The problem states:

• There is a two-digit number between 10 and 100.
• The sum of its digits is 9.
• If the number is multiplied by 7, it becomes four times the number obtained by reversing the digits of the original number.

Let the two-digit number be $10a + b$, where $a$ is the tens digit and $b$ is the units digit. We are given two conditions:

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

2. When the number is multiplied by 7, it equals four times the number obtained by reversing its digits: $7(10a + b) = 4(10b + a)$

Solving the system of equations:

1. From the first condition, $a + b = 9$, we get: $a = 9 - b$

2. Substituting $a = 9 - b$ into the second condition: $7(10(9 - b) + b) = 4(10b + (9 - b))$ Simplifying: $7(90 - 9b + b) = 4(10b + 9 - b)$ $7(90 - 8b) = 4(9b + 9)$ $630 - 56b = 36b + 36$ $630 - 36 = 36b + 56b$ $594 = 92b$ $b = \frac{594}{92} = 6.46$

It seems there is a mistake somewhere in the simplification. Let me correct and recompute it.The solution to the system of equations is:

• $a = 3$
• $b = 6$

Thus, the original number is $10a + b = 10(3) + 6 = 36$.

Therefore, the number is 36.

Would you like more details on the steps? Here are 5 related questions to expand on this topic:

1. How do we handle similar problems involving reversed digits?
2. What is the significance of writing numbers as $10a + b$ for two-digit problems?
3. How can we generalize this method for numbers with more digits?
4. How does solving systems of equations help in solving word problems like this?
5. Can we solve this problem using trial and error instead of algebraic methods?

Tip: When solving word problems involving digits, it’s often helpful to express the number in terms of powers of 10, such as $10a + b$ for a two-digit number.