A specific two-digit integer is equal to four times the sum of its constituent digits. When the position of these digits is reversed, the resulting integer exceeds the initial integer by 18. What is the original integer?

The correct option is 24.

Step-by-step Explanation:

Let the tens digit of the original two-digit integer be $x$ and the units digit be $y$.

The value of the original integer can be written as:

$\text{Original Number}=10x+y$

Condition 1: The two-digit integer is equal to four times the sum of its constituent digits.

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

$10x+y=4x+4y$

$10x-4x=4y-y$

$6x=3y$

$y=2x$   --- (Equation 1)

Condition 2: When the position of these digits is reversed, the resulting integer exceeds the initial integer by 18.

The reversed integer is represented as $10y+x$.

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

$9y-9x=18$

Dividing the entire equation by 9:

$y-x=2$   --- (Equation 2)

Solving the equations:

Substitute $y=2x$ from Equation 1 into Equation 2:

$2x-x=2$

$x=2$

Now, substitute $x=2$ back into Equation 1 to find $y$:

$y=2\left(2\right)=4$

Therefore, the original integer is $10x+y=10\left(2\right)+4=24$.