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If x and y are positive integers such that \(5*x^3 = 6*y^4\), what is the minimum possible value of \(x+y\)?
A. 1260
B. 1240
C. 1080
D. 1060
E. None of these
Are You Up For the Challenge: 700 Level Questions
A. 1260
B. 1240
C. 1080
D. 1060
E. None of these
Are You Up For the Challenge: 700 Level Questions
21 Feb 2021, 07:37
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5 * x^3 = 6 * y^4
=> x^3 = (6y * y^3)/ 5
=> x = y * cube root over (6y/5)
Given the conditions, cube root over (6y/5) needs to be a positive integer. So, the least value of y = 6*6*5 = 180
=> x = 180 * cube root over (6*180)/5 = 180 * 6 = 1080
So, x+y = 1080+180 = 1260 (Option: A)
=> x^3 = (6y * y^3)/ 5
=> x = y * cube root over (6y/5)
Given the conditions, cube root over (6y/5) needs to be a positive integer. So, the least value of y = 6*6*5 = 180
=> x = 180 * cube root over (6*180)/5 = 180 * 6 = 1080
So, x+y = 1080+180 = 1260 (Option: A)
08 Jan 2021, 07:02
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Bunuel
If x and y are positive integers such that \(5*x^3 = 6*y^4\), what is the minimum possible value of \(x+y\)?
A. 1260
B. 1240
C. 1080
D. 1060
E. None of these
A. 1260
B. 1240
C. 1080
D. 1060
E. None of these
\(5*x^3 = 6*y^4\)...
This means that y is surely a multiple of 5 and x is a multiple of 6.
Let \(x=5^a*6^b\), and \(y = 5^c*6^d\)
So \(5*x^3 = 6*y^4\)...
=> \(5*(5^a*6^b)^3 = 6*(5^c*6^d)^4\)...
\(5^{(1+3a)}*6^{3b}=5^{4c}*6^{(1+4d)}\)..
Let us equate the power and get the minimum possible values of the variables.
\(5^{(1+3a)}=5^{4c}.......1+3a=4c\).....The minimum value is when a=1 and c=1.
\(6^{3b}=6^{(1+4d)}......3b=1+4d\)....Both b and d are integers, and minimum values are when d=2, and b becomes 3.
\(x=5^a*6^b=5^1*6^3=1080\), and \(y = 5^c*6^d=5^1*6^2=180\)
\(x+y=1080+180=1260\)
A
