B5. Given is the sequence of numbers $a_{0}, a_{1}, a_{2}, \ldots, a_{2020}$. It is given that $a_{0}=0$. Moreover, for every $k=1,2, \ldots, 2020$ it holds that

$$
a_{k}= \begin{cases}a_{k-1} \cdot k & \text { if } k \text { is divisible by } 8, \\ a_{k-1}+k & \text { if } k \text { is not divisible by } 8 .\end{cases}
$$

What are the last two digits of $a_{2020}$?

## C-Problems

For the C-Problems, not only the answer is important; a reasoning should also be provided that shows your answer is correct. Each correctly worked out C-Problem is worth 10 points. Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper for each problem as well.

NOTE: Solve each C-Problem on a separate sheet of paper and also submit the corresponding scratch paper per problem.
B5. 02 We see that $a_{2000}=a_{1999} \cdot 2000$ ends in three zeros. We can now determine the last digits of $a_{2001}, a_{2002}$ up to $a_{2020}$ using the formulas. It is only important to keep the last two digits, the rest we will ignore.

| $a_{2001}$ | $a_{2002}$ | $a_{2003}$ | $a_{2004}$ | $a_{2005}$ | $a_{2006}$ | $a_{2007}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $0+1=1$ | $1+2=3$ | $3+3=6$ | $6+4=10$ | $10+5=15$ | $15+6=21$ | $21+7=28$ |


| $a_{2008}$ | $a_{2009}$ | $a_{2010}$ | $a_{2011}$ | $a_{2012}$ | $a_{2013}$ | $a_{2014}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $28 \cdot 8=224$ | $24+9=33$ | $33+10=43$ | $43+11=54$ | $54+12=66$ | $66+13=79$ | $79+14=93$ |


| $a_{2015}$ | $a_{2016}$ | $a_{2017}$ | $a_{2018}$ | $a_{2019}$ | $a_{2020}$ |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $93+15=108$ | $8 \cdot 16=128$ | $28+17=45$ | $45+18=63$ | $63+19=82$ | $82+20=102$ |  |

We see that $a_{2020}$ ends in the digits 02.

## C-questions