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Math Challenge II-A Number Theory

 
 
Picture of Neo Liang
Math Challenge II-A Number Theory
by Neo Liang - Monday, July 29, 2019, 11:47 AM
 

Lecture 8, problem 8.16

I got 1 pair, but the answer says 2. 1/x+1/y=1/5.   5(x+y)=xy.  xy-5(x+y)+25=25. (x-5)(y-5)=25. 25=1*25=5*5=25*1=(-1)*(-25)=(-5)*(-5)=(-25)*(-1) This means that the solutions to the equation are (6,30), (10,10), (30, 6), (4, -20), (0, 0), (-20, 4), but 0<x<y, so the only solution is (6, 30). This means that the answer is 1.

 
Picture of David Reynoso
Re: Math Challenge II-A Number Theory
by David Reynoso - Wednesday, July 31, 2019, 10:46 PM
 

You are correct. The way the problem appears on the handout there is only one solution. We intended for the problem to say $0 < x \leq y$, which would lead to the posted solution. The solution should be fixed now. Thanks for letting us know!

 
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