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# x^x * y^y = 24^3 / 2

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Intern
Joined: 09 Sep 2016
Posts: 35
Location: Georgia
GPA: 3.75
WE: Analyst (Investment Banking)
x^x * y^y = 24^3 / 2  [#permalink]

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26 Feb 2017, 12:04
3
9
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:07) correct 33% (02:21) wrong based on 164 sessions

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if $$x^x * y^y = \frac{24^3}{2}$$ , and x and y are integers such that x < y, what is the value of y - x?

a) -1
b) 1
c) 3
D) 5
e) 7

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Marshall & McDonough Moderator
Joined: 13 Apr 2015
Posts: 1677
Location: India
Re: x^x * y^y = 24^3 / 2  [#permalink]

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27 Feb 2017, 23:13
x^x * y^y = ((2^3)^3 * 3^3)/2 = 2^8 * 3^3 = (2^2)^4 * 3^3 = 4^4 * 3^3

x < y --> x = 3 and y = 4 --> y - x = 1

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Re: x^x * y^y = 24^3 / 2  [#permalink]

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28 Feb 2017, 01:11
1
1
giobas wrote:
if $$x^x * y^y = \frac{24^3}{2}$$ , and x and y are integers such that x < y, what is the value of y - x?

a) -1
b) 1
c) 3
D) 5
e) 7

Hit Kudos if you liked the question

Since this is a PS question, we know that there will be only one value for y - x.
Since x and y are integers, divide out the 2 on the right hand side.

$$x^x * y^y = \frac{24*24*24}{2}$$

$$x^x * y^y = 12*24*24$$

$$x^x * y^y = 4*3*8*3*8*3$$

$$x^x * y^y = 4^4*3^3$$

Since x < y, x = 3 and y = 4

So y - x = 4 - 3 = 1

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Karishma
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Re: x^x * y^y = 24^3 / 2  [#permalink]

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28 Feb 2017, 09:28
giobas wrote:
if $$x^x * y^y = \frac{24^3}{2}$$ , and x and y are integers such that x < y, what is the value of y - x?

a) -1
b) 1
c) 3
D) 5
e) 7

Hit Kudos if you liked the question

$$x^x * y^y = \frac{3^3*2^9}{2}$$

Or, $$x^x * y^y = 3^3*2^8$$

Or, $$x^x * y^y =3^3* 4^4$$

So, $$x = 4 & y = 3$$

Thus, the value of $$y - x = 4 - 3 = 1$$

Hence, answer must be (B) 1
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Intern
Joined: 10 Jan 2017
Posts: 2
x^x * y^y = 24^3 / 2  [#permalink]

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28 Feb 2017, 12:34
It is NOT mentioned that x and y are positive integers!

So, after simplifying the eq, x can be -4 (x < y)

y - x = 3 - (-4) = 7 (Option E)
Veritas Prep GMAT Instructor
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Re: x^x * y^y = 24^3 / 2  [#permalink]

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01 Mar 2017, 01:04
Love2gmail wrote:
It is NOT mentioned that x and y are positive integers!

So, after simplifying the eq, x can be -4 (x < y)

y - x = 3 - (-4) = 7 (Option E)

You are given that x < y.

If x is 3 and y is -4, this condition is not met.
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Karishma
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x^x * y^y = 24^3 / 2  [#permalink]

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01 Mar 2017, 12:06
VeritasPrepKarishma wrote:
Love2gmail wrote:
It is NOT mentioned that x and y are positive integers!

So, after simplifying the eq, x can be -4 (x < y)

y - x = 3 - (-4) = 7 (Option E)

You are given that x < y.

If x is 3 and y is -4, this condition is not met.

x can be less than y in 2 cases

Case 1:

x^x * y^y = 3^3 * 4^4

which gives us y - x = 4 - 3 = 1

Case 2:

x^x * y^y = (-4)^4 * 3^3

here, it satisfies x < y

and power 4 of a -ve will be a +ve

which gives us y - x = 3 - (-4) = 7

we need extra information that x & y are positive integers.
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Re: x^x * y^y = 24^3 / 2  [#permalink]

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07 Mar 2017, 17:12
giobas wrote:
if $$x^x * y^y = \frac{24^3}{2}$$ , and x and y are integers such that x < y, what is the value of y - x?

a) -1
b) 1
c) 3
D) 5
e) 7

Let’s simplify the given equation:

(x^x)(y^y) = 24^3/2

(x^x)(y^y) = [(2^3 * 3)^3]/2

(x^x)(y^y) = (2^9)(3^3)/2

(x^x)(y^y) = (2^8)(3^3)

(x^x)(y^y) = ((2^2)^4)(3^3)

(x^x)(y^y) = (4^4)(3^3)

Since x < y, x = 3 and y = 4, and y - x = 4 - 3 = 1.

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Re: x^x * y^y = 24^3 / 2   [#permalink] 07 Mar 2017, 17:12
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