Question

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.

Answer 1

​​Let the digits at units and tens places be $x$ and $y,$ respectively.

$\therefore$ Product of digits $=xy=14$

$\Rightarrow y=\frac{14}{x}\dots \dots \left(i\right)$

so, the two digit number will be $(10y+x)$

when the digits will be reversed, then the reversed number will be $10x+y$

According to the question, we have

Original No. $+45=$ Reversed No.

$\Rightarrow (10y+x)+45=10x+y$

$\Rightarrow 9y-9x=-45$

$\Rightarrow y-x=-5\dots \dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we get

$\frac{14}{x}-x=-5$

$\Rightarrow \frac{14-x^2}{x}=-5$

$\Rightarrow 14-x^2=-5x$

$\Rightarrow x^2-5x-14=0$

$\Rightarrow x^2-(7-2)x-14=0$

$\Rightarrow x^2-7x+2x-14=0$

$\Rightarrow x(x-7)+2(x-7)=0$

$\Rightarrow (x-7)(x+2)=0$

$\Rightarrow x-7=0orx+2=0$

$\Rightarrow x=7$ or $x=-2$

$\Rightarrow x=7$ ($\because$ the digit cannot be negative)

Putting $x=7$ in equation $\left(i\right),$ we get: $y=2$

so, Required number $=10\times 2+7=27$