A Tim number is a five-digit positive integer with the property that it is a multiple of 15 , its hundreds digit is 3, and its tens digit is equal to the sum of its first (leftmost) three digits. How many Tim numbers are there?
Suppose $A B C D E$ is a Tim number. That is, $A, B, C, D$, and $E$ are the digits of the number and since it has five digits, $A \neq 0$.

Since $A B C D E$ is a multiple of 15 , it must be a multiple of both 3 and 5 .

Since $A B C D E$ is a multiple of 5 , either $E=0$ or $E=5$.

Since $A B C D E$ is a multiple of $3, A+B+C+D+E$ is a multiple of 3 .

We are also given that $C=3$ and $D=A+B+C=A+B+3$.

Combining $C=3$ and $D=A+B+3$ with the fact that $A+B+C+D+E$ is a multiple of 3 , we have that $A+B+3+A+B+3+E=2 A+2 B+E+6$ is a multiple of 3 .

Since 6 is a multiple of 3 , this means $2 A+2 B+E$ is a multiple of 3 .

We are looking for triples $(A, B, E)$ with the property that $E=0$ or $E=5$ and $2 A+2 B+E$ is a multiple of 3. However, we must keep in mind that $A+B+3=D$ is a digit, which means $A+B+3 \leq 9$ or $A+B \leq 6$.

We will consider two cases: $E=0$ and $E=5$. We note that once we choose values for $A$ and $B$, the value of $D$ will be determined.

$E=0$ : In this case, we need $2 A+2 B$ to be a multiple of 3 , so $A+B$ is a multiple of 3 .

Combined with the requirement that $A+B \leq 6$, this means $A+B=3$ or $A+B=6$. We also need to remember that $A \neq 0$, which means $(A, B)$ is one of $(1,2),(2,1),(3,0),(1,5),(2,4)$, $(3,3),(4,2),(5,1)$, and $(6,0)$ for a total of nine possibilities.

$\underline{E=5}$ : In this case, we need $2 A+2 B+5$ to be a multiple of 3 . For the same reason as in the previous case, $A+B \leq 6$.

If $A+B=1$, then $2 A+2 B+5=7$ which is not a multiple of 3 .

If $A+B=2$, then $2 A+2 B+5=9$ which is a multiple of 3 .

If $A+B=3$, then $2 A+2 B+5=11$ which is not a multiple of 3 .

If $A+B=4$, then $2 A+2 B+5=13$ which is not a multiple of 3 .

If $A+B=5$, then $2 A+2 B+5=15$ which is a multiple of 3 .

If $A+B=6$, then $2 A+2 B+5=17$ which is not a multiple of 3 .

Therefore, we must have that $A+B=2$ or $A+B=5$. Keeping in mind that $A \neq 0$, we get that $(A, B)$ can be $(1,1),(2,0),(1,4),(2,3),(3,2),(4,1)$, or $(5,0)$ for a total of seven possibilities.

This gives a total of $9+7=16$ Tim numbers, which are listed below. One can check that all 16 of these integers are indeed Tim numbers.

12360, 15390, 21360, 24390, 30360, 33390, 42390, 51390,

$60390,11355,14385,20355,23385,32385,41385,50385$

ANSWER: 16

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