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15 Sep 2019, 10:50
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65% (02:08) correct
35%
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If \(x-y=\frac{x}{y}\), where x and y are positive integers. Find the number of solutions possible for the given equation?
A. 0
B. 1
C. 2
D. 4
E. 8
A. 0
B. 1
C. 2
D. 4
E. 8
15 Sep 2019, 11:00
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nick1816
If \(x-y=\frac{x}{y}\), where x and y are positive integers. Find the number of solutions possible for the given equation?
A. 0
B. 1
C. 2
D. 4
E. 8
A. 0
B. 1
C. 2
D. 4
E. 8
Given: \(x-y=\frac{x}{y}\), where x and y are positive integers.
Asked: Find the number of solutions possible for the given equation?
\(x-y=\frac{x}{y}\)
xy - y^2 = x
x(y-1) - y^2 = 0
x = y^2/(y-1)
Since x & y are positive integers; \(y \neq 0\)
If y = 2; x = 4/1 = 4
If y = 3; x = 9/2 = 4.5
In general y/(y-1) is NOT an integer except for y = 2
Only solution (x,y) = (4,2)
IMO B
15 Sep 2019, 11:06
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x-y=x/y
xy-y^2=x
xy-x=y^2
x(y-1)=y^2
x= y^2/(y-1) ........(1)
y=1 x= n.d
y=2 , x= 4
y=3 , x= 9/2
y=4 , x= 16/3
y=5 , x= 25/4
so only integer value we got is x=4, y=2
option B (1 Solution)
xy-y^2=x
xy-x=y^2
x(y-1)=y^2
x= y^2/(y-1) ........(1)
y=1 x= n.d
y=2 , x= 4
y=3 , x= 9/2
y=4 , x= 16/3
y=5 , x= 25/4
so only integer value we got is x=4, y=2
option B (1 Solution)
