To solve this problem, we need to fill in the boxes with digits from 0 through 9 such that the equation becomes valid. Given the constraint that each digit can be used at most once, it's important to check all possible combinations of digits to ensure the equation holds.
First, let's describe the general structure of the equation:
[tex]\[
\text{left\_num} \% ~ \text{of} \quad \text{X Y} = \text{A B C}
\][/tex]
Here:
- [tex]\(\text{left\_num}\)[/tex] is a single digit.
- [tex]\(\text{X}\)[/tex] and [tex]\(\text{Y}\)[/tex] form a two-digit number.
- [tex]\(\text{A}\)[/tex], [tex]\(\text{B}\)[/tex], and [tex]\(\text{C}\)[/tex] form a three-digit number, which is the result.
The goal is to find digits for [tex]\(\text{left\_num}\)[/tex], [tex]\(X\)[/tex], [tex]\(Y\)[/tex], [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] such that:
[tex]\[
\text{left\_num} \% ~ \text{of} \quad XY = ABC
\][/tex]
In mathematical terms, this can be written as:
[tex]\[
\frac{\text{left\_num}}{100} \times (10X + Y) = 100A + 10B + C
\][/tex]
After evaluating all possible permutations of the digits 0 through 9 while ensuring each digit is used at most once, it turns out that no combination satisfies this equation.
Therefore, based on this thorough investigation of the permutations, we conclude that:
No solution found.
This means it's not possible to fill those boxes with digits from 0 to 9 under given constraints to make the equation true.
