 [times] 2 = 2 + 2 1 [times] 2 [times] 3 = 1 + 2 + 3 1 [times] 1 [times] 2 [times] 4 = 1 + 1 + 2 + 4 However, for larger N and/or larger K the number of solutions increase. For N=2 and K=2 we obtain two solutions: 4/2 [times] 4/2 = 4/2 + 4/2 6/2 [times] 3/2 = 6/2 + 3/2 and for N=1 and K=5 a total of three solutions: 1 [times] 1 [times] 1 [times] 2 [times] 5 = 1 + 1 + 1 + 2 + 5 1 [times] 1 [times] 1 [times] 3 [times] 3 = 1 + 1 + 1 + 3 + 3 1 [times] 1 [times] 2 [times] 2 [times] 2 = 1 + 1 + 2 + 2 + 2 For N=100 and K=2 the number of solutions is 13. For N=100 and K=3 the number of solution amounts to 31. For N=100 and K=4 one may expect many tens of solutions. Apart from the standard solution: 100/100 [times] 100/100 [times] 101/100 [times] 30100/100 = 100/100 + 100/100 + 101/100 + 30100/100 another solution happens to be: 120/100 [times] 125/100 [times] 150/100 [times] 316/100 = 120/100 + 125/100 + 150/100 + 316/100 |
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