Determine all possible pairs (x,y) of
positive integers, with x ≤ y, that satisfy this equation:
x*y = 160 + 90*gcd(x,y)
let gcd(x,y)=g
and x=a*g and y=b*g then we have
a*b*g^2=160+90*g
abg^290g=160
g(abg90)=160
so g must be a divisor of 160
160=2^5*5
so the possible values of g are:
1,2,4,5,8,10,16,20,32,40,80,160
now we can eliminate those for which
(160/g)+90 is not divisible by g
this leaves us with:
1 and 2
g=1 gives us ab=250
for each possible value of a,b with a<=b we get
a solution with x=g*a=a and y=g*b=b
giving us the solutions:
(1,250),(2,125),(5,50),(10,25)
g=2 gives us ab=85
for each possible value of a,b we get
a solution with x=g*a=2a and y=g*b=2b
giving us the solutions:
(2,170),(10,34)
thus all solutions are:
(1,250),(2,125),(5,50),(10,25),(2,170),(10,34)

Posted by Daniel
on 20131003 16:17:54 