, 2022

Solution:

Let's first solve the problem, ignoring the possibility that the $a_{i}$ and $b_{i}$ can be zero. Call a positive integer $s$ an $A$-sum if $s=\sum_{i=1}^{m} a_{i}$ for some nonnegative integer $m$ (in particular, 0 is always an $A$-sum). Define the term $B$-sum similarly. Let $E$ be the expected value of the smallest positive integer that is both an $A$-sum and a $B$-sum.

The first key observation to make is that if $s$ is both an $A$-sum and a $B$-sum, then the distance to the next number that is both an $A$-sum and a $B$-sum is $E$. To see this, note that if
$$s=\sum_{i=1}^{m} a_{i}=\sum_{j=1}^{n} b_{i}$$
the distance to the next number that is both an $A$-sum and a $B$-sum is the minimal positive integer $t$ so that there exist $m^{\prime}$ and $n^{\prime}$ so that
$$t=\sum_{i=1}^{m^{\prime}} a_{m+i}=\sum_{j=1}^{n^{\prime}} b_{n+i}$$
This is the same question of which we defined $E$ to be the answer, but with renamed variables, so the expected value of $t$ is $E$. As a result, we conclude that the expected density of numbers that are both $A$-sums and $B$-sums is $\frac{1}{E}$.

We now compute this density. Note that since the expected value of $a_{i}$ is $\frac{101}{2}$, the density of $A$-sums is $\frac{2}{101}$. Also, the density of $B$-sums is $\frac{2}{101}$. Moreover, as $n$ goes to infinity, the probability that $n$ is an $A$-sum approaches $\frac{2}{101}$ and the probability that $n$ is a $B$-sum approaches $\frac{2}{101}$. Thus, the density of numbers that are simultaneously $A$-sums and $B$-sums is $\frac{4}{101^{2}}$, so $E=\frac{101^{2}}{4}$.

We now add back the possibility that some of the $a_{i}$ and $b_{i}$ can be 0. The only way this changes our answer is that the $s$ we seek can be 0, which happens if and only if $a_{1}=b_{1}=0$. Thus our final answer is
$$\frac{1}{101^{2}} \cdot 0+\frac{101^{2}-1}{101^{2}} \cdot \frac{101^{2}}{4}=\frac{101^{2}-1}{4}=2550$$