How many positive three-digit integers with a $5$ in the units place are divisible by $15$?
Odd multiples of 5 have a 5 in the units place, while even multiples of 5 have a 0 in the units place. Multiples of 15 are multiples of 5, so we look at how many three-digit multiples of 15 are odd to find how many have a 5 in the units place. The three-digit multiples of 15 range from 105 to 990, or $15\times7$ to $15\times 66$. So there are $66-7+1=60$ three-digit multiples of 15. Half of them will be odd, so there are 30 odd, three-digit multiples of 15. There are $\boxed{30}$ positive, three-digit integers with a 5 in the units place that are divisible by 15.
30
Level 4
Number Theory
Problem: How many positive three-digit integers with a $5$ in the units place are divisible by $15$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the three-digit integers with a $5$ in the units place, we note that these integers can be expressed in the form $100a + 10b + 5$, where $a$ is a digit from $1$ to $9$ (since it is a three-digit number) and $b$ is a digit from $0$ to $9$. 

Next, we need to determine the conditions for divisibility by $15$. A number is divisible by $15$ if it is divisible by both $3$ and $5$. Since all numbers in our form end with $5$, they are already divisible by $5$. 

Now, we check for divisibility by $3$. A number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of the digits for our number is $a + b + 5$. 

We need to find the values of $a$ and $b$ such that $a + b + 5$ is divisible by $3$. Rearranging gives us $a + b \equiv -5 \equiv 1 \pmod{3}$. 

Next, we analyze the possible values of $a$ and $b$. The digit $a$ can take values from $1$ to $9$, and for each value of $a$, we will find the corresponding values of $b$ that satisfy the condition.

1. If $a = 1$, then $1 + b \equiv 1 \pmod{3}$ implies $b \equiv 0 \pmod{3}$. Possible values for $b$ are $0, 3, 6, 9$ (4 values).
2. If $a = 2$, then $2 + b \equiv 1 \pmod{3}$ implies $b \equiv -1 \equiv 2 \pmod{3}$. Possible values for $b$ are $2, 5, 8$ (3 values).
3. If $a = 3$, then $3 + b \equiv 1 \pmod{3}$ implies $b \equiv -2 \equiv 1 \pmod{3}$. Possible values for $b$ are $1, 4, 7$ (3 values).
4. If $a = 4$, then $4 + b \equiv 1 \pmod{3}$ implies $b \equiv -3 \equiv 0 \pmod{3}$. Possible values for $b$ are $0, 3, 6, 9$ (4 values).
5. If $a = 5$, then $5 + b \equiv 1 \pmod{3}$ implies $b \equiv -4 \equiv 2 \pmod{3}$. Possible values for $b$ are $2, 5, 8$ (3 values).
6. If $a = 6$, then $6 + b \equiv 1 \pmod{3}$ implies $b \equiv -5 \equiv 1 \pmod{3}$. Possible values for $b$ are $1, 4, 7$ (3 values).
7. If $a = 7$, then $7 + b \equiv 1 \pmod{3}$ implies $b \equiv -6 \equiv 0 \pmod{3}$. Possible values for $b$ are $0, 3, 6, 9$ (4 values).
8. If $a = 8$, then $8 + b \equiv 1 \pmod{3}$ implies $b \equiv -7 \equiv 2 \pmod{3}$. Possible values for $b$ are $2, 5, 8$ (3 values).
9. If $a = 9$, then $9 + b \equiv 1 \pmod{3}$ implies $b \equiv -8 \equiv 1 \pmod{3}$. Possible values for $b$ are $1, 4, 7$ (3 values).

Now, we count the total combinations:
- For $a = 1$: 4 values of $b$.
- For $a = 2$: 3 values of $b$.
- For $a = 3$: 3 values of $b$.
- For $a = 4$: 4 values of $b$.
- For $a = 5$: 3 values of $b$.
- For $a = 6$: 3 values of $b$.
- For $a = 7$: 4 values of $b$.
- For $a = 8$: 3 values of $b$.
- For $a = 9$: 3 values of $b$.

Adding these gives: 
4 + 3 + 3 + 4 + 3 + 3 + 4 + 3 + 3 = 30.

Thus, the total number of positive three-digit integers with a $5$ in the units place that are divisible by $15$ is $30$.<|END_OF_LATENT|> The final answer is $30$.