Question

How many three-digit positive integers x are there, such that subtracting the sum of digits of x from x gives a three-digit number whose digits are all the same?​

Answer 1

Step-by-step explanation:Let $x$ be a three-digit positive integer, and let $a$, $b$, and $c$ be the digits of $x$. Then we want $x - (a + b + c)$ to be a three-digit number whose digits are all the same. This means that $x - (a + b + c) = 111$, $222$, $\ldots$, or $888$.

If $x - (a + b + c) = 111$, then we have $x = a + b + c + 111$. Since $x$ is a three-digit number, $1 \leq a \leq 9$, $0 \leq b \leq 9$, and $0 \leq c \leq 9$. Thus, there are $9 \cdot 10 \cdot 10 = 900$ possible values of $x$ that give a three-digit number whose digits are all $1$.

Similarly, we have $x = a + b + c + 222$, $x = a + b + c + 333$, $\ldots$, and $x = a + b + c + 888$. Each of these equations gives $900$ possible values of $x$.

Therefore, there are a total of $8 \cdot 900 = \boxed{7200}$ three-digit positive integers $x$ that satisfy the conditions of the problem.