What is the remainder when the sum $1 + 7 + 13 + 19 + \cdots + 253 + 259$ is divided by $6$?
First of all, we see that every term has a remainder of $1$ when divided by $6.$ Now, we just need to find how many terms there are. The nth term can be given by the expression $6n - 5.$ Therefore, we set $259 = 6n - 5$ to find $n = 44,$ thus there are $44$ terms in our sum. Thus, the remainder of the sum is the same as the remainder of $44$ when divided by $6,$ which is $\boxed{2}.$
2
Level 4
Number Theory
Problem: What is the remainder when the sum $1 + 7 + 13 + 19 + \cdots + 253 + 259$ is divided by $6$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder when the sum of the series is divided by 6, first identify the series. The series consists of numbers starting from 1 and increasing by 6 each time: 1, 7, 13, 19, ..., 259. This is an arithmetic series where the first term \( a = 1 \) and the common difference \( d = 6 \).

Next, determine the number of terms in the series. The \( n \)-th term of an arithmetic series can be expressed as:
\[
a_n = a + (n-1)d
\]
Setting \( a_n = 259 \):
\[
259 = 1 + (n-1) \cdot 6
\]
Solving for \( n \):
\[
258 = (n-1) \cdot 6 \\
n-1 = 43 \\
n = 44
\]
Thus, there are 44 terms in the series.

Now, calculate the sum of the series using the formula for the sum of an arithmetic series:
\[
S_n = \frac{n}{2} \cdot (a + a_n)
\]
Substituting the known values:
\[
S_{44} = \frac{44}{2} \cdot (1 + 259) = 22 \cdot 260 = 5720
\]

Next, find the remainder of 5720 when divided by 6. To do this, calculate \( 5720 \mod 6 \). First, simplify \( 5720 \) by dividing it by 6:
\[
5720 \div 6 = 953 \quad \text{(integer part)}
\]
Now, multiply back to find the product:
\[
953 \cdot 6 = 5718
\]
Subtract this from 5720 to find the remainder:
\[
5720 - 5718 = 2
\]

Thus, the remainder when the sum \( 1 + 7 + 13 + 19 + \cdots + 253 + 259 \) is divided by 6 is \( 2 \).<|END_OF_LATENT|> The final answer is $2$.