Two Digit Number Solution

Let the two-digit number be represented algebraically. Let the tens digit be $t$ and the units digit be $u$. The number can be written as $10t + u$. The sum of its digits is $t + u$.

Formulating Number Equations

We are given two conditions that can be translated into equations:

• Condition 1: The number is five times the sum of its digits. $10t + u = 5(t + u)$
• Condition 2: If 9 is added to the number, the digits interchange their positions. $10t + u + 9 = 10u + t$

Simplifying Digit Equations

Let's simplify the equations derived from the conditions:

• From Condition 1: $10t + u = 5t + 5u$ $10t - 5t = 5u - u$ $5t = 4u \quad (Equation A)$
• From Condition 2: $10t - t + u - 10u = -9$ $9t - 9u = -9$ Divide by 9: $t - u = -1$ $u - t = 1 \quad (Equation B)$

Finding the Number Digits

Now, we solve the system of two linear equations (Equation A and Equation B) to find the values of $t$ and $u$.

From Equation B, we can express $u$ in terms of $t$: $u = t + 1$

Substitute this expression for $u$ into Equation A:

Now, substitute the value of $t$ back into the expression for $u$ (from Equation B):

So, the tens digit ($t$) is 4 and the units digit ($u$) is 5. The number is 45.

Calculating Sum of Digits

The question asks for the sum of the digits of the number.

Sum of digits = $t + u = 4 + 5 = 9$.

The sum of the digits of the number is 9.