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LCM and GCD Crossed Value Derivation (Posted on 2024-01-11) Difficulty: 3 of 5
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

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Some Thoughts A step towards a proof | Comment 2 of 3 |
I also found the same solutions:
(x,y) = (9,1); sum = 10
(x,y) = (65,6); sum = 70

This is not a full proof, just a special case:
Suppose y is prime and x = ky
Then lcm = x = ky ; gcd = y

x + 3y - 5 = 2*LCM(x,y) - 11*GCD(x,y) becomes
ky + 3y - 5 = 2ky - 11y
(k + 3 - 2k + 11)y = 5
y = 5/(14 - k)
For y to be an integer, the denominator must be either 1 or 5;
so k can only be 13 or 9;
so y can only be 5 or 1;
so y=5 and x=13y --> (x,y) = (65,5)   or
or y=1 and x=9y  --> (x,y) = (9,1)  
(x+y) can only be 10 or 70; so the maximum value is 70


  Posted by Larry on 2024-01-11 09:40:25
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