Multiplying and adding
(All calculations are in base 10.)
Let x and y be distinct positive integers, with x < y .
If the condition holds that x y − ( x + y ) = 5 0 , what is the sum total of all possible values of x ?
The answer is 6.
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3 solutions
Add 1 on both sides of the equation, to get x y − x − y + 1 = 5 1 , which factors down to ( x − 1 ) ( y − 1 ) = 5 1 . The factors of 51 are 1, 3, 17, and 51. Therefore, the solutions are:
x = 2 , y = 5 2
x = 4 , y = 1 8
x = 1 8 , y = 4
x = 5 2 , y = 2
Only x = 2 and x = 4 satisfy x < y .
2 + 4 = 6
x has to be less than 8 (smallest possibility is 8 * 9, but 72 - 17 = 55, which is too large) and larger than 1.
Checking the possibilities for x = 2, 3, 4, 5, 6, 7, we discover 2 solutions: x = 2 (y = 52) and x = 4 (y=18).
2 + 4 = 6.
It's worth noting that ∂ y ∂ ( x y − ( x + y ) ) = x − 1 > 0 for x > 1 . Thus, given fixed x , x y − ( x + y ) is strictly increasing. So, after testing ( x , y ) = ( 8 , 9 ) and finding 5 5 , we can indeed conclude that x < 8 .
Solving for y gives y = x − 1 5 1 + 1 . So x − 1 = 1 , 3 , 1 7 , or 5 1 , but only the first two give x < y . The answer is 2 + 4 = 6 .