
C4. We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there?


Solution. If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.

- In order to obtain $n=7 m+1$, arrange the kids in a circle and let each kid send a message to the first $4 m$ kids to its right and hence receive a message from the first $4 m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
- In order to obtain $n=7 m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7 m-1$ kids in a circle and let each kid on the circle send a message to the first $4 m-1$ kids to its right and hence receive a message from the first $4 m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.

There are 26 two-digit numbers with remainder 0 or 1 modulo 7 . (All numbers of the form $7 m$ and $7 m+1$ with $2 \leqslant m \leqslant 14$.)
