Q10

What is the number of pairs of natural numbers ( x , y ) (x,y) in which x > y x>y and 5 x + 6 y = 1 ? \frac { 5 }{ x } +\frac { 6 }{ y } =1?

This question is a part of my set NMTC 2015 .
5 3 0 7 2 4 6 1

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1 solution

Cleres Cupertino
Sep 2, 2015

5 x + 6 y = 1 5 y + 6 x = x y x y 5 y 6 x + 30 = 30 \frac { 5 }{ x } +\frac { 6 }{ y } =1 \Leftrightarrow 5y+6x=xy \Leftrightarrow xy-5y-6x+30=30

( x 5 ) ( y 6 ) = 30 = 1 × 2 × 3 × 5 \Leftrightarrow (x-5)(y-6)=30=1 \times 2 \times 3 \times 5

1) x 5 = 1 x = 6 ; y 6 = 30 y = 36 x-5=1 \Leftrightarrow x=6; \quad y-6=30 \Leftrightarrow y=36

2) x 5 = 2 x = 7 ; y 6 = 15 y = 21 x-5=2 \Leftrightarrow x=7; \quad y-6=15 \Leftrightarrow y=21

3) x 5 = 3 x = 8 ; y 6 = 10 y = 16 x-5=3 \Leftrightarrow x=8; \quad y-6=10 \Leftrightarrow y=16

4) x 5 = 5 x = 10 ; y 6 = 6 y = 12 x-5=5 \Leftrightarrow x=10; \quad y-6=6 \Leftrightarrow y=12

5) x 5 = 6 x = 11 ; y 6 = 5 y = 11 x-5=6 \Leftrightarrow x=11; \quad y-6=5 \Leftrightarrow y=11

6) x 5 = 10 x = 15 ; y 6 = 3 y = 9 x-5=10 \Leftrightarrow \boxed{x=15}; \quad y-6=3 \Leftrightarrow y=9

7) x 5 = 15 x = 20 ; y 6 = 2 y = 8 x-5=15 \Leftrightarrow \boxed{x=20}; \quad y-6=2 \Leftrightarrow y=8

8) x 5 = 30 x = 35 ; y 6 = 1 y = 7 x-5=30 \Leftrightarrow \boxed{x=35}; \quad y-6=1 \Leftrightarrow y=7

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