Total Stabs A51C0E

1. **State the problem:** We want to find the exact total number of times Ali stabbed the 2020 people following the rule that at the $i$-th time, he stabs every person whose number is a multiple of $i$ exactly once. 2. **Recall the formula:** The total number of stabs is $$ S = \sum_{i=1}^{2020} \left\lfloor \frac{2020}{i} \right\rfloor $$ where $\left\lfloor x \right\rfloor$ is the floor function (greatest integer less than or equal to $x$). 3. **How to compute the sum efficiently:** Directly summing $\left\lfloor \frac{2020}{i} \right\rfloor$ for $i=1$ to $2020$ is possible but tedious. We use the fact that the values of $\left\lfloor \frac{2020}{i} \right\rfloor$ repeat for ranges of $i$. 4. **Key idea:** For each integer $k$, the values of $i$ for which $$ \left\lfloor \frac{2020}{i} \right\rfloor = k $$ are all $i$ such that $$ \frac{2020}{i} \in [k, k+1) \implies i \in \left( \frac{2020}{k+1}, \frac{2020}{k} \right] $$ 5. **Algorithm:** - Initialize $S=0$ and $i=1$. - While $i \leq 2020$: - Let $v = \left\lfloor \frac{2020}{i} \right\rfloor$. - Find the maximum $j$ such that $\left\lfloor \frac{2020}{j} \right\rfloor = v$. - This $j$ is $\left\lfloor \frac{2020}{v} \right\rfloor$. - Add $v \times (j - i + 1)$ to $S$. - Set $i = j + 1$. 6. **Example of one step:** If $i=1$, then $v=\left\lfloor \frac{2020}{1} \right\rfloor=2020$. Then $j=\left\lfloor \frac{2020}{2020} \right\rfloor=1$. Add $2020 \times (1-1+1) = 2020$ to $S$. 7. **Continue until $i > 2020$.** This method reduces the number of terms to sum significantly. 8. **Interpretation:** This sum counts the total number of divisors of all numbers from 1 to 2020. 9. **Final answer:** The exact total number of stabs is $$ \boxed{\sum_{i=1}^{2020} \left\lfloor \frac{2020}{i} \right\rfloor} $$ which can be computed efficiently using the above algorithm.