Let each of x and y is a nonnegative integer such that:
x + 3y - 5 = 2*LCM(x,y) - 11*GCD(x,y)
Find the maximum possible value of x+y
Let g=gcd(x,y) Then there must be f and h such that x=f*g and y=g*h with f coprime to h.
Also with two numbers we have the property that x*y = lcm(x,y)*gcd(x,y), this implies lcm(x,y)=f*g*h.
Plug everything into the problem's equation:
fg + 3gh - 5 = 2fgh - 11g.
Note that everything is a factor of g, except the constant 5. So we can infer that g is a factor of 5; but the only positive factors of 5 are 1 and 5 so we have two cases: g=1 and g=5.
Case 1: g=1
Then the equation reduces to f + 3h - 5 = 2fh - 11.
Rearrange this into 4fh - 2f - 6h = 12
Then factor into (2h-1)*(2f-3) = 15
For positive h then 2h-1 will also be positive, so we need positive factorizations of 9, which are 1*15, 3*5, 5*3 and 15*1.
Subcase 1-1: 2h-1=1 and 2f-3=15.
Then h=1 and f=9, which makes x=9 and y=1.
This checks out so one answer is x+y=10.
Subcase 1-2: 2h-1=3 and 2f-3=5.
Then h=2 and f=4, but this violates the condition that f and h are coprime.
Subcase 1-3: 2h-1=5 and 2f-3=3.
Then h=3 and f=3, but this violates the condition that f and h are coprime.
Subcase 1-4: 2h-1=15 and 2f-3=1.
Then h=8 and f=2, but this violates the condition that f and h are coprime.
Case 2: g=5
Then the equation reduces to f + 3h - 1 = 2fh - 11.
Rearrange this into 4fh - 2f - 6h = 20
Then factor into (2h-1)*(2f-3) = 23
For positive h then 2h-1 will also be positive, so we need positive factorizations of 23, which are 1*23 and 23*1.
Subcase 2-1: 2h-1=1 and 2f-3=23.
Then h=1 and f=13, which makes x=65 and y=5.
This checks out so one answer is x+y=70.
Subcase 2-2: 2h-1=23 and 2f-3=1.
Then h=12 and f=2, but this violates the condition that f and h are coprime.
So then there are a total of two (x,y) pairs that satisfy the problem: (9,1) and (65,5), with x+y=10 and x+y=70 respectively. So then the maximum possible value of x+y is 70.
Solution
