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If x and y are positive integers such that 5*x^3 = 6*y^4, what is the 08 Jan 2021, 06:22
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A
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40% (02:52) correct 60% (02:23) wrong based on 132 sessions

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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

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Abhibarua
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If x and y are positive integers such that 5*x^3 = 6*y^4, what is the 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)
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If x and y are positive integers such that 5*x^3 = 6*y^4, what is the 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
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\(5*x^3 = 6*y^4\)...
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
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If x and y are positive integers such that 5*x^3 = 6*y^4, what is the 21 Feb 2021, 08:39
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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

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5 x^3 = 6 y^4
For the equality to hold, y must be a multiple of 5: y = 5p; and x must be a multiple of 6: x = 6q
Substituting:
5 * 6^3 * q^3 = 6 * 5^4 * p^4
=> 6^2 * q^3 = 5^3 * p^4

If q=5, then p^4 = 6^2 --- not possible since p should be an integer
Thus, q = a multiple of 5 and should also have some power of 6 so that on the LHS, you get 6 raised to the power that is a multiple of 4

The smallest number that fits is q = 5 * 6^2, since then we have:
6^2 * (5 * 6^2)^3 = 5^3 * p^4

=> 6^8 * 5^3 = 5^3 * p^4
=> p = 6^2

Thus: x = 6q = 6 * (5 * 6^2) = 1080
y = 5p = 5 * (6^2) = 180

Thus: x + y = 1260

Answer A

Also, Abhibarua - that is a nice method :)
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