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Re: x and y are positive integers [#permalink]
2
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) = 1216 = xy. 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) = 254 = xy. 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) = 324 = xy. 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]
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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Re: x and y are positive integers [#permalink]
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) = 1216 = xy. 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) = 254 = xy. 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) = 324 = xy. 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]
1
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]
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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