The cost of five water bottles is \$13, rounded to the nearest dollar, and the cost of six water bottles is \$16, also rounded to the nearest dollar. If all water bottles cost the same integer number of cents, compute the number of possible values for the cost of a water bottle.

[i]Proposed by Eugene Chen[/i]
1. Let the cost of a water bottle be \( x \) cents.
2. Given that the cost of five water bottles is approximately $13, we can write:
   \[
   5x \approx 13 \text{ dollars}
   \]
   Since the cost is rounded to the nearest dollar, we have:
   \[
   12.5 \leq 5x < 13.5
   \]
   Converting dollars to cents (1 dollar = 100 cents), we get:
   \[
   1250 \leq 5x < 1350
   \]
   Dividing the entire inequality by 5:
   \[
   250 \leq x < 270
   \]

3. Similarly, for the cost of six water bottles being approximately $16, we write:
   \[
   6x \approx 16 \text{ dollars}
   \]
   Since the cost is rounded to the nearest dollar, we have:
   \[
   15.5 \leq 6x < 17.5
   \]
   Converting dollars to cents:
   \[
   1550 \leq 6x < 1750
   \]
   Dividing the entire inequality by 6:
   \[
   \frac{1550}{6} \leq x < \frac{1750}{6}
   \]
   Simplifying the fractions:
   \[
   258.33\overline{3} \leq x < 291.66\overline{6}
   \]

4. Combining the two inequalities:
   \[
   250 \leq x < 270
   \]
   and
   \[
   258.33\overline{3} \leq x < 291.66\overline{6}
   \]
   The intersection of these two ranges is:
   \[
   258.33\overline{3} \leq x < 270
   \]

5. Since \( x \) must be an integer, we consider the integer values within this range:
   \[
   259 \leq x \leq 269
   \]

6. The number of possible integer values for \( x \) is:
   \[
   269 - 259 + 1 = 11
   \]

The final answer is \(\boxed{11}\).