### Instruction: In base $10$, the number $2013$ ends in the digit $3$.  In base $9$, on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$.  For how many values of $b$ does the base-$b$-representation of $2013$ end in the digit $3$?### Response: To solve this problem, I need to find out what it means for a number to end in a certain digit in a different base.
I know that in base $10$, the last digit of a number tells me how many ones it has, the second-to-last digit tells me how many tens, and so on.
In base $9$, the last digit tells me how many ones, the second-to-last digit tells me how many nines, and so on.
So, in general, the last digit of a number in base $b$ tells me how many units of $b^0 = 1$ it has.
That means that the last digit of a number in any base is the same as the remainder when I divide that number by the base.
For example, $2013$ divided by $9$ gives me a quotient of $223$ and a remainder of $6$, which is why $2013$ ends in $6$ in base $9$.
So, the question is asking me for how many values of $b$ does $2013$ divided by $b$ have a remainder of $3$.
This is equivalent to asking for how many values of $b$ does $2013 - 3$ divided by $b$ have a remainder of $0$.
In other words, how many values of $b$ are factors of $2010$?
To find the factors of $2010$, I can prime factorize it as $2 \times 3 \times 5 \times 67$.
Any factor of $2010$ must be a product of some combination of these prime factors.
For example, $6$ is a factor of $2010$ because it is $2 \times 3$, and $335$ is a factor of $2010$ because it is $5 \times 67$.
There are $2^4 = 16$ possible combinations of these prime factors, so there are $16$ factors of $2010$.
However, not all of them are valid values of $b$.
The base $b$ has to be greater than $3$, because otherwise $3$ would not be a valid digit in base $b$.
So, I have to exclude the factors of $2010$ that are less than or equal to $3$, which are $1$, $2$, and $3$.
That leaves me with $16 - 3 = 13$ possible values of $b$ that satisfy the condition.