A student read the book with $480$ pages two times. If he in the second time for every day read $16$ pages more than in the first time and he finished it $5$ days earlier than in the first time. For how many days did he read the book in the first time?
1. Let \( x \) be the number of pages the student reads per day the first time.
2. Let \( y \) be the number of days the student reads the book the first time. Then, we have:
   \[
   y = \frac{480}{x}
   \]
3. The student reads \( x + 16 \) pages per day the second time and finishes the book in \( y - 5 \) days. Thus:
   \[
   y - 5 = \frac{480}{x + 16}
   \]
4. Substitute \( y = \frac{480}{x} \) into the second equation:
   \[
   \frac{480}{x} - 5 = \frac{480}{x + 16}
   \]
5. Clear the fractions by multiplying through by \( x(x + 16) \):
   \[
   480(x + 16) - 5x(x + 16) = 480x
   \]
6. Simplify and expand the equation:
   \[
   480x + 7680 - 5x^2 - 80x = 480x
   \]
7. Combine like terms:
   \[
   -5x^2 - 80x + 7680 = 0
   \]
8. Divide the entire equation by -5 to simplify:
   \[
   x^2 + 16x - 1536 = 0
   \]
9. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 16 \), and \( c = -1536 \):
   \[
   x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-1536)}}{2 \cdot 1}
   \]
   \[
   x = \frac{-16 \pm \sqrt{256 + 6144}}{2}
   \]
   \[
   x = \frac{-16 \pm \sqrt{6400}}{2}
   \]
   \[
   x = \frac{-16 \pm 80}{2}
   \]
10. This gives us two solutions:
    \[
    x = \frac{64}{2} = 32 \quad \text{and} \quad x = \frac{-96}{2} = -48
    \]
11. Since the number of pages read per day cannot be negative, we discard \( x = -48 \) and accept \( x = 32 \).

12. Calculate the number of days the student read the book the first time:
    \[
    y = \frac{480}{32} = 15
    \]

The final answer is \(\boxed{15}\) days.