Problem 2011

a) If Ats thought of numbers $1$ and $21$ and Pets of numbers $10$ and $12$, then both get the sum $22$ and the products will be $21$ and $120$, respectively, both of which give the remainder $21$ when divided by $99$.

b) Let the numbers Ats chose be $a$ and $b$, the ones Pets chose $c$ and $d$. According to the conditions stated in the problem, the numbers $(a + b) - (c + d)$ and $ab - cd$ are both divisible by $101$. Let $(a + b) - (c + d) = 101k$; then $a = 101k - b + c + d$, from where
$$\begin{aligned} ab - cd &= (101k - b + c + d)b - cd = 101kb - b^2 + bc + bd - cd \\ &= 101kb - (c - b)(d - b). \end{aligned}$$
Hence also the product $(c - b)(d - b)$ is divisible by $101$. As $101$ is a prime number, it has to divide either the factor $c - b$ or the factor $d - b$. W.l.o.g., let $c - b$ be divisible by $101$. As all the numbers are on the interval from $1$ to $101$, it means that $c = b$. But then $(a + b) - (c + d) = a - d$, which due to divisibility by $101$ means that $a = d$. Therefore Ats and Pets must have chosen the same numbers.