AMC 10 · 2002 · #11
Grade 5 arithmeticPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We want the smallest disk count, so tool #14 (Extreme Principle) says push each disk to hold as much as it can and check the boundary of 1.44 MB: which files can share a disk and which cannot. The files come in three sizes, so tool #7 (Identify Subproblems) splits the job into three groups. Start with the biggest files, because they are the hardest to pair up, and see what each big file is forced to do. Then fit the medium files, then the small ones, filling any leftover room. Counting the disks each group forces gives both a working plan and a proof that fewer disks is impossible.
Count the files of each size
Subtract the named groups: 30-3-12 leaves 15 small files of 0.4 MB, beside 3 big (0.8 MB) and 12 medium (0.7 MB).
"The rest" just means everything not already counted, so subtract the named groups from the total.
4.NBT.B.4Identify SubproblemsEach big file needs its own disk
Two bigs (1.6) or big-plus-medium (1.5) exceed 1.44; only 0.8+0.4=1.2 fits, so the 3 big files take 3 disks, each with one small file.
The biggest files are the pickiest, so pin them down first and give each the only partner that fits.
The biggest files are the pickiest, so pin them down first and give each the only partner that fits.
▸ Why?
Two big files together already overflow one disk, so each one has to occupy a disk of its own.
▸ Why?
Whatever room is left on that disk caps what else can join it, so the fit is decided by one comparison.
Pair up the medium files
Two mediums make 0.7+0.7=1.4, within 1.44, but three overflow and none fits the 0.24 MB left on a big disk, so 12 ÷ 2=6 disks.
Two mediums fill a disk almost exactly, and a third never fits, so pairing is the tightest they can pack.
5.NBT.A.3Identify SubproblemsGroup the leftover small files
That leaves 15-3=12 small files; three of them make 1.2 MB but four make 1.6, so pack three per disk: 12 ÷ 3=4 disks.
Fit as many small files as stay under the limit, which is three, then count how many groups of three you need.
5.NBT.B.7Identify SubproblemsAdd up the disks
Add the disks each group forced: 3+6+4=13. Every group was packed as tightly as the limit allows, so no plan uses fewer — choice (B).
Once each group is packed as tightly as it can be, adding the group totals gives the true minimum.
4.NBT.B.4Extreme PrinciplePack the biggest, pickiest files first and fill each disk as tightly as the limit allows — then adding up the forced groups gives the fewest disks.
- Count the files of each size
- Each big file needs its own disk
- Pair up the medium files
- Group the leftover small files
- Add up the disks
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