If the digits of a two-digit number are interchanged, the number so formed is 18 more than the original number. The sum of digits is 8. The thrice value of original number is

Let unit’s digit $=x$

Ten’s digit $=y$

$\therefore$ The original number $=1\left(x\right)+10\left(y\right)$

$=x+10y$

After inter changing the digits

The new number is $=10x+y$

As per the problem $10x+y=x+10y+18$

$9x-9y=18$

$x-y=2\cdots \cdots \left(1\right)$

Given the sum of digits

$x+y=8\cdots \cdots \left(2\right)$

By solving is (2)

$2x=10\Rightarrow x=5$

$x-y=2$

$5-y=2\Rightarrow y=3$

$\therefore$ The original number $=35$

Thrice value of original $=3\times 35=105$