10,11Auto: Voomenkov C.I.

At the New Year's Eve party, several married couples arrived, each with 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families to take them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio. How many children could there have been at this party in total?
Let there be $p$ married couples and $d$ children at the evening. Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and after her selection, a father could be chosen from one of the $p-2$ remaining couples. Therefore, the total number of trios is

$d(p-1)(p-2)=3630$. Since $d \leq 10 p$, we get $3630 \leq 10 p^{3}$, which means $p^{3} \geq 363>73$. Therefore, $p \geq 8$.

The number $3630=2 \cdot 3 \cdot 5 \cdot 11^{2}$ has two divisors $p-1$ and $p-2$, differing by 1. If one of these divisors is divisible by 11, then the other gives a remainder of 1 or 10 when divided by 11. Then it is coprime with 11, and thus divides $2 \cdot 3 \cdot 5=30$ and is at least 10. It is not hard to see that this divisor can only be 10; then $p-2=10, p-1=11$ and $d=3630: 110=33$.

If both numbers $p-2$ and $p-1$ are not divisible by 11, then the number $2 \cdot 3 \cdot 5=30$ is divisible by their product, which contradicts $p \geq 8$.

## Answer

33 children.