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If x^xy^y = 24^3/2, and x and y are integers such that x < y, what is

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If x^x*y^y = 24^3/2, and x and y are integers such that x < y, what [#permalink] New post   26 Feb 2017, 12:04
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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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B
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KarishmaB
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Re: If x^x*y^y = 24^3/2, and x and y are integers such that x < y, what [#permalink] New post   28 Feb 2017, 01:11
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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

Answer (B)
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Re: If x^x*y^y = 24^3/2, and x and y are integers such that x < y, what [#permalink] New post   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)
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Re: If x^x*y^y = 24^3/2, and x and y are integers such that x < y, what [#permalink] New post   01 Mar 2017, 01:04
Expert Reply
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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Re: If x^x*y^y = 24^3/2, and x and y are integers such that x < y, what [#permalink] New post   07 Mar 2017, 17:12
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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.

Answer: B
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Re: If x^xy^y = 24^3/2, and x and y are integers such that x < y, what is [#permalink] New post   17 Mar 2020, 07:54
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Bunuel wrote:
If \(x^xy^y = \frac{24^3}{2}\), 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


\(x^x*y^y=\frac{(2^3*3)^{3}}{2}\)

\(x^x*y^y=(2^9*3^3)/2\)
\(x^x*y^y=(2^8*3^3)\)
\(x^x*y^y=(4^4*3^3)\)
x = 3 y =4
y-x=1
Ans: B
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Re: If x^xy^y = 24^3/2, and x and y are integers such that x < y, what is [#permalink] New post   17 Mar 2020, 22:22
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Bunuel wrote:
If \(x^xy^y = \frac{24^3}{2}\), 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


Solution



    • Prime factorization of \(24 = 2^3*3\)
    • \(x^xy^y = \frac{(2^3*3)^3}{2} = \frac{2^9*3^3}{2} = 2^8*3^3 = 4^4*3^3\)
    • Now comparing both sides of the above equation, we get,
      o \(x = 3\) and \(y = 4\) (since x < y)
    • Therefore, \(y-x = 4-3 = 1\)
Thus, the correct answer is Option B.
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Re: If x^xy^y = 24^3/2, and x and y are integers such that x < y, what is [#permalink] New post   27 Oct 2023, 01:26
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Re: If x^xy^y = 24^3/2, and x and y are integers such that x < y, what is [#permalink]
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