A two-digit number is 4 times the sum of its digits.

Solution:

Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.

The number is 4 times the sum of the two digits. Thus, we have

$10 y+x=4(x+y)$

$\Rightarrow 10 y+x=4 x+4 y$

$\Rightarrow 4 x+4 y-10 y-x=0$

$\Rightarrow 3 x-6 y=0$

$\Rightarrow 3(x-2 y)=0$

$\Rightarrow x-2 y=0$

After interchanging the digits, the number becomes $10 x+y$.

If 18 is added to the number, the digits are reversed. Thus, we have

$(10 y+x)+18=10 x+y$

$\Rightarrow 10 x+y-10 y-x=18$

$\Rightarrow 9 x-9 y=18$

$\Rightarrow 9(x-y)=18$

$\Rightarrow x-y=\frac{18}{9}$

$\Rightarrow x-y=2$

So, we have the systems of equations

$x-2 y=0$,

$x-y=2$

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

$(x-y)-(x-2 y)=2-0$

$\Rightarrow x-y-x+2 y=2$

$\Rightarrow y=2$

Substituting the value of y in the first equation, we have

$x-2 \times 2=0$

$\Rightarrow x-4=0$

$\Rightarrow x=4$

Hence, the number is $10 \times 2+4=24$.