We are tasked with finding all three-digit numbers such that when the number and the sum of its digits are added, the resulting number is 387.

Let the three-digit number be represented as $N$. We can express $N$ as:

$N = 100a + 10b + c$

where:

• $a$ is the hundreds digit,
• $b$ is the tens digit,
• $c$ is the units digit.

The sum of the digits of $N$ is $a + b + c$. According to the problem, the sum of the number and the sum of its digits equals 387, so we have the equation:

$N + (a + b + c) = 387$

Substitute $N = 100a + 10b + c$ into the equation:

$(100a + 10b + c) + (a + b + c) = 387$

Simplify the equation:

$100a + 10b + c + a + b + c = 387$

Combine like terms:

$(100a + a) + (10b + b) + (c + c) = 387$

This simplifies to:

$101a + 11b + 2c = 387$

We now need to find values of $a$, $b$, and $c$ (which are digits, so $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$) that satisfy this equation.

Solving the equation:

1. Divide the equation by 2: $101a + 11b + 2c = 387$ This equation can be solved by trying values of $a$, and finding corresponding ( b ,.after