C1. We consider sequences of integers $a_{1}, a_{2}, \ldots, a_{n}$ consisting of $n$ integers. For a given $k \leqslant n$, we can divide the numbers in the sequence into $k$ groups as follows: $a_{1}$ goes into the first group, $a_{2}$ into the second group, and so on until $a_{k}$ which goes into the $k$-th group. Then $a_{k+1}$ goes back into the first group, $a_{k+2}$ into the second group, and so forth. We call the sequence $k$-divisible if this division has the property that the sum of the numbers in each group is the same.

For example, the sequence $1,2,3,4,-2,6,13,12,17,8$ is 4-divisible, because

$$
1+(-2)+17=2+6+8=3+13=4+12 \text {. }
$$

However, this sequence is not 3-divisible, because the sums of $1+4+13+8, 2+(-2)+12$, and $3+6+17$ are not all the same.

(a) Give a sequence of 6 different numbers that is both 2-divisible and 3-divisible.

(b) Give a sequence of 7 different numbers that is simultaneously 2-divisible, 3-divisible, and 4-divisible.

(c) Find the largest number $k \leqslant 99$ for which there exists a $k$-divisible sequence of 99 different numbers. (Thus, provide an example of such a sequence and prove that such a sequence does not exist for larger $k$.)
C1. (a) An example of a sequence that meets the criteria is 5,7,6,3,1,2. This sequence consists of six different numbers and is 2-divisible, since $5+6+1=7+3+2$. It is also 3-divisible, since $5+3=7+1=6+2$.

This is an example of a good solution. There are many more possible sequences. Below, we show how we arrived at this sequence.

We are looking for a sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ that is 2-divisible and 3-divisible. Therefore, the following must hold:

$$
a_{1}+a_{4}=a_{2}+a_{5}=a_{3}+a_{6} \quad \text { and } \quad a_{1}+a_{3}+a_{5}=a_{2}+a_{4}+a_{6}
$$

If we choose $a_{4}=-a_{1}, a_{5}=-a_{2}$, and $a_{6}=-a_{3}$, then the first two equations are already satisfied. The third equation then gives $a_{1}+a_{3}-a_{2}=a_{2}-a_{1}-a_{3}$, or $a_{1}+a_{3}=a_{2}$. We choose $a_{1}=1, a_{3}=2$, and thus $a_{2}=3$. We then get the sequence $1,3,2,-1,-3,-2$ consisting of six different numbers. We can optionally increase all six numbers by 4 to get a sequence with only positive numbers: 5, 7, 6, 3, 1, 2.

(b) An example of a sequence that meets the criteria is $8,17,26,27,19,10,1$. This sequence of seven different numbers is 2-divisible, since $8+26+19+1=17+27+10$. It is 3-divisible, as $8+27+1=17+19=26+10$. It is also 4-divisible, since $8+19=17+10=26+1=27$. This is an example of a good solution. There are many more possible sequences. Below, we show how we arrived at this sequence.

We are looking for a sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ that is 2-, 3-, and 4-divisible. Therefore, the following must hold:

$$
\begin{aligned}
& a_{1}+a_{3}+a_{5}+a_{7}=a_{2}+a_{4}+a_{6}, \\
& a_{1}+a_{4}+a_{7}=a_{2}+a_{5}=a_{3}+a_{6}, \\
& a_{1}+a_{5}=a_{2}+a_{6}=a_{3}+a_{7}=a_{4} .
\end{aligned}
$$

What immediately stands out is that in equation (1), $a_{1}+a_{5}$ and $a_{3}+a_{7}$ appear on the left side, and $a_{2}+a_{6}$ and $a_{4}$ appear on the right side. If the sequence is 4-divisible, these four are all equal. We thus see that a 4-divisible sequence is automatically 2-divisible.

From the equations in (3), we get

$$
a_{1}=a_{4}-a_{5}, \quad a_{2}=a_{4}-a_{6}, \quad a_{3}=a_{4}-a_{7}
$$

If we substitute these into the equations (2), we get

$$
2 a_{4}+a_{7}-a_{5}=a_{4}+a_{5}-a_{6}=a_{4}+a_{6}-a_{7} .
$$

By subtracting $a_{4}$ from each, we find

$$
a_{4}+a_{7}-a_{5}=a_{5}-a_{6}=a_{6}-a_{7} .
$$

We thus find that

$$
\begin{aligned}
& a_{4}=\left(a_{5}-a_{6}\right)-\left(a_{7}-a_{5}\right)=2 a_{5}-a_{6}-a_{7}, \\
& a_{5}=\left(a_{6}-a_{7}\right)+a_{6}=2 a_{6}-a_{7}
\end{aligned}
$$

We have thus expressed $a_{1}, a_{2}, a_{3}$, $a_{4}$, and $a_{5}$ in terms of $a_{6}$ and $a_{7}$. Every solution can be found by a suitable choice of $a_{6}$ and $a_{7}$ such that all seven numbers are different. We try $a_{6}=10$ and $a_{7}=1$ and find successively:

$$
\begin{gathered}
a_{5}=2 \cdot 10-1=19, \quad a_{4}=2 \cdot 19-10-1=27, \quad a_{3}=27-1=26, \\
a_{2}=27-10=17 \quad \text { and } \quad a_{1}=27-19=8 .
\end{gathered}
$$

Thus, we find a solution.

(c) The largest $k$ for which there exists a $k$-divisible sequence of 99 different numbers is $k=50$. An example of such a sequence is

$$
1,2, \ldots, 48,49,100,99,98, \ldots, 52,51
$$

The 99 numbers in the sequence are indeed different, and we see that $1+99=2+98=\ldots=48+52=49+51=100$, so this sequence is 50-divisible.

Suppose now that $k>50$ and that we have a sequence $a_{1}, a_{2}, \ldots, a_{99}$ that is $k$-divisible. Consider the group that contains $a_{49}$. Since $49-k99$, this group can contain no other numbers besides $a_{49}$. Next, consider the group that contains $a_{50}$. Since $50-k99$, this group can contain no other numbers besides $a_{50}$. The numbers $a_{49}$ and $a_{50}$ thus both belong to a group alone and must therefore be equal. However, this is not allowed, as the numbers must be different.