Second Round, March 2019

a. An example of a correct sequence is $5,7,6,3,1,2$. This sequence consists of six distinct numbers and is 2-composite since $5+6+1=7+3+2$. It is also 3-composite since $5+3=7+1=6+2$.

b. A possible solution is $8,17,26,27,19,10,1$. This sequence consists of seven distinct integers and is 2-composite since $8+26+19+1=17+27+10$. It is 3-composite since $8+27+1=17+19=26+10$. It is also 4-composite since $8+19=17+10=26+1=27$.

c. The largest $k$ for which a $k$-composite sequence of 99 distinct integers exists, is $k=50$. An example of such a sequence is $$1,2,\dots ,48,49,100,99,98,\dots ,52,51.$$ The 99 integers in the sequence are indeed distinct and we see that $1+99=2+98=\cdots =48+52=49+51=100$, so this sequence is 50-composite.

Now suppose that $k>50$ and that we have a $k$-composite sequence $a_1,a_2,\dots ,a_{99}$. Consider the group that contains the number $a_{49}$. Since $49-k<0$ and $49+k>99$, this group cannot contain any other number beside $a_{49}$. Next, consider the group containing the number $a_{50}$. Since $50-k<0$ and $50+k>99$, this group cannot contain any other number beside $a_{50}$. Hence, the numbers $a_{49}$ and $a_{50}$ each form a group by themselves and must therefore have the same value. But this is not allowed since the 99 numbers in the sequence had to be distinct.