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 January 20th, 2012, 03:42 PM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 132 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Romanian Baccalaoureat-Question [color=#000000]Define a relation * on $\mathbb{R}$, such that $x*y=(x-3)(y-3)+3$ which satisfies the associative property. Compute $\hspace{370pt}\fbox{\Large\sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011}}$ .[/color]
 January 20th, 2012, 04:09 PM #2 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Romanian Baccalaoureat-Question I don't think this question is entirely clear. Is this to be taken to mean: $a_1^{\frac{1}{3}}*a_2^{\frac{1}{3}}*\ldots*a_{2011 }^{\frac{1}{3}}$ Where: $a_i=i$ (Implying that the 2nd two is just a typo?) Or are you going for some wicked sequence that goes like this: $\{1,2,2,3,3,3, \ldots\}$ where each number is repeated its own amount of times? I presume that's not the case, as that'd be on the scale of a Euler problem from project Euler. xD Edit: Offtopic, but you might be interested in this: http://projecteuler.net/ It is primarily computer-esque and programmer-friendly problems, but they are quite interesting.
 January 20th, 2012, 06:20 PM #3 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Romanian Baccalaoureat-Question Okay, so I know that * is commutative as well: $x*y=(x-3)(y-3)+3=(y-3)(x-3)+3=y*x$ OMG, I just made an awesome result: Define the following operator $\alpha$ such that: $\alpha_{i=1}^{n}u_i=u_1*u_2*u_3*\, \ldots \,*u_n$ $\alpha_{i=1}^{n}u_i=\left(\prod_{i=1}^{n}(u_i-3)\right)+3$ This follows cleverly from the definition of *: \begin{align} u_1*u_2&=(u_1-3)(u_2-3)+3\\ (u_1*u_2)*u_3&=(((u_1-3)(u_2-3)+3)-3)(u_3-3)+3\\ &=(u_1-3)(u_2-3)(u_3-3)+3\\ \text{ad infinitum. . .} \end{align} You can get a quick answer to the original problem by making the various substitutions: $(u_i=i^{\frac{1}{3}}, \, n=2011) \Rightarrow (\alpha_{i=1}^{2011}i^{\frac{1}{3}}=\left(\prod_{i =1}^{2011}\left(i^{\frac{1}{3}}-3\right)\right)+3)$ W|A can evaluate that product: http://www.wolframalpha.com/input/?i=%5 ... 7D%7D-3%29 So, $\alpha_{i=1}^{2011}i^{\frac{1}{3}}=3$ Wow, my number sense is retarded. I just realized that the product is 0 because at $n=27$, the product produces a 0. From that point on, it is forever 0. Hence the above result. edit: This is all assuming you made a typo.
January 20th, 2012, 06:48 PM   #4
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Re: Romanian Baccalaoureat-Question

Quote:
 Originally Posted by CherryPi Okay, so I know that * is commutative as well: $x*y=(x-3)(y-3)+3=(y-3)(x-3)+3=y*x$ OMG, I just made an awesome result: Define the following operator $\alpha$ such that: $\alpha_{i=1}^{n}u_i=u_1*u_2*u_3*\, \ldots \,*u_n$ $\alpha_{i=1}^{n}u_i=\left(\prod_{i=1}^{n}(u_i-3)\right)+3$ This follows cleverly from the definition of *: \begin{align} u_1*u_2&=(u_1-3)(u_2-3)+3\\ (u_1*u_2)*u_3&=(((u_1-3)(u_2-3)+3)-3)(u_3-3)+3\\ &=(u_1-3)(u_2-3)(u_3-3)+3\\ \text{ad infinitum. . .} \end{align} You can get a quick answer to the original problem by making the various substitutions: $(u_i=i^{\frac{1}{3}}, \, n=2011) \Rightarrow (\alpha_{i=1}^{2011}i^{\frac{1}{3}}=\left(\prod_{i =1}^{2011}\left(i^{\frac{1}{3}}-3\right)\right)+3)$ W|A can evaluate that product: http://www.wolframalpha.com/input/?i=%5 ... 7D%7D-3%29 So, $\alpha_{i=1}^{2011}i^{\frac{1}{3}}=3$ Wow, my number sense is retarded. I just realized that the product is 0 because at $n=27$, the product produces a 0. From that point on, it is forever 0. Hence the above result. edit: This is all assuming you made a typo.
[color=#000000]
In some parts I can't follow you.

Hint: use the fact that $x*3=3*x=3$ for all $x\in\mathbb{R}$.[/color]

 January 21st, 2012, 06:52 AM #5 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Romanian Baccalaoureat-Question The reasoning of CherryPi was correct. Here the solution using the hint. $\sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{27}*\;\ldots\;*\sqrt[3]{2011}=\{ \sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011}\}*\sqrt[3]{27}$ $=\{ \sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011} \} * 3 =\; 3$
January 21st, 2012, 10:56 AM   #6
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Re: Romanian Baccalaoureat-Question

Quote:
 Originally Posted by ZardoZ In some parts I can't follow you.
It just relies on looking at what the operation * does.

When you perform the operation $n$ times, you result in a product with just a 3 on the outside. That's what I meant here:

Quote:
 Originally Posted by CherryPi $\alpha_{i=1}^{n}u_i=\left(\prod_{i=1}^{n}(u_i-3)\right)+3$ This follows cleverly from the definition of *: \begin{align} u_1*u_2&=(u_1-3)(u_2-3)+3\\ (u_1*u_2)*u_3&=(((u_1-3)(u_2-3)+3)-3)(u_3-3)+3\\ &=(u_1-3)(u_2-3)(u_3-3)+3\\ \text{ad infinitum. . .} \end{align}
Look closely: You can see that
$u_1*u_2*u_3*\,\ldots\,*u_{(n-2)}*u_{(n-1)}*u_n=\left(\prod_{i=1}^{n}(u_i-3)\right)+3$
It's just a pattern that follows from the operation's definition. (The "ad infinitum" part was particularly important. It meant that, if you kept doing this process over and over, you'd get the precise result involving the product and the addition of the 3 on the outside. It was also me being a bit lazy. :P Pardon me.)

 January 21st, 2012, 03:13 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,476 Thanks: 886 Math Focus: Elementary mathematics and beyond Re: Romanian Baccalaoureat-Question That's pretty good, CherryPi. I won't say I understand it in fine detail, but I think I do get the general idea, which is more than what I had when I first read the topic.
 January 21st, 2012, 04:06 PM #8 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Romanian Baccalaoureat-Question Thanks, Greg. I think it was easier to solve this generally, but it was kinda unilluminating until I actually related the general solution back to the problem.

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