A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
Let the digits at units and tens places be $x$ and $y$, respectively.
$\therefore x y=14$
$\Rightarrow y=\frac{14}{x} \quad \ldots(\mathrm{i})$
According to the question:
$(10 y+x)+45=10 x+y$
$\Rightarrow 9 y-9 x=-45$
$\Rightarrow y-x=-5$ ........(ii)
From (i) and (ii), we get:
$\frac{14}{x}-x=-5$
$\Rightarrow \frac{14-x^{2}}{x}=-5$
$\Rightarrow 14-x^{2}=-5 x$
$\Rightarrow x^{2}-5 x-14=0$
$\Rightarrow x^{2}-(7-2) x-14=0$
$\Rightarrow x^{2}-7 x+2 x-14=0$
$\Rightarrow x(x-7)+2(x-7)=0$
$\Rightarrow(x-7)(x+2)=0$
$\Rightarrow x-7=0$ or $x+2=0$
$\Rightarrow x=7$ or $x=-2$
$\Rightarrow x=7 \quad(\because$ the digit cannot be negative $)$
Putting $x=7$ in equation (i), we get:
$y=2$
$\therefore$ Required number $=10 \times 2+7=27$
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