37
A two-digit number is such that the product of its digits is $12$. When $36$ is added to this number, the digits interchange their places. Find the number.
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Let unit digit be $y$
and ten's digit = $x$
hence, the two digit number = $10x + y$, (1/2)
ATQ
$xy = 12$ ..............(i) (1)
$10x + y + 36 = 10y + x$
$x - y + 4 = 0$ ....... (ii) (1)
From (i) and (ii)
$x^2 + 4x - 12 = 0$ (1/2)
$(x + 6)(x - 2) = 0$ (1/2)
Hence, $x = 2$
and $y = 6$ (1/2)
$\therefore$ Number = $26$ (1/2)
and ten's digit = $x$
hence, the two digit number = $10x + y$, (1/2)
ATQ
$xy = 12$ ..............(i) (1)
$10x + y + 36 = 10y + x$
$x - y + 4 = 0$ ....... (ii) (1)
From (i) and (ii)
$x^2 + 4x - 12 = 0$ (1/2)
$(x + 6)(x - 2) = 0$ (1/2)
Hence, $x = 2$
and $y = 6$ (1/2)
$\therefore$ Number = $26$ (1/2)