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Let $x$ and $y$ be two natural numbers such that $x y=12(x+y)$ and $x \leq y$. Then the total number of pairs $(x, y)$ is
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The correct answer is:
8
$x y-12 x-12 y=0 \Rightarrow(x-12)(y-12)=144$
Now 144 can be factorised into two factors $x$ and $y$ where $x \leq y$ and the factors are (1,144),(2,72),(3,48),(4,36),(6,24),(8,18)
(9,16),(12,12)
Thus there are eight solutions.
Now 144 can be factorised into two factors $x$ and $y$ where $x \leq y$ and the factors are (1,144),(2,72),(3,48),(4,36),(6,24),(8,18)
(9,16),(12,12)
Thus there are eight solutions.
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