Question

The sum of a two-digit number and the number obtained by reversing the digits is 110. If the tens digit of the number is 6 more than the units digit, then find the number.

Hint:

For the digit problem, always remember that $x$ and $y$ must be single-digit whole numbers ($0-9$). If you get a fraction, re-check your equations!

Answer 1

Step 1: Understanding the Concept:
A two-digit number with tens digit $x$ and units digit $y$ is expressed as $10x + y$. When digits are reversed, the number becomes $10y + x$. We will use the given conditions to form a system of linear equations.
Step 2: Forming Equations:
Let the tens digit be $x$ and the units digit be $y$.
Original number $= 10x + y$; Reversed number $= 10y + x$.
Condition 1: $(10x + y) + (10y + x) = 110$
$11x + 11y = 110 \implies x + y = 10$ \quad ---(Eq. 1)
Condition 2: Tens digit is 6 more than units digit.
$x = y + 6 \implies x - y = 6$ \quad ---(Eq. 2)
Step 3: Solving the Equations:
Adding Eq. 1 and Eq. 2:
$(x + y) + (x - y) = 10 + 6$
$2x = 16 \implies x = 8$
Substituting $x = 8$ in Eq. 1:
$8 + y = 10 \implies y = 2$
Step 4: Final Answer:
The digits are 8 and 2. Thus, the number is 82.