Let the two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit. We are given the following information:

1. The product of the digits is 18: $x \cdot y = 18$

2. After interchanging the digits, the new number becomes $10y + x$, and it is 27 less than the original number: $10y + x = (10x + y) - 27$

Now, we have two equations:

1. $x \cdot y = 18$
2. $10y + x = 10x + y - 27$

Step 1: Simplify the second equation.

Rearranging the second equation: $10y + x = 10x + y - 27$ Subtracting $y$ and $x$ from both sides: $9y - 9x = -27$ Dividing by 9: $y - x = -3 \quad \text{or} \quad x = y + 3$

Step 2: Substitute $x = y + 3$ into the first equation.

Substituting into $x \cdot y = 18$: $(y + 3) \cdot y = 18$ Expanding: $y^2 + 3y = 18$ Rearranging: $y^2 + 3y - 18 = 0$

Step 3: Solve the quadratic equation.

Solve using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 3$, and $c = -18$: $y = \frac{-3 \pm \sqrt{3^2 - 4(1)(-18)}}{2(1)}$ $y = \frac{-3 \pm \sqrt{9 + 72}}{2}$ $y = \frac{-3 \pm \sqrt{81}}{2}$ $y = \frac{-3 \pm 9}{2}$ Thus, $y = \frac{-3 + 9}{2} = 3$ or $y = \frac{-3 - 9}{2} = -6$.

Since $y$ is a digit, we discard $y = -6$ and take $y = 3$.

Step 4: Find $x$.

Using $x = y + 3$, we get: $x = 3 + 3 = 6$

Step 5: Conclusion.

The original number is $10x + y = 10(6) + 3 = 63$.

Verification:

• The product of the digits $6 \cdot 3 = 18$ is correct.
• Interchanging the digits gives $36$, and $63 - 36 = 27$, which is also correct.

Thus, the number is $\boxed{63}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if the product of the digits was different, say 24?
2. How would the problem change if the new number was 30 less instead of 27?
3. Can you derive a general formula for any two-digit number given similar conditions?
4. What if the digits of the original number were reversed and added instead of subtracted?
5. How can the quadratic formula be applied in other digit-related problems?

Tip: Always check your solution by plugging it back into the original problem to ensure accuracy!