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15 Dec 2016, 04:38
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C
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Difficulty:
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Question Stats:
77% (01:27) correct
23%
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If x and y are positive integers 1800x = y^3, what is the minimum possible value of x?
A. 30
B. 18
C. 15
D. 6
E. 5
A. 30
B. 18
C. 15
D. 6
E. 5
15 Dec 2016, 04:57
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Bunuel
\(y^3 = 2^3*3^2*5^2*X\)
\(y = (2^3*3^2*5^2*X)^{\frac{1}{3}} = 2*(3^2*5^2*X)^{\frac{1}{3}}\)
For expression under the radical to be an integer the min value of X should be \(3*5 = 15\)
Answer C
General Discussion
15 Dec 2016, 06:46
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Taken that both x & y are integers then to conform to basic conditions extracted value of 1800x from the root ^1/3 should be an integer.
y=(1800x)^1/3;
To find out the acceptable value of "x" it is simply required to perform prime factorization of 1800 which should be as follows:
1800 = 2^3 * 3^2*5^2
So y= (2^3 * 3^2*5^2*x) ^ 1/3
2^3 can easility be extracted from the root ^1/3
3^2*5^2 cannot be extracted, as it should be added 1 power to both of them in order to extract from the root ^1/3.
Thus, the acceptable value of "x" should be 5*3 = 15. Answer C.
y=(1800x)^1/3;
To find out the acceptable value of "x" it is simply required to perform prime factorization of 1800 which should be as follows:
1800 = 2^3 * 3^2*5^2
So y= (2^3 * 3^2*5^2*x) ^ 1/3
2^3 can easility be extracted from the root ^1/3
3^2*5^2 cannot be extracted, as it should be added 1 power to both of them in order to extract from the root ^1/3.
Thus, the acceptable value of "x" should be 5*3 = 15. Answer C.
