smc

First, let $R(n)$ be the sum of the $n$th row. Now, with some observation and math instinct, we can guess that $R(n)=2^n-n$. Now we try to prove it by induction, $R(1)=2^n-n=2^1-1=1$ (works for base case) $R(k)=2^k-k$ $R(k+1)=2^{k+1}-(k+1)=2(2^k)-k-1$ By definition from the question, the next row is always$:$ Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1) Which gives us $:$ $2(2^k-k)+(k+1)-2=2(2^k)-k-1$ Hence, proven Last, simply substitute $n=2023$, we get $R(2023)=2^{2023}-2023$ Last digit of $2^{2023}$ is $8$, $8-3=\boxed{5}$