Exponents/Powers

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Exponents/Powers

by swerve » Sat May 09, 2020 1:31 pm

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Which of the following, if multiplied by \(x^3 y^2 z\), would yield a product that is both a perfect cube and a perfect power of five?

A. \(x^5y^{15}z^{15}\)
B. \(x^{12} y^{13} z^{14}\)
C. \(x^2 y z^2\)
D. \(x^2 y^3 z^4\)
E. \(y z^2\)

The OA is B

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Re: Exponents/Powers

by GMATGuruNY » Sat May 09, 2020 2:36 pm
The prompt should indicate that x, y and z are PRIME NUMBERS.
swerve wrote:
Sat May 09, 2020 1:31 pm
Which of the following, if multiplied by \(x^3 y^2 z\), would yield a product that is both a perfect cube and a perfect power of five?

A. \(x^5y^{15}z^{15}\)
B. \(x^{12} y^{13} z^{14}\)
C. \(x^2 y z^2\)
D. \(x^2 y^3 z^4\)
E. \(y z^2\)
For an integer to be a perfect cube, every exponent in its prime-factorization must be a MULTIPLE OF 3.
For an integer to be a perfect power of 5 -- also known as a perfect fifth -- every exponent in its prime-factorization must be a MULTIPLE OF 5.
Implication:
When the correct answer is multiplied by x³y²z, every exponent in the resulting product must be a multiple of both 3 and 5.
In other words, every exponent in the resulting product must be a MULTIPLE OF 15.
Only B works:
x¹³y¹²z¹⁴ * x²y³z = x¹⁵y¹⁵z¹⁵.

The correct answer is B.
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