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Topic: MATHS STUFF (Read 636 times) |
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NICOLA
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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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Sir Col
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Re: MATHS STUFF
« Reply #1 on: Jan 1st, 2005, 3:19pm » |
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[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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| « Last Edit: Jan 6th, 2005, 11:46am by Sir Col » |
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Nicola
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You are a complete loser lmao
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ThudnBlunder
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Re: MATHS STUFF
« Reply #3 on: Jan 1st, 2005, 3:48pm » |
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Quote:| 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... |
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b = 1.11
c = 1.17
d = 2.13
Exact change please!
on Jan 1st, 2005, 3:41pm, Nicola wrote:| You are a complete loser lmao |
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You should do your own homework!
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| « Last Edit: Jan 1st, 2005, 10:55pm by ThudnBlunder » |
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Sir Col
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Re: MATHS STUFF
« Reply #4 on: Jan 1st, 2005, 4:18pm » |
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on Jan 1st, 2005, 3:48pm, 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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| « Last Edit: Jan 1st, 2005, 4:59pm by Sir Col » |
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Sir Col
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Re: MATHS STUFF
« Reply #5 on: Jan 1st, 2005, 4:32pm » |
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on Jan 1st, 2005, 3:41pm, Nicola wrote:| You are a complete loser lmao |
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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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ThudnBlunder
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Re: MATHS STUFF
« Reply #6 on: Jan 2nd, 2005, 12:23am » |
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on Jan 1st, 2005, 4:18pm, Sir Col wrote:| The two variable version, ab=a+b is fairly trivial: b=a/(a-1), or writing a=(m+n)/n, b=(m+n)/m. |
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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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| « Last Edit: Jan 2nd, 2005, 7:41am by ThudnBlunder » |
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Sir Col
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Re: MATHS STUFF
« Reply #7 on: Jan 2nd, 2005, 3:17am » |
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Actually I wasn't completely wrong either. 
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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ThudnBlunder
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Re: MATHS STUFF
« Reply #8 on: Jan 2nd, 2005, 6:48am » |
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Quote:| 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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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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| « Last Edit: Jan 2nd, 2005, 9:26pm by ThudnBlunder » |
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Sir Col
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Re: MATHS STUFF
« Reply #9 on: Jan 2nd, 2005, 7:49am » |
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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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ThudnBlunder
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Re: MATHS STUFF
« Reply #10 on: Jan 2nd, 2005, 8:00am » |
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on Jan 2nd, 2005, 7:49am, Sir Col wrote:| Oops! Sorry, I didn't appreciate that we were talking money. |
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I was never talking anything else. Have you [completely] lost it?
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| « Last Edit: Jan 2nd, 2005, 12:09pm by ThudnBlunder » |
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JocK
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Re: MATHS STUFF
« Reply #11 on: Jan 3rd, 2005, 2:01am » |
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on Jan 1st, 2005, 2:46pm, NICOLA wrote:
What mathematical principle is being
displayed in this problem?
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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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| « Last Edit: Jan 3rd, 2005, 2:43am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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