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A 2-digit number is obtained by either multiplying the sum of the digits by $7$ and then adding $3$ or by multiplying the difference of the digits by $19$ and then subtracting $1$. It is given that the digit at ten's place is greater than that of unit's place. Find the 2-digit number.
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Let the unit's place digit be $y$ and ten's digit be $x$.
So, number be $10x + y$
Therefore, $10x + y = 7 (x + y) + 3$
$\Rightarrow x - 2y = 1$ --- (1)
Also, $10x + y = 19 (x - y) - 1$
$\Rightarrow -9x + 20y = -1$ --- (2)
Solving (1) and (2), we get
$x = 9, y = 4$
$\therefore$ the required number is $94$.
So, number be $10x + y$
Therefore, $10x + y = 7 (x + y) + 3$
$\Rightarrow x - 2y = 1$ --- (1)
Also, $10x + y = 19 (x - y) - 1$
$\Rightarrow -9x + 20y = -1$ --- (2)
Solving (1) and (2), we get
$x = 9, y = 4$
$\therefore$ the required number is $94$.