#1

Number Theory Level 3

Consider the equation 5 x + 7 y = 5 7 \frac{5}{x} + \frac{7}{y} = \frac{5}{7} for x > 0 ; y > 0 x > 0 ; y > 0 . Suppose the number of integer solution is A with the set of integer pairs ( x i , y i ) (x_i,y_i) being ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . . ( x A , y A ) (x_1,y_1),(x_2,y_2),....(x_A,y_A) . Find ∑ i = 1 A x i + ∑ i = 1 A y i \sum_{i=1}^{A} x_i + \sum_{i=1}^{A} y_i .

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1490 360 No integer solution possible 910

Sayan Chaudhuri by Sayan Chaudhuri , 23, India

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1 solution

Krutarth Patel
Nov 21, 2015

5 x + 7 y = 5 7 ⇒ 35 x + 49 y = 5 x y ⇒ 5 x y − 35 x − 49 y + 343 = 343 ⇒ ( x − 7 ) ( 5 y − 49 ) = 343 \begin{aligned} \frac{5}{x} + \frac{7}{y} & = \frac{5}{7} \\ \Rightarrow 35x + 49y & = 5xy \\ \Rightarrow 5xy - 35x - 49y + 343 & = 343 \\ \Rightarrow (x - 7)(5y - 49) & = 343 \end{aligned}

343 343 can be written as a product of 2 2 integers in 8 8 ways:

( ± 1 , ± 343 ) , ( ± 343 , ± 1 ) , ( ± 7 , ± 49 ) , ( ± 49 , ± 7 ) (\pm{1}, \pm{343}), (\pm{343}, \pm{1}), (\pm{7}, \pm{49}), (\pm{49}, \pm{7}) .

  1. x − 7 = 1 , 5 y − 49 = 343 ⇒ x = 8 , y = 392 5 x - 7 = 1, 5y - 49 = 343 \Rightarrow x = 8, y = \frac{392}{5}

  2. x − 7 = 343 , 5 y − 49 = 1 ⇒ x = 350 , y = 10 x - 7 = 343, 5y - 49 = 1 \Rightarrow x = 350, y = 10

  3. x − 7 = 7 , 5 y − 49 = 49 ⇒ x = 14 , y = 98 5 x - 7 = 7, 5y - 49 = 49 \Rightarrow x = 14, y = \frac{98}{5}

  4. x − 7 = 49 , 5 y − 49 = 7 ⇒ x = 56 , y = 56 5 x - 7 = 49, 5y - 49 = 7 \Rightarrow x = 56, y = \frac{56}{5}

  5. x − 7 = − 1 , 5 y − 49 = − 343 ⇒ x = 6 , y = − 294 5 x - 7 = -1, 5y - 49 = -343 \Rightarrow x = 6, y = \frac{-294}{5}

  6. x − 7 = − 343 , 5 y − 49 = − 1 ⇒ x = − 336 , y = 48 5 x - 7 = -343, 5y - 49 = -1 \Rightarrow x = -336, y = \frac{48}{5}

  7. x − 7 = − 7 , 5 y − 49 = − 49 ⇒ x = 0 , y = 0 x - 7 = -7, 5y - 49 = -49 \Rightarrow x = 0, y = 0

  8. x − 7 = − 49 , 5 y − 49 = − 7 ⇒ x = − 42 , y = 42 5 x - 7 = -49, 5y - 49 = -7 \Rightarrow x = -42, y = \frac{42}{5}

Of the above 8 cases only x = 350 , y = 10 x = 350, y = 10 satisfy x > 0 , y > 0 x > 0, y > 0 .

Hence, ∑ i = 1 A x i + ∑ i = 1 A y i = 350 + 10 = 360 \sum_{i=1}^{A} x_i + \sum_{i=1}^{A} y_i = 350 + 10 = \boxed{360}

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Great usage of Simon's Favorite Factoring Trick.

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