For how many digits $C$ is the positive three-digit number $1C3$ a multiple of 3?
$1C3$ is a multiple of 3 if and only if the sum of the digits, $1+C+3$, is a multiple of 3. Plugging in each possible digit for $C$ gives us that $1+C+3$ is a multiple of 3 only for $C=2, 5, 8$. Thus, $1C3$ is a multiple of 3 for $\boxed{3}$ digits $C$.
3
Level 2
Prealgebra
Problem: For how many digits $C$ is the positive three-digit number $1C3$ a multiple of 3?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many digits \( C \) make the number \( 1C3 \) a multiple of 3, we need to apply the rule of divisibility for 3. A number is a multiple of 3 if the sum of its digits is a multiple of 3. 

First, we identify the digits of the number \( 1C3 \). The digits are \( 1 \), \( C \), and \( 3 \). 

Next, we calculate the sum of these digits:
\[
1 + C + 3 = C + 4
\]

Now, we need to find the values of \( C \) such that \( C + 4 \) is a multiple of 3. We can express this condition mathematically:
\[
C + 4 \equiv 0 \mod 3
\]

This can be rearranged to:
\[
C \equiv -4 \mod 3
\]
Since \(-4\) is equivalent to \(2\) modulo \(3\) (because \(-4 + 6 = 2\)), we have:
\[
C \equiv 2 \mod 3
\]

Now, we need to find the digits \( C \) that satisfy this condition. The possible digits for \( C \) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \). We will check which of these digits are congruent to \( 2 \mod 3 \):

- \( C = 0 \): \( 0 \mod 3 \) (not valid)
- \( C = 1 \): \( 1 \mod 3 \) (not valid)
- \( C = 2 \): \( 2 \mod 3 \) (valid)
- \( C = 3 \): \( 0 \mod 3 \) (not valid)
- \( C = 4 \): \( 1 \mod 3 \) (not valid)
- \( C = 5 \): \( 2 \mod 3 \) (valid)
- \( C = 6 \): \( 0 \mod 3 \) (not valid)
- \( C = 7 \): \( 1 \mod 3 \) (not valid)
- \( C = 8 \): \( 2 \mod 3 \) (valid)
- \( C = 9 \): \( 0 \mod 3 \) (not valid)

The valid values of \( C \) that satisfy \( C \equiv 2 \mod 3 \) are \( 2, 5, \) and \( 8 \). 

Thus, there are a total of \( 3 \) digits \( C \) for which the number \( 1C3 \) is a multiple of 3. 

The final answer is \( 3 \).<|END_OF_LATENT|> The final answer is $3$.