What is the remainder when $1 + 2 + 3 + 4 + \dots + 9 + 10$ is divided by 9?
I notice that the sum $1 + 2 + 3 + 4 + \dots + 9 + 10$ is an arithmetic series with the first term $a = 1$, the common difference $d = 1$, and the number of terms $n = 10$. I recall that the formula for the sum of an arithmetic series is $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term of the series. In this case, $a = 1$ and $l = 10$, so I plug them into the formula and simplify: $S_{10} = \frac{10}{2}(1 + 10) = 5 \cdot 11 = 55$. Now I need to find the remainder when $55$ is divided by $9$. I could use long division, but I also know a shortcut: I can find the remainder by adding the digits of $55$ and dividing that by $9$. The digits of $55$ add up to $5 + 5 = 10$, and the remainder when $10$ is divided by $9$ is $1$ (since $10 = 9 \cdot 1 + 1$). Therefore, the remainder when $55$ is divided by $9$ is also $1$. 