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# The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse

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The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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Updated on: 25 Jul 2017, 19:39
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Question Stats:

62% (01:40) correct 38% (03:00) wrong based on 659 sessions

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The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

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Originally posted by carcass on 25 Jul 2017, 10:38.
Last edited by Bunuel on 25 Jul 2017, 19:39, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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26 Jul 2017, 07:05
18
12
$$\sqrt{63 - 36\sqrt{3}} = x + y \sqrt{3}$$

Square both sides:

$$( \sqrt{63 - 36\sqrt{3}} )^2 = (x + y \sqrt{3} )^2 \\ 63 - 36 \sqrt{3} = x^2 + 3y^2 + 2xy \sqrt{3}$$

On the right, the integer part is the x^2 + 3y^2, so that equals 63, and the 2xy√3 is equal to -36√3, so 2xy = -36, and xy = -18.
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##### General Discussion
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Joined: 04 Jan 2015
Posts: 2438
Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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26 Jul 2017, 08:10
7
1
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

$$\sqrt{63-36\sqrt{3}}$$ can be written as

=$$\sqrt{63-2*6*3\sqrt{3}}$$

=$$\sqrt{6^2 + 3\sqrt{3}^2 - 2*6*3\sqrt{3}}$$

=$$\sqrt{(6 - 3\sqrt{3})^2}$$

=$$6 - 3\sqrt{3}$$

Comparing it with $$x + y \sqrt{3}$$

We can say the value of x = 6 and y = -3

Thus, the product of x*y = -18

And the correct answer is Option A.

Thanks,
Saquib
Quant Expert
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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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26 Jul 2017, 09:05
6
4
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

Here is what comes to my mind when I see this question:

$$\sqrt{63-36\sqrt{3}} = x + y \sqrt{3}$$

Now, this is a PS question so I will have a unique value for xy. All I need to do is find one set of values which satisfy this equation.
There is nothing I can compare while there is the sqrt on the left hand side. So let's square both sides.

$$63 - 36\sqrt{3} = x^2 + 3y^2 + 2xy\sqrt{3}$$

The co-efficient of the irrational term has to be equal to on both sides. So
-36 = 2xy
xy = -18

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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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09 Aug 2017, 12:34
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

We can equate the two expressions:

√(63 - 36√3) = x + y√3

Squaring both sides we have:

63 - 36√3 = (x + y√3)^2

63 - 36√3 = (x + y√3)(x + y√3)

63 - 36√3 = x^2 + 2xy√3 + 3y^2

We can equate the terms containing √3 on each side of the equation. We see that:

-36√3 = 2xy√3

-18 = xy

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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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19 Nov 2017, 07:35
EgmatQuantExpert wrote:
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

$$\sqrt{63-36\sqrt{3}}$$ can be written as

=$$\sqrt{63-2*6*3\sqrt{3}}$$

=$$\sqrt{6^2 + 3\sqrt{3}^2 - 2*6*3\sqrt{3}}$$

=$$\sqrt{(6 - 3\sqrt{3})^2}$$

=$$6 - 3\sqrt{3}$$

Comparing it with $$x + y \sqrt{3}$$

We can say the value of x = 6 and y = -3

Thus, the product of x*y = -18

And the correct answer is Option A.

Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

Elegant solution!
But a bit hard to deduct under time pressure
Manager
Joined: 12 Nov 2016
Posts: 137
Concentration: Entrepreneurship, Finance
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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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19 Nov 2017, 07:48
1
VeritasPrepKarishma wrote:
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

Here is what comes to my mind when I see this question:

$$\sqrt{63-36\sqrt{3}} = x + y \sqrt{3}$$

Now, this is a PS question so I will have a unique value for xy. All I need to do is find one set of values which satisfy this equation.
There is nothing I can compare while there is the sqrt on the left hand side. So let's square both sides.

$$63 - 36\sqrt{3} = x^2 + 3y^2 + 2xy\sqrt{3}$$

The co-efficient of the irrational term has to be equal to on both sides. So
-36 = 2xy
xy = -18

Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule? Not in basic GMAT quant rules like GMAT official quantitative review...
I do have difficulties with this concept you have two sides of the equations with two terms on one side and 3 terms on the other side and then it looks like you arbitrarily pick a couple, how about other terms 63 and X^2 and 3y^2? Do they have to be equal and why?
Manager
Joined: 12 Nov 2016
Posts: 137
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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02 Jan 2018, 10:34
Erjan_S wrote:
VeritasPrepKarishma wrote:
carcass wrote:
The number $$\sqrt{63-36\sqrt{3}}$$ can be expressed as $$x + y \sqrt{3}$$ for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27

Here is what comes to my mind when I see this question:

$$\sqrt{63-36\sqrt{3}} = x + y \sqrt{3}$$

Now, this is a PS question so I will have a unique value for xy. All I need to do is find one set of values which satisfy this equation.
There is nothing I can compare while there is the sqrt on the left hand side. So let's square both sides.

