What is the remainder when $1 + 2 + 3 + 4 + \dots + 9 + 10$ is  divided by 8?
Notice that we can pair many of these terms: \[1+7=2+6=3+5=8,\]so the remainder we want is the same as the remainder when $4+9+10$ is divided by 8.  We also see that this is the remainder when  \[4+1+2=7\]is divided by 8, so the answer is $\boxed{7}$.
7
Level 1
Number Theory
Problem: What is the remainder when $1 + 2 + 3 + 4 + \dots + 9 + 10$ is  divided by 8?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder when the sum of the first ten natural numbers is divided by 8, we first need to calculate the sum. The sum of the first n natural numbers can be calculated using the formula n(n + 1)/2. Here, n is 10. 

Calculating the sum:
10(10 + 1)/2 = 10 * 11 / 2 = 110.

Next, we need to determine the remainder when 110 is divided by 8. This involves performing the division of 110 by 8.

Calculating the division:
110 ÷ 8 = 13 with a remainder. To find the exact remainder, we multiply the quotient by 8 and subtract from 110:
13 * 8 = 104.

Now, we subtract this product from the original sum:
110 - 104 = 6.

Thus, the remainder when 110 is divided by 8 is 6. However, we need to ensure that we have correctly interpreted the problem. The final answer must be verified against the original question.

Revisiting the calculation, we can also check the sum of the numbers directly:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.

Now, we find the remainder of 55 when divided by 8:
55 ÷ 8 = 6 with a remainder. Again, we calculate:
6 * 8 = 48.

Subtracting gives:
55 - 48 = 7.

Thus, the correct remainder when the sum of the first ten natural numbers is divided by 8 is indeed 7.<|END_OF_LATENT|> The final answer is $7$.