3. Find the natural numbers of the form $\overline{ab}$ which, when divided by 36, give a remainder that is a perfect square.

(7 points)

## E:14178 from GM 11/2011
3. SOLUTION

$\overline{a b}=36 \cdot \mathrm{c}+\mathrm{r} .0 \leq \mathrm{r}<36$ and $\mathrm{r}$ is a perfect square

$\Rightarrow \mathrm{r} \in\{0,1,4,9,16,25\}$

...... $1 \mathrm{p}$

$36 \cdot \mathrm{c} \leq \overline{a b} \Rightarrow \mathrm{c} \in\{0,1,2\}$

We obtain the numbers: $16,25,36,37,40,45,52,61,72,73,76,81,88,97$. $4 p$