15
In a $2$-digit number, the digit at the unit's place is $5$ less than the digit at the ten's place. The product of the digits is $36$. Find the number.
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Let digit at ten's place be $x$
then digit at unit's place $= x - 5$
$x(x - 5) = 36$
$\Rightarrow x^2 - 5x - 36 = 0$
$(x-9)(x + 4) = 0$
$x \ne -4$ so, $x = 9$
$\therefore$ Required number is $94$
then digit at unit's place $= x - 5$
$x(x - 5) = 36$
$\Rightarrow x^2 - 5x - 36 = 0$
$(x-9)(x + 4) = 0$
$x \ne -4$ so, $x = 9$
$\therefore$ Required number is $94$