2011 AMC 10A Problem 2

We want the fewest small bottles that together hold at least $500$ milliliters.

1. Estimate first: $35 \times 10 = 350$, which is too little, so the answer is more than $10$.
2. Try $14$ bottles: $35 \times 14 = 490$ milliliters. That is $10$ milliliters short of $500$, so $14$ is not enough.
3. Try $15$ bottles: $35 \times 15 = 525$ milliliters. That is at least $500$, so $15$ bottles fill the large bottle (with some left over).

So the smallest number that works is $15$.

The answer is $\boxed{\textbf{(E)}\ 15}$.

"Completely fill" means the total has to reach the target. If $500 \div 35$ is not a whole number, the leftover part still needs a whole extra bottle, so you always round up. Whether you round up or down comes from the wording of the question, not from how big the remainder is.

We need the smallest whole number $k$ with $35k \ge 500$.

Compute $35 \cdot 14 = 490$, which is less than $500$, so $14$ bottles do not suffice. Then $35 \cdot 15 = 525 \ge 500$, so $15$ bottles are enough.

Equivalently, $\frac{500}{35} = 14\frac{2}{7}$, and since a fraction of a bottle cannot be bought, round up to $15$.

The answer is $\boxed{\textbf{(E)}\ 15}$.