Let x and y be two natural numbers such that xy = 12(x + y) and x le y. Then the total number of pairs (x, y) is

Question

Let x and y be two natural numbers such that xy = 12(x + y) and x le y. Then the total number of pairs (x, y) is (A) 8 (B) 6 (C) 4 (D) 16

Tardigrade Admin · Accepted Answer

xy - 12x - 12y = 0 ⇒ (x - 12) (y - 12) = 144. Now 144 can be factorised into two factors x and y where x le 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.

Q. 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

8

6

4

16

$x y - 12 x - 12 y = 0$
$\Rightarrow (x - 12) (y - 12) = 144$ .
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.

Q. Let x and y be two natural numbers such that xy=12(x+y) and x≤y. Then the total number of pairs (x,y) is

8

6

4

16

xy−12x−12y=0 ⇒(x−12)(y−12)=144. Now 144 can be factorised into two factors x and y where x≤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.

Q. 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

$x y - 12 x - 12 y = 0$
$\Rightarrow (x - 12) (y - 12) = 144$ .
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.