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x and y are positive integers

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phoenixio
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x and y are positive integers [#permalink] New post   26 Aug 2016, 07:38
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x and y are positive integers such that x < y. If \(6\sqrt{6}= x\sqrt{y}\), then xy could equal

A)36
B)48
C)54
D)96
E)108
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E
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phoenixio
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Re: x and y are positive integers [#permalink] New post   26 Aug 2016, 09:59
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6\(\sqrt{6}\)= 2*3\(\sqrt{6}\)= 3\(\sqrt{24}\)
which means x=3 and y= 24, therefore xy=72

or

6\(\sqrt{6}\)= 2*3\(\sqrt{6}\)= 2\(\sqrt{54}\)
which means x=2 and y= 54, therefore xy=108 (option E)
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yasir9909
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Re: x and y are positive integers [#permalink] New post   26 Aug 2016, 09:35
Could you please share detailed solution of this question?
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GreenlightTestPrep
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Re: x and y are positive integers [#permalink] New post   29 Aug 2016, 14:31
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phoenixio wrote:
x and y are positive integers such that x < y. If \(6\sqrt{6}= x\sqrt{y}\), then xy could equal

A)36
B)48
C)54
D)96
E)108


First, notice that 6√6 = (√36)(√6) = √216
How many different squares are "hiding" in 216?

Well, 216 = (1)(216), and 1 is a perfect square.
So, we can write: 6√6 = √216 = (√1)(√216) = 1√216 = x√y. So, xy = (1)(216) = 216 (NOT AMONG THE ANSWER CHOICES)

Also, 216 = (4)(54), and 4 is a perfect square.
So, we can write: 6√6 = √216 = (√4)(√54) = 2√54 = x√y. So, xy = (2)(54) = 108 ...AMONG THE ANSWER CHOICES!

Also, 216 = (9)(24), and 9 is a perfect square.
So, we can write: 6√6 = √216 = (√9)(√24) = 3√24 = x√y. So, xy = (3)(24) = 72 (NOT AMONG THE ANSWER CHOICES)

Answer:
Show: ::
E


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Re: x and y are positive integers [#permalink] New post   27 Jan 2019, 05:28
phoenixio wrote:
x and y are positive integers such that x < y. If \(6\sqrt{6}= x\sqrt{y}\), then xy could equal

A)36
B)48
C)54
D)96
E)108


Here's another approach:
GIVEN: (6)(√6) = (x)(√y)
Rewrite 6 as √36 to get: (√36)(√6) = (x)(√y)
Rewrite x as √(x²) to get: (√36)(√6) = √(x²)(√y)
Simplify both sides to get: √216 = √(x²y)
From this, we can conclude that 216 = x²y

If x and y are positive integers, what are some possible values of x and y?
To help us with this, let's find the prime factorization of 216
216 = (2)(2)(2)(3)(3)(3)

So, we can write: (2)(2)(2)(3)(3)(3) = x²y
Now it's a matter of looking for possible values of x and y that meet the above condition.

Here's one possibility: 216 = (9)(24) = (3²)(24)
In other words, x = 3 and y = 24
In this case, xy = (3)(24) = 72. 72 is NOT among the answer choices.
KEEP LOOKING!

Here's another possibility: 216 = (4)(54) = (2²)(54)
In other words, x = 2 and y = 54
In this case, xy = (2)(54) = 108.
108 IS among the answer choices!

Answer: E

Cheers,
Brent
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Paul121
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Re: x and y are positive integers [#permalink] New post   31 Dec 2022, 13:09
In all your solutions, I couldn't find why it cannot be A.
(x^2)*y = 36*6 = 216
Here we can get x = 6, y= 6.
6 * 6 * 6 = 216.
And xy=36
I don't see that x and y have to be different values.



GreenlightTestPrep wrote:
phoenixio wrote:
x and y are positive integers such that x < y. If \(6\sqrt{6}= x\sqrt{y}\), then xy could equal

A)36
B)48
C)54
D)96
E)108


First, notice that 6√6 = (√36)(√6) = √216
How many different squares are "hiding" in 216?

Well, 216 = (1)(216), and 1 is a perfect square.
So, we can write: 6√6 = √216 = (√1)(√216) = 1√216 = x√y. So, xy = (1)(216) = 216 (NOT AMONG THE ANSWER CHOICES)

Also, 216 = (4)(54), and 4 is a perfect square.
So, we can write: 6√6 = √216 = (√4)(√54) = 2√54 = x√y. So, xy = (2)(54) = 108 ...AMONG THE ANSWER CHOICES!

Also, 216 = (9)(24), and 9 is a perfect square.
So, we can write: 6√6 = √216 = (√9)(√24) = 3√24 = x√y. So, xy = (3)(24) = 72 (NOT AMONG THE ANSWER CHOICES)

Answer:
Show: ::
E


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GreenlightTestPrep
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Re: x and y are positive integers [#permalink] New post   31 Dec 2022, 13:27
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Paul121 wrote:
In all your solutions, I couldn't find why it cannot be A.
(x^2)*y = 36*6 = 216
Here we can get x = 6, y= 6.
6 * 6 * 6 = 216.
And xy=36
I don't see that x and y have to be different values.


The condition that x < y implies x and y must be different.
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Paul121
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Re: x and y are positive integers [#permalink] New post   31 Dec 2022, 13:32
1
I missed that part, sorry and thanks for the quick answer.

GreenlightTestPrep wrote:
Paul121 wrote:
In all your solutions, I couldn't find why it cannot be A.
(x^2)*y = 36*6 = 216
Here we can get x = 6, y= 6.
6 * 6 * 6 = 216.
And xy=36
I don't see that x and y have to be different values.


The condition that x < y implies x and y must be different.
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Re: x and y are positive integers [#permalink] New post   02 Sep 2024, 06:53
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Re: x and y are positive integers [#permalink]
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