# 2002 AMC 12A Problems/Problem 9

The following problem is from both the 2002 AMC 12A #9 and 2002 AMC 10A #11, so both problems redirect to this page.

## Problem

Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$

## Solution

A $0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.

We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each.

Thus our answer is $3+10 = \boxed{\textbf{(B) }13 }$.

## Solution 2

Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.

We are left with $12$ files($0.7$ MB), where the optimal number of files per disks is $2$, so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$.

~quantumpsiinverted