### Instruction: Find $1_6 + 2_6 + 3_6 + \cdots + 45_6$. Express your answer in base $6$.### Response: I notice that this is a sum of an arithmetic sequence, where the first term is $1_6$, the common difference is $1_6$, and the last term is $45_6$.
To find the number of terms in the sequence, I can use the formula $a_n = a_1 + (n-1)d$, where $a_n$ is the last term, $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference.
Plugging in the values, I get $45_6 = 1_6 + (n-1)1_6$, which simplifies to $44_6 = (n-1)1_6$.
To solve for $n$, I need to divide both sides by $1_6$, but I don't know how to do that in base $6$.
To make things easier, I can convert everything to base $10$, where I'm more familiar with the operations.
To convert a number from base $6$ to base $10$, I can multiply each digit by its corresponding power of $6$ and add them up.
For example, $45_6 = 4 \times 6^1 + 5 \times 6^0 = 24 + 5 = 29_{10}$.
Similarly, $1_6 = 1 \times 6^0 = 1_{10}$ and $44_6 = 4 \times 6^1 + 4 \times 6^0 = 24 + 4 = 28_{10}$.
So, in base $10$, the equation becomes $28 = (n-1)1$, which is easy to solve.
Adding $1$ to both sides, I get $29 = n$.
This means there are $29$ terms in the sequence.
To find the sum of the sequence, I can use another formula: $S_n = \frac{n}{2}(a_1 + a_n)$, where $S_n$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term.
Plugging in the values, I get $S_{29} = \frac{29}{2}(1 + 45_6)$.
But before I can simplify this, I need to convert $45_6$ back to base $10$.
I already did this before, so I know it's $29_{10}$.
So, in base $10$, the sum is $S_{29} = \frac{29}{2}(1 + 29) = \frac{29}{2}(30) = \frac{870}{2} = 435$.
But the question wants the answer in base $6$, not base $10$.
So, I need to convert $435_{10}$ to base $6$.
To do this, I can repeatedly divide by $6$ and write down the remainders.
For example, $435 \div 6 = 72$ with a remainder of $3$.
Then, $72 \div 6 = 12$ with a remainder of $0$.
Then, $12 \div 6 = 2$ with a remainder of $0$.
Finally, $2 \div 6 = 0$ with a remainder of $2$.
The remainders form the digits of the base $6$ number, starting from the last one.
So, $435_{10} = 2003_6$.