Let $A, B$ and $C$ be non-zero digits such that $(A B)^{2}=$ $C A B$, that is, the two-digit number $A B$ squared gives the three-digit number $C A B$. Determine the value of $A+B+C$.

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Starting from the equality $(A B)^{2}=C A B$ and denoting the two-digit number $A B$ by $x$, we have $x^{2}=(A B)^{2}=C A B=C \cdot 100+x$, that is, $x^{2}-x=C \cdot 100$. Therefore, the product $x(x-1)=x^{2}-x$ is divisible by 100. Considering the factorization $100=2^{2} \cdot 5^{2}$, we divide the solution into three cases, according to the highest power of 5 that divides $x$.

1st Case: $5^{2}$ divides $x$.

Since $x$ is a two-digit number, the possible values of $x$ are 25, 50, and 75. We construct a table.

| $x$ | $x-1$ | $x \cdot(x-1)$ |
| :---: | :---: | :---: |
| 25 | 24 | 600 |
| 50 | 49 | 2450 |
| 75 | 74 | 5550 |

Therefore, the number $x \cdot(x-1)$ is a multiple of 100 only if $x=25$. Since $25^{2}=625$, in this case we have $C=6$.

2nd Case: 5 divides $x$, but $5^{2}$ does not divide $x$.

Then, necessarily, 5 divides $x-1$, since 5 divides $x \cdot(x-1)$. But this is an impossibility, because 5 cannot divide both consecutive numbers $x-1$ and $x$ (remember that numbers divisible by 5 end in 0 or 5). Thus, this case is excluded.

3rd Case: 5 does not divide $x$.

Then $5^{2}$ divides $x-1$. As in Case 1, we have $x-1=25, 50$ or 75 and the possible values of $x$ are 26, 51, and 76. We construct a table.

| $x$ | $x-1$ | $x \cdot(x-1)$ |
| :---: | :---: | :---: |
| 26 | 25 | 650 |
| 51 | 50 | 2550 |
| 76 | 75 | 5700 |

The product $x \cdot(x-1)$ is a multiple of 100 only if $x=76$, but this case is also excluded, since $76^{2}=5776$ has more than three digits.

Thus, the only possibility is $A=2, B=5$ and $C=6$, with sum $A+B+C=13$.