$$63 - 36\sqrt{3} = x^2 + 3y^2 + 2xy\sqrt{3}$$

The co-efficient of the irrational term has to be equal to on both sides. So
-36 = 2xy
xy = -18

Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule? Not in basic GMAT quant rules like GMAT official quantitative review...
I do have difficulties with this concept you have two sides of the equations with two terms on one side and 3 terms on the other side and then it looks like you arbitrarily pick a couple, how about other terms 63 and X^2 and 3y^2? Do they have to be equal and why?
VeritasPrepKarishma
Bunuel
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1323
The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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02 Jan 2018, 14:58
6
1
Erjan_S wrote:
Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule?

You'd never need to mathematically prove that's true on the GMAT, but it is true. I can explain, but test takers don't need to understand how to do this in order to solve GMAT questions:

We arrived at this:

$$63 - 36\sqrt{3} = x^2 + 3y^2 + 2xy \sqrt{3}$$

x and y are integers, so x^2 + 3y^2 is just some integer, and 2xy is also some integer. So let's just let a = x^2 + 3y^2, and b = 2xy, to make this all easier to look at. Then a and b are integers and we have

$$63 - 36\sqrt{3} = a + b \sqrt{3}$$

Now the number √3 is what is called an 'irrational number', which means it is impossible to write √3 as some fraction c/d where c and d are both integers (I can prove that √3 is irrational separately if anyone is interested). Let's take the equation above, and write it with √3 on one side:

\begin{align} 63 - 36\sqrt{3} &= a + b \sqrt{3} \\ 63 - a &= b \sqrt{3} + 36 \sqrt{3} \\ 63 - a &= \sqrt{3} (b + 36) \\ \frac{63 - a}{b + 36} &= \sqrt{3} \end{align}

But there's something wrong here: 63-a and b+36 are both integers. If this equation were right, then we would have just written √3 as a fraction using two integers, but we know that's impossible to do since √3 is an irrational number. So there must be something wrong with our solution, and the only thing that might be wrong is that we might have divided by zero in the second last line. So b+36 must be equal to 0 (and similarly 63-a must be equal to 0). From there we find that b = -36, so 2xy = -36, and xy = -18.

That's how you can prove that in similar equations, the rational parts and irrational parts must be equal. So if you saw, say, this equation:

5 + 10√2 = e + f + (m + p)√2

where all the letters are integers, then it would need to be true that e+f = 5, and m +p = 10.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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Location: Pune, India
Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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02 Jan 2018, 23:06
3
Erjan_S wrote:
Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule? Not in basic GMAT quant rules like GMAT official quantitative review...
I do have difficulties with this concept you have two sides of the equations with two terms on one side and 3 terms on the other side and then it looks like you arbitrarily pick a couple, how about other terms 63 and X^2 and 3y^2? Do they have to be equal and why?
VeritasPrepKarishma
Bunuel

Ian has already explained very well why this will be true.
I will just add why it makes intuitive sense to me.

An irrational number is one which cannot be constructed from ratio of integers. So I cannot have an expression with just integers equal to an irrational number.
$$\sqrt{3}$$ cannot be (ab + c)/d etc etc where all are integers. There are no building blocks for irrational numbers among integers.

So if I have a $$\sqrt{3}$$ on the left hand side, I must have one on the right hand side too.
If I have 4 $$\sqrt{3}$$s on the left hand side, I must have 4 $$\sqrt{3}$$s on the right hand side too. That is why I can equate the co-efficients.
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The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse  [#permalink]

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07 Jan 2019, 06:34
I think it's important to note that once we get to here:

$$63 - 36 \sqrt{3} = x^2 + 3y^2 + 2xy \sqrt{3}$$

The reason we can set $$-36 \sqrt{3}$$ equal to $$2xy \sqrt{3}$$ is because we know from the question that $$x$$ and $$y$$ are integers. If we didn't know that $$x$$ and $$y$$ were integers, then we would have more algebra to do and several more possibilities for potential solutions. For example, $$x = 3$$ and $$y = \sqrt{3} - 3$$ would be correct solutions to the equation.

Also, I think EgmatQuantExpert posted a very clever solution. However, I think parentheses are helpful around $$3\sqrt{3}$$ in line 3. I have added those below:

1. $$\sqrt{63-36\sqrt{3}}$$ can be written as

2. $$\sqrt{63-2*6*3\sqrt{3}}$$

3. $$\sqrt{6^2 + (3\sqrt{3})^2 - 2*6*3\sqrt{3}}$$

4. $$\sqrt{(6 - 3\sqrt{3})^2}$$

5. $$6 - 3\sqrt{3}$$
The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse &nbs [#permalink] 07 Jan 2019, 06:34
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