Professor Piraldo knows many multiplication tricks. One day, Juquinha asked what $62 \cdot 68$ equals, and the professor quickly answered 4216. After a few minutes of excitement from the class, he decided to explain this trick. The trick works when multiplying two two-digit numbers that have the same tens digit and the sum of the units is 10. In the example, both have 6 in the tens place and $2+8=10$. He explained that the result can have up to 4 digits, the product of the units defines the last 2 digits, and the first 2 digits, when there are two since the result can have three digits in total, are the result of the tens digit multiplied by its successor.

(a) Using the trick, calculate $45 \cdot 45$.

(b) Using the trick, calculate $71 \cdot 79$.

(c) Now let's prove that the trick works. Consider two two-digit numbers $\overline{a b}=10 a+b$ and $\overline{a c}=10 a+c$, with $b+c=10$. Show that the product $b \cdot c$ determines the last 2 digits and $a(a+1)$ determines the first 2 digits of the product $\overline{a b} \cdot \overline{a c}$.
Solution

(a) Using the trick, the last two digits are given by $5 \cdot 5=25$ and the first two by $4 \cdot(4+1)=20$. Therefore, $45 \cdot 45=2025$.

(b) In this case, we must use the trick carefully, as the last two digits are given by $1 \cdot 9=9$. Therefore, the number ends in 09. The first two digits are $7 \cdot(7+1)=$ 56. Using the trick, we conclude that $71 \cdot 79=5609$.
(c) Let's develop the multiplication using the decimal expansion of each number:

$$
\begin{aligned}
\overline{a b} \cdot \overline{a c} & =(10 a+b)(10 a+c) \\
& =100 a^{2}+10 a c+10 a b+b c \\
& =100 a^{2}+10 a(b+c)+b c \\
& =100 a^{2}+100 a+b c \\
& =100 a(a+1)+b c
\end{aligned}
$$

Since $b c<9 \cdot 9=81$, the result above has only two digits. Moreover, since $100 a(a+1)$ is a number ending in 00, the digits in the tens and units place of the product $\overline{a b} \cdot \overline{a c}$ coincide with the result of $b c$ and those in the hundreds and thousands place with the result of $a(a+1)$.