Exercise 2. How many 8-digit numbers are there whose decimal notation is of the form ab2019cd with a $>0$, and which are divisible by 360?
Solution to Exercise 2: To start, a solution number must be divisible by 10, so we must have $\mathrm{d}=0$, and the number written as ab2019c must be divisible by 36, hence by 4. The number written as ab20100 is divisible by 100, hence by 4, so the number written as 9c must also be divisible by 4, so c is 2 or 6.

Finally, the number must be divisible by 36, hence by 9, so the sum of its digits must be divisible by 9, so $\mathrm{a}+\mathrm{b}+2+0+1+9+\mathrm{c}$ is divisible by 9, so $\mathrm{a}+\mathrm{b}+\mathrm{c}+3$ is divisible by 9, so $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is 6 or 15 or 24 (the sum of three digits cannot exceed 27).

If $\mathrm{c}=2$, then $a+b+c \leqslant 2+9+9=20$, so $a+b+c$ is 6 or 15, so $a+b$ is 4 or 13, which gives the following solutions:

$$
\text { 13201920, 22201920,31201920, 40201920, }
$$

$$
49201920,58201920,67201920,76201920,85201920,94201920 .
$$

If $\mathrm{c}=6$, since $\mathrm{a}>0$, we have $\mathrm{a}+\mathrm{b}+\mathrm{c}>6$ so $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is 15 or 24, so $\mathrm{a}+\mathrm{b}$ is 9 or 18, which gives the following solutions:

18201960, 27201960, 36201960, 45201960, 54201960, 63201960, 72201960, 81201960, 90201960,

99201960.

In total, we have 20 solutions.