### 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 consecutive numbers in base $6$.
I wonder if there is a formula for this kind of sum, similar to the one for base $10$.
I recall that the formula for the sum of the first $n$ natural numbers in base $10$ is $S_n = \frac{n(n+1)}{2}$.
I wonder if this formula works in base $6$ as well.
I try a few examples to test the formula.
For $n = 1$, I get $S_1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$.
This matches $1_6$.
For $n = 2$, I get $S_2 = \frac{2(2+1)}{2} = \frac{3}{1} = 3$.
This matches $1_6 + 2_6 = 3_6$.
For $n = 3$, I get $S_3 = \frac{3(3+1)}{2} = \frac{12}{2} = 6$.
But this is not a valid base $6$ number.
I need to convert it to base $6$.
I remember that to convert a base $10$ number to base $6$, I need to repeatedly divide by $6$ and write the remainders in reverse order.
For $6$, I get $6 \div 6 = 1$ with remainder $0$.
So the base $6$ representation is $10_6$.
This matches $1_6 + 2_6 + 3_6 = 10_6$.
So far, the formula seems to work, but I need to be careful with the conversion step.
I decide to use the formula to find the answer to the problem.
I need to find the value of $n$ such that $n_6 = 45_6$.
I can convert $45_6$ to base $10$ by multiplying each digit by its place value and adding them up.
I get $4 \times 6^1 + 5 \times 6^0 = 24 + 5 = 29$.
So $n = 29$.
I plug in $n = 29$ into the formula and get $S_{29} = \frac{29(29+1)}{2} = \frac{29(30)}{2} = \frac{870}{2} = 435$.
This is the base $10$ answer, but I need to convert it to base $6$.
I use the same method as before to convert $435$ to base $6$.
I get $435 \div 6 = 72$ with remainder $3$, $72 \div 6 = 12$ with remainder $0$, $12 \div 6 = 2$ with remainder $0$, and $2 \div 6 = 0$ with remainder $2$.
I write the remainders in reverse order and get $2003_6$.