2021/22 AHSMC Part 1 Problem 9

Each child can bring $0, 1, 2, \ldots, 10$ toys. Since there are $100$ children, at least $\bigl\lceil 100/11 \bigr\rceil = \lceil 9.09 \rceil = 10$ must bring the same number of toys. This shows $n \ge 10$. To show that $n = 10$ is the answer, suppose that $10$ children brought $0$ toys each, $10$ more children brought one toy each, and so on until the last $10$ children brought $9$ toys each. This accounts for all $100$ children, and in this situation the statement is false for any $n > 10$. The answer is (B).