Let's denote the two-digit number as $\text{AB}$, where $A$ is the tens digit and $B$ is the units digit.

Given:

1. The sum of the digits is 9: $A + B = 9$.
2. When 27 is subtracted from the number, the digits are reversed: $10A + B - 27 = 10B + A$.

Step 1: Set up the equations

From the information given, we have two equations:

1. $A + B = 9$
2. $10A + B - 27 = 10B + A$

Step 2: Simplify the second equation

We can rearrange the second equation to isolate terms involving $A$ and $B$:

$10A + B - 27 = 10B + A$ $10A - A + B - 10B = 27$ $9A - 9B = 27$

This simplifies to:

$A - B = 3$

Step 3: Solve the system of equations

Now, we have the system:

1. $A + B = 9$
2. $A - B = 3$

We can solve these equations by adding them together:

$(A + B) + (A - B) = 9 + 3$ $2A = 12 \implies A = 6$

Now, substitute $A = 6$ into the first equation:

$6 + B = 9 \implies B = 3$

Step 4: Find the number

The number is formed by $A$ in the tens place and $B$ in the units place:

$\text{Number} = 10A + B = 10(6) + 3 = 63$

Conclusion

The number is 63.

Would you like further details on any part of this solution or have additional questions?

Here are 5 related questions:

1. What would happen if the digits sum to a different number?
2. How would the solution change if a different number were subtracted?
3. Can we generalize this problem to a number with more digits?
4. What if the digits were reversed by a different operation instead of subtraction?
5. How can we verify that 63 is indeed the correct answer?

Tip: Always double-check your final solution by substituting it back into the original problem conditions.