Finding the Unit Digit of the Two-Digit Number

The problem asks for the unit digit of a two-digit number 'N' that satisfies a condition based on its digits.

Algebraic Representation of the Number

Let the two-digit number be $N$. We can represent $N$ using its tens digit, let's call it $a$, and its units digit, let's call it $b$. The value of the number is given by:

$N = 10a + b$

Constraints: $a$ must be an integer from 1 to 9 (since it's a two-digit number), and $b$ must be an integer from 0 to 9.

Setting up the Digit Sum and Product Equation

The problem states that the sum of the digits plus the product of the digits equals the number $N$. We can write this as an equation:

Sum of digits + Product of digits = $N$

$(a + b) + (a \times b) = 10a + b$

Solving the Equation for Digits

Let's simplify the equation step-by-step:

1. Start with the equation: $a + b + ab = 10a + b$.
2. Subtract $b$ from both sides: $a + ab = 10a$.
3. Subtract $a$ from both sides: $ab = 9a$.
4. Since $a$ is the tens digit, $a \neq 0$. We can safely divide both sides by $a$: $\frac{ab}{a} = \frac{9a}{a}$.
5. This simplifies to: $b = 9$.

The result $b = 9$ tells us that the unit digit must be 9.

Conclusion: The Unit Digit

The equation $b=9$ satisfies the condition for any tens digit $a$ from 1 to 9. For example, if $N=39$, $a=3, b=9$. Sum ($3+9=12$) + Product ($3 \times 9=27$) = $12 + 27 = 39$, which is $N$.

Therefore, the unit digit of the number $N$ is 9.