Problem 1192

1. Count the numbers divisible by 5:
$$\left\lfloor \frac{2015}{5} \right\rfloor = 403$$
So, there are 403 numbers less than or equal to 2015 that are divisible by 5.

2. Count the numbers divisible by 13:
$$\left\lfloor \frac{2015}{13} \right\rfloor = 155$$
So, there are 155 numbers less than or equal to 2015 that are divisible by 13.

3. Count the numbers divisible by both 5 and 13 (i.e., divisible by their least common multiple, 65):
$$\left\lfloor \frac{2015}{65} \right\rfloor = 31$$
So, there are 31 numbers less than or equal to 2015 that are divisible by 65.

4. Apply the Principle of Inclusion and Exclusion (PIE):
The number of integers less than or equal to 2015 that are divisible by 5 or 13 is given by:
$$403 + 155 - 31 = 527$$

5. Count the numbers divisible by both 5 and 13 (i.e., divisible by 65) again to exclude them:
Since we need the numbers divisible by 5 or 13 but not both, we subtract the numbers divisible by 65 from the total:
$$527 - 31 = 496$$