### Instruction: Find $1_6 + 2_6 + 3_6 + \cdots + 45_6$. Express your answer in base $6$.### Response: I know that adding numbers in different bases is similar to adding in base $10$, except that I have to carry over whenever I exceed the base.
For example, $5_6 + 2_6 = 11_6$, because $5 + 2 = 7$, which is one group of six and one leftover.
I also know that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
So, I can use this formula to find the sum of the first $45$ base $6$ numbers, and then convert the result to base $6$.
To find the sum of the first $45$ base $6$ numbers, I need to convert $45_6$ to base $10$ first.
I can do this by multiplying each digit by its place value and adding them up.
So, $45_6 = 4 \times 6^1 + 5 \times 6^0 = 24 + 5 = 29_{10}$.
Now, I can plug this into the formula: $\frac{29(30)}{2} = 435_{10}$.
This is the sum of the first $45$ base $6$ numbers in base $10$.
To convert it back to base $6$, I need to divide it by $6$ and write down the remainders.
So, $435 \div 6 = 72$ 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 $2$.
To get the base $6$ representation, I read the remainders from bottom to top.
So, $435_{10} = 2003_6$.
Therefore, the answer is $2003_6$.