### Instruction: Find $1_6 + 2_6 + 3_6 + \cdots + 45_6$. Express your answer in base $6$.### Response: I want to find the sum of the first $45$ natural numbers in base $6$.
I know that the formula for the sum of the first $n$ natural numbers is $n(n+1)/2$, so I can try to apply that formula to this problem.
But before I do that, I need to convert $45_6$ to base $10$, so I can use the formula.
I remember that $45_6 = 4 \cdot 6^1 + 5 \cdot 6^0 = 24 + 5 = 29$ in base $10$.
So the problem is equivalent to finding $1 + 2 + 3 + \cdots + 29$ in base $10$ and then converting the answer back to base $6$.
I can use the formula now: $1 + 2 + 3 + \cdots + 29 = 29(29+1)/2 = 435$ in base $10$.
To convert $435$ to base $6$, I need to divide it by $6$ repeatedly and write down the remainders.
I get $435 \div 6 = 72$ 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 $2$.
So $435 = 2003_6$ in base $6$.
That means the final answer is $2003_6$.