Let's denote the two-digit number as $10a + b$, where $a$ is the tens digit and $b$ is the units digit. We are given two key pieces of information:

1. The sum of the digits is 9: $a + b = 9$

2. Nine times the number is equal to twice the number obtained by reversing the digits: $9 \times (10a + b) = 2 \times (10b + a)$

Step 1: Set up the equations

From the first condition, we have: $a + b = 9$

From the second condition: $9(10a + b) = 2(10b + a)$ Expanding both sides: $90a + 9b = 20b + 2a$

Step 2: Simplify the equation

Rearrange the terms to group $a$ and $b$ on opposite sides: $90a - 2a = 20b - 9b$ $88a = 11b$

Step 3: Solve for $a$ and $b$

Simplify the equation by dividing both sides by 11: $8a = b$

Step 4: Substitute into the sum equation

Substitute $b = 8a$ into the first equation $a + b = 9$: $a + 8a = 9$ $9a = 9$ $a = 1$

Now, substitute $a = 1$ back into $b = 8a$: $b = 8 \times 1 = 8$

Step 5: Find the number

The number is $10a + b = 10(1) + 8 = 18$.

Step 6: Verify the solution

• The sum of the digits of 18 is $1 + 8 = 9$, which satisfies the first condition.
• Reversing the digits gives 81, and checking the second condition: $9 \times 18 = 162 \quad \text{and} \quad 2 \times 81 = 162$ Both sides are equal, confirming the solution.

The number is $\boxed{18}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if the digits sum up to a different number, like 8 or 10?
2. Can you derive a formula for similar problems involving sums of digits?
3. How would the solution change if the multiplier was a different number, like 8 instead of 9?
4. What if the number was three digits long? How would the approach change?
5. Could this type of problem have more than one solution?

Tip: Always verify your solution by substituting the found values back into the original problem conditions.