The sum of the digits of a two-digit number is 10. If 18 is subtracte…

The correct answer is option 2: 73

Let the two-digit number be 10x+y where $x$ is the tens digit and $y$ is the units digit.

The sum of the digits is 10: x +y =10       (1)

If 18 is subtracted, the digits in the resulting number are equal:

10x + y -18 = 10 z + z $$\text{where }z\text{ is the common digit in the resulting number.}$$

Simplifying,

10x + y -18 =11 z 

10x + y = 11z + 18                                  (2)

From equation (1),

y = 10 - x

Substituting y in equation (2)

10x + 10 - x = 11z + 18

9x + 10 = 11z + 18

9x = 11z +8                           (3)

Finding integer solutions:

• Since $x$ and z are digits ($z$
• For z= 5, 9x = 55 + 8
• x=63/9=7
• When x=7, then y =10-7 Thus y= 3

This means the number is: 73

10x + y = 10*7 + 3 = 73

Verification:

1. Sum of the digits: 7 + 3 = 10

2. Subtracting 18: 73-18 =55 and the digits in 55 are equal