What is the modulo $6$ remainder of the sum $1+2+3+4+\dots+98+99+100?$
I notice that the sum is an arithmetic series with a common difference of $1$ and $100$ terms. In this case, $n = 100$, $a_1 = 1$, and $a_n = 100$, so the sum is $S_{100} = \frac{100}{2}(1 + 100) = 50 \cdot 101 = 5050$. To find the modulo $6$ remainder of $5050$, I can use the division algorithm to write $5050 = 6q + r$, where $q$ is the quotient and $r$ is the remainder. I can divide $5050$ by $6$ using long division or a calculator and get $q = 841$ and $r = 4$. This means that $5050 \equiv 4 \pmod{6}$, so the modulo $6$ remainder of the sum is $4$. 