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Bunuel
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If (90x)^(1/3) and (y/75)^(1/2) are positive integers, and both x and 06 Jul 2020, 04:51
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C
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25% (medium)

Question Stats:

83% (02:31) correct 17% (02:36) wrong based on 388 sessions

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If \(\sqrt[3]{90x}\) and \(\sqrt{\frac{y}{75}}\) are positive integers, and both x and y are integers, what is the least possible value of \(xy\) ?

A. 20500
B. 21500
C. 22500
D. 23500
E. 24500
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C
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If (90x)^(1/3) and (y/75)^(1/2) are positive integers, and both x and 06 Jul 2020, 04:57
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³√90x and √y/75 are positive integers

Least value of ³√90x = (3*3*2*5*x)^1/3
X has to be 3*2*2*5*5 for ³√90x to be an integer.

Value of X = 300
And Y has to be 75

Value of xy = 300*75 = 22500

Answer should be C

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If (90x)^(1/3) and (y/75)^(1/2) are positive integers, and both x and 06 Jul 2020, 05:18
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(90x)^1/3 : (2*3*3*5x)^1/3 : x has to be : 4x3x25 = 300 for integer value
√y/75 : y= 75 for integer value

xy least value : 300 x 75
= 22500

IMO-C
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If (90x)^(1/3) and (y/75)^(1/2) are positive integers, and both x and 06 Jul 2020, 05:41
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Bunuel
If \(\sqrt[3]{90x}\) and \(\sqrt{\frac{y}{75}}\) are positive integers, and both x and y are integers, what is the least possible value of \(xy\) ?

A. 20500
B. 21500
C. 22500
D. 23500
E. 24500
Show more

\(\sqrt[3]{90x}\) is an Integer i.e. 90x must be a PERFECT CUBE

CONCEPT: For a number to be a perfect cube, all prime factors must have exponents which are multiples of 3

i.e. \(90x = 2*3^2*5*x\)

ie.. x must be minimum \(2^2*3*5^2\)


\(\sqrt{\frac{y}{75}}\) is an Integer i.e. \(\frac{y}{75}\) must be a PERFECT SQUARE

CONCEPT: For a number to be a perfect square, all prime factors must have exponents which are multiples of 2

i.e. \(\frac{y}{75} = 1\)

ie.. y must be minimum \(75\)

\(x*y_{min} = 2^2*3*5^2*75 = 22500\)



Answer: Option C
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If (90x)^(1/3) and (y/75)^(1/2) are positive integers, and both x and 12 Aug 2025, 04:05
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