parmenides51 wrote: The cells of a $2021\times 2021$ table are filled with numbers using the following rule. The bottom left cell, which we label with coordinate $(1, 1)$, contains the number $0$. For every other cell $C$, we consider a route from $(1, 1)$ to $C$, where at each step we can only go one cell to the right or one cell up (not diagonally). If we take the number of steps in the route and add the numbers from the cells along the route, we obtain the number in cell $C$. For example, the cell with coordinate $(2, 1)$ contains $1 = 1 + 0$, the cell with coordinate $(3, 1)$ contains $3 = 2 + 0 + 1$, and the cell with coordinate $(3, 2)$ contains $7 = 3 + 0 + 1 + 3$. What is the last digit of the number in the cell $(2021, 2021)$? Insight: We can compute the numbers for the first few cells and deduce a pattern. Solution: Suppose $C_n$ is the value in cell $(x,y)$, where $n = x+y-2$. Then, we have $$C_n = n + \sum_{0\le i<n}{C_i}$$We claim that $C_n = 2^n - 1$ and prove by induction. With this proposition, $C_0 = 2^0 - 1 = 0$, which is true. Suppose the proposition holds for $C_k = 2^k - 1$. Then, we shall prove $C_{k+1} = 2^{k+1} - 1$. By definition, $$C_{k+1} = (k+1) + \sum_{0\le i<k+1}{C_i} = \sum_{0\le i<k}{C_i} + k + C_k + 1 = (2^k - 1) + (2^k - 1) + 1 = 2^{k+1} - 1$$which completes our proof. $\blacksquare$ Hence, the number in the cell $(2021, 2021)$ is $C_{4040} = 2^{4040} - 1$. Its last digit is $2^{4040} - 1 \equiv 6 - 1 \equiv \boxed{5} (\mod 10) $.