I notice that the sum of the first 10 natural numbers is an arithmetic series, which means I can use a formula to find it without adding up all the terms individually. The formula for the sum of an arithmetic series with first term $a_1$, last term $a_n$, and common difference $d$ is $S_n = \frac{n}{2}(a_1 + a_n)$. In this case, $a_1 = 1$, $a_n = 10$, $d = 1$, and $n = 10$, so I can plug these values into the formula and get $S_{10} = \frac{10}{2}(1 + 10) = 5 \times 11 = 55$. Now I have to find the remainder when 55 is divided by 8. I know that 8 goes into 48 exactly 6 times, so 48 is a multiple of 8. That means 55 is 7 more than a multiple of 8, so the remainder is 7. 
What is the remainder when $1 + 2 + 3 + 4 + \dots + 9 + 10$ is  divided by 8?