Let's break down the equation:

$ABC + ABC + ABC = CCC$

This can be rewritten as:

$3 \times ABC = CCC$

Where $ABC$ is a three-digit number represented by $100A + 10B + C$, and $CCC$ is a number where all three digits are $C$, meaning $CCC = 111 \times C$.

Now, let's express the equation:

$3 \times (100A + 10B + C) = 111C$

Expanding both sides:

$300A + 30B + 3C = 111C$

Subtracting $3C$ from both sides:

$300A + 30B = 108C$

Dividing the whole equation by 3 to simplify:

$100A + 10B = 36C$

Now, we need to find values for $A$, $B$, and $C$ such that this equation holds, and $A$, $B$, and $C$ are unique digits between 0 and 9.

Let's try different values of $C$:

• For $C = 1$, $36C = 36$. But $100A + 10B = 36$ has no valid digit combinations for $A$ and $B$.
• For $C = 2$, $36C = 72$. This gives $100A + 10B = 72$, so $A = 0$ and $B = 7$.

Thus, $ABC = 072$ and $CCC = 222$.

Finally, let's check if this satisfies the original equation:

$072 + 072 + 072 = 222$

So, the solution is $CCC = 222$.

Would you like more details or have any questions?

Here are some follow-up questions to deepen your understanding:

1. How would the equation change if $CCC$ were $333$?
2. Can we find another solution if we allow $A$, $B$, or $C$ to repeat?
3. Why does the equation work out for $C = 2$ but not for higher values?
4. How would the problem differ if we changed the multiplier from 3 to 4?
5. What happens if $A$, $B$, and $C$ were allowed to be non-unique?

Tip: When breaking down problems, isolating one variable at a time can simplify complex equations effectively.