AMC 10 · 2002 · #11
Grade 5 arithmeticJamal 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?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Thirty files must be stored on disks that each hold $1.44$ MB. Three files are $0.8$ MB, twelve are $0.7$ MB, and the remaining files are $0.4$ MB. A file cannot be split across two disks. Find the smallest number of disks that can hold all thirty files.
Givens: Each disk holds at most $1.44$ MB; $3$ files are $0.8$ MB each; $12$ files are $0.7$ MB each; The remaining files are $0.4$ MB each; A single file cannot be split between two disks; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Unknowns: The smallest number of disks that can store all $30$ files
Understand
Restated: Thirty files must be stored on disks that each hold $1.44$ MB. Three files are $0.8$ MB, twelve are $0.7$ MB, and the remaining files are $0.4$ MB. A file cannot be split across two disks. Find the smallest number of disks that can hold all thirty files.
Givens: Each disk holds at most $1.44$ MB; $3$ files are $0.8$ MB each; $12$ files are $0.7$ MB each; The remaining files are $0.4$ MB each; A single file cannot be split between two disks; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems
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.
Execute — Answer: B
4.NBT.B.4 Step 1 Count the files of each size
- There are $30$ files in all.
- Take away the $3$ big files and the $12$ medium files to find how many small ones are left: $30-3-12=15$.
- So there are $3$ files of $0.8$ MB, $12$ files of $0.7$ MB, and $15$ files of $0.4$ MB.
💡 "The rest" just means everything not already counted, so subtract the named groups from the total.
5.NBT.B.7 Step 2 Each big file needs its own disk
- Look at a $0.8$ MB file.
- Two big files together are $0.8+0.8=1.6$ MB, which is more than $1.44$, so no disk holds two of them.
- A big file with a medium file is $0.8+0.7=1.5$ MB, still more than $1.44$, so those cannot share either.
- The only file that fits alongside a big one is a small file: $0.8+0.4=1.2$ MB, which is under $1.44$.
- So each of the $3$ big files takes its own disk, and we drop one small file onto each.
- That uses $3$ disks and $3$ of the small files.
💡 The biggest files are the pickiest, so pin them down first and give each the only partner that fits.
5.NBT.A.3 Step 3 Pair up the medium files
- Now the $12$ medium files.
- Two of them are $0.7+0.7=1.4$ MB, which is at most $1.44$, so two fit on a disk.
- Three would be $2.1$ MB, far too much.
- A medium file cannot join a big file's disk either: that disk already holds $0.8+0.4=1.2$ MB, leaving only $0.24$ MB, and $0.7>0.24$.
- So the medium files must pair off two per disk on fresh disks: $12\div 2=6$ disks.
- Each such disk holds $1.4$ MB, leaving only $0.04$ MB, so no small file can squeeze in.
💡 Two mediums fill a disk almost exactly, and a third never fits, so pairing is the tightest they can pack.
5.NBT.B.7 Step 4 Group the leftover small files
- Three small files went onto the big-file disks, so $15-3=12$ small files remain.
- Three of them are $3\times 0.4=1.2$ MB, which is under $1.44$, but four are $4\times 0.4=1.6$ MB, too much.
- So pack them three to a disk: $12\div 3=4$ disks.
💡 Fit as many small files as stay under the limit, which is three, then count how many groups of three you need.
4.NBT.B.4 Step 5 Add up the disks
- Total the disks each group forced: $3$ for the big files, $6$ for the medium files, and $4$ for the leftover small files.
- That is $3+6+4=13$.
- Every step was forced or filled a disk as tightly as possible, so no plan uses fewer.
- The smallest number of disks is $13$, which is choice (B).
💡 Once each group is packed as tightly as it can be, adding the group totals gives the true minimum.
4.NBT.B.4 There are $30$ files in all. Take away the $3$ big files and the $12$ medium fil 5.NBT.B.7 Look at a $0.8$ MB file. Two big files together are $0.8+0.8=1.6$ MB, which is m 5.NBT.A.3 Now the $12$ medium files. Two of them are $0.7+0.7=1.4$ MB, which is at most $1 5.NBT.B.7 Three small files went onto the big-file disks, so $15-3=12$ small files remain. 4.NBT.B.4 Total the disks each group forced: $3$ for the big files, $6$ for the medium fil Review
Reasonableness: Check the total size: $3(0.8)+12(0.7)+15(0.4)=2.4+8.4+6.0=16.8$ MB. Dividing by $1.44$ gives about $11.7$, so at least $12$ disks are needed just by volume, and $12$ disks would hold $12\times 1.44=17.28$ MB — enough room by volume alone. That is exactly why (A) $12$ is the trap: the no-splitting rule wastes space (the medium-file disks leave $0.04$ MB unused each), forcing a $13$th disk. Our packing is also legal to double-check: $3$ disks at $1.2$ MB, $6$ disks at $1.4$ MB, $4$ disks at $1.2$ MB, all at or under $1.44$, storing $3+12+12=27$ small/medium/big slots plus the $3$ paired smalls $=30$ files. And $13$ is truly minimal: the big files force $3$ disks and the medium files force $6$ more that cannot take any small files, so the twelve leftover smalls need $\lceil 12/3\rceil=4$ additional disks no matter what.
Alternative: Instead of counting disks group by group, argue by wasted space. The best a disk can do with these sizes is $1.4$ MB (two mediums) or $1.2$ MB (big plus small, or three smalls); no combination reaches $1.44$. So every disk wastes at least $0.04$ MB. With total content $16.8$ MB, even a perfect $1.44$ packing needs $\lceil 16.8/1.44\rceil=12$ disks, and the forced waste on the medium and big files pushes the requirement to $13$. Both routes land on (B).
CCSS standards used (min grade 5)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Finding the $15$ small files via $30-3-12$ and totaling the disks $3+6+4=13$.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Computing disk loads like $0.8+0.4=1.2$, $0.8+0.8=1.6$, and $3\times 0.4=1.2$.)5.NBT.A.3Read, write, and compare decimals to thousandths (Comparing sums such as $1.4$, $1.5$, $1.6$, and $2.1$ against the $1.44$ MB disk limit.)
⭐ Pack 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.
⭐ Pack 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.
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