## Romanian Baccalaoureat-Question

IMO, IMC, USAMO, TST, Putnam, post your own math challenges to the group.

### Romanian Baccalaoureat-Question

Define a relation * on $\mathbb{R}$, such that $x*y=(x-3)(y-3)+3$ which satisfies the associative property. Compute

$\hspace{370pt}\fbox{\Large\sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011}}$

.
Last edited by ZardoZ on Sat Jan 21, 2012 3:54 pm, edited 1 time in total.

ZardoZ
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### 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.
Erimess wrote:I get students who have a number of units that were sold and a unit sales price, and don't know how to get the total sales amount. They need a "formula."

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CherryPi
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### 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\\
\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.
Erimess wrote:I get students who have a number of units that were sold and a unit sales price, and don't know how to get the total sales amount. They need a "formula."

My Math Forum: Where you learn more math than you did in school.
CherryPi
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### Re: Romanian Baccalaoureat-Question

CherryPi wrote: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\\
\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.

In some parts I can't follow you.

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

ZardoZ
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### Re: Romanian Baccalaoureat-Question

The reasoning of CherryPi was correct. Here the solution using the hint.

$\sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{27}*\;\ldots\;*\sqrt[3]{2011}=\{ \sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011}\}*\sqrt[3]{27}$
$=\{ \sqrt[3]{1}*\sqrt[3]{2}*\sqrt[3]{3}*\;\ldots\;*\sqrt[3]{2011} \} * 3 =\; 3$
wnvl

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### Re: Romanian Baccalaoureat-Question

ZardoZ wrote: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:

CherryPi wrote:$\alpha_{i=1}^{n}u_i=\left(\prod_{i=1}^{n}(u_i-3)\right)+3$
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\\
\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. Pardon me.)
Erimess wrote:I get students who have a number of units that were sold and a unit sales price, and don't know how to get the total sales amount. They need a "formula."

My Math Forum: Where you learn more math than you did in school.
CherryPi
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### 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.
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greg1313
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### 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.
Erimess wrote:I get students who have a number of units that were sold and a unit sales price, and don't know how to get the total sales amount. They need a "formula."

My Math Forum: Where you learn more math than you did in school.
CherryPi
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Posts: 245
Joined: Wed Jul 06, 2011 6:01 am