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 Divisibility with lcm and gcd (Posted on 2016-08-15)
Each of X and Y is a positive integer such that:
lcm(X,Y) + gcd(X,Y) = X+Y

Is one of X and Y always divisible by the other?

 No Solution Yet Submitted by K Sengupta No Rating

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 Answer and reasoning | Comment 1 of 4
Consider two numbers X and Y, whose GCD is g. Then X = gx and Y = gy, and x and y are relatively prime. Then the LCM of X and Y is gxy.

The conditions of the problem require LCM(X,Y) + GCD(X,Y) = X+Y, or:
gxy + g = gx + gy and (dividing by the non-zero g)
xy + 1 = x + y

subtracting y + 1 from both sides and collecting terms gives

y(x-1) = (x-1)

if x = 1, then this is true for all y, otherwise we can divide by (x-1) and conclude that y = 1. Either way, at least one of x and y = 1.

Assume that y = 1. Then  Y = gy = g and X = gx and so Y divides X. Similarly, if we take x = 1 then X = g and X divides Y, and if x = y = 1 then X = Y = g and X and Y each divide the other.

In all cases, then, if the equation holds then one of the two numbers divides the other.

 Posted by Paul on 2016-08-15 13:30:32

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