2013 Excellence in Mathematics Contest
Team Project
School Name:
Group Members:
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Reference Sheet
Formulas and Facts
You may need to use some of the following formulas and facts in working through this project. You may not need
to use every formula or each fact.
A  bh
Area of a rectangle
C  2l  2w
Perimeter of a rectangle
A   r2
Area of a circle
y2  y1
x2  x1
Slope
m
Circumference of a circle
1
A  bh
2
Area of a triangle
12 inches = 1 foot
5280 feet = 1 mile
3 feet = 1 yard
16 ounces = 1 pound
2.54 centimeters = 1 inch
100¢ = $1
1 kilogram = 2.2 pounds
1 ton = 2000 pounds
1 gigabyte = 1000 megabytes
1 mile = 1609 meters
1 gallon = 3.8 liters
1 square mile = 640 acres
1 sq. yd. = 9 sq. ft
1 cu. ft. of water = 7.48 gallons
1 ml = 1 cu. cm.
C  2 r
V   r 2h
Volume of cylinder
Lateral SA = 2  r  h
Lateral surface area of cylinder
V  lwh
Volume of rectangular prism
 b  b 2  4ac
2a
Quadratic Formula
x
4
V   r3
3
Volume of a sphere
tan  
sin 
cos 
TEAM PROJECT
2013 Excellence in Mathematics Contest
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The Team Project is a group activity in which the students are presented an open ended, problem situation
relating to a specific theme. The team members are to solve the problems and write a narrative about the theme
which answers all the mathematical questions posed. Teams are graded on accuracy of mathematical content,
clarity of explanations, and creativity in their narrative.
Part 1 – What is a Byte?
Before we discuss the answer to this question, we have to go back to the smallest unit of data that a computer uses.
This is called a bit. The word “bit” comes from putting the words binary digit together. Computers convert all of
its inputs, outputs, and processes to a binary system. This means that everything in a computer uses the symbols 0
or 1. A bit can be used to represent two states of information, such as “Yes or No”, “On or Off”, “0 or 1.”
The next unit of data is the byte which is equivalent to 8 bits. A byte can represent 28 = 256 states of information
such as numbers or a combination of numbers and letters. One byte could be equal to one character. Ten bytes
could be equal to a word. One hundred bytes would equal an average sentence.
As you are probably well aware, units of data that a computer uses are represented with terms such as kilobyte,
megabyte, gigabyte, terabyte, etc.
In the 1970’s and 1980’s, computer data was saved to 5-¼-inch floppy disks that could initially hold 100 kilobytes
of data. Later in the 1980’s and into the 1990’s, the standard way to save data
was with the 3-½ inch disk (that wasn’t as floppy). It could hold 1.44
megabytes of data. Now, we have flash drives (a.k.a thumb drives or
jump drives) that can hold several gigabytes of data. We also have
portable disk drives that can hold up to several terabytes of data.
These prefixes (kilo, mega, giga, and tera) tell us how many bytes of
data the storage device can hold. Traditionally, the prefixes
have the following meanings:




kilo – one thousand
mega – one million
giga – one billion
tera – one trillion
These prefixes give you some idea of the size of these storage devices. But to really wrap our heads around the
amount of data that can be stored is a challenge. The goal of this project is to help you to make sense of these
quantities.
Part 2 – How Much Data?
As mentioned in Part 1, computers use the binary system and so the actual data storage sizes are based on powers
of 2. For example, one byte is equivalent to 23 = 8 bits.
1. Please consider the table and extend the pattern to complete the table. Note that you may need to be flexible in
expressing an equivalent number when dealing with the very large numbers. That is, not everything has to be
based on 1 byte.
Name
Equivalent Number
1 byte
20 = 1 byte
1 kilobyte
210 = 1,024 bytes
1 megabyte
220 = 1,048,576 bytes
1 gigabyte
230 bytes = 1,024 megabytes
1 terabyte
240 bytes = 1,024 gigabytes
1 petabyte
250 bytes = 1,024 terabytes
1 exabyte
260 bytes = 1,024 petabytes
1 zetabyte
270 bytes = 1,024 exabytes
1 yottabyte
280 bytes = 1,024 zetabytes
1 brontobyte
290 bytes = 1,024 yottabytes
Note that we can
be flexible in
terms of how
they relate the
equivalent
number…
Source: www.whatsabyte.com
2. An average 10 megapixel JPEG image will be about 4.5 megabytes. How many pictures can you store on a 1
gigabyte storage device?
1 gigabyte = 1024 megabytes
1 picture
1024 MB 
 227.5
4.5 MB
About 227 pictures can be stored.
3. How many pictures can you store on a 1terabyte storage device?
1 terabyte = 1024 gigabytes
227.5 picture
1024 GB  233, 016.8
GB
About 233,016 pictures can be stored.
Part 3 – Music Downloads
While music file size depends on the length of the song, a typical size is 4
megabytes. In this part of the project, we will determine the time needed to
download music files as a way to make sense of the size of the storage device.
1. How many music files (assuming each is 4 megabytes in size) will fit on a 8
gigabyte storage device?
8 GB  8 1024 MB  8192 MB
1 song
 2048 songs
4 MB
An 8 gig storage device can store about 2048 songs.
8192 MB 
2. Apple iTunes is reported to be able to download music at a rate of 128 kbps (kilobytes per second). How long
will it take to fill the 8 gigabyte storage device (use your results from #1 above).
8192 MB  8192 1024 KB  8,388, 608 KB
1 second
 65,536 seconds
128 KB
65,536 seconds  1092 minutes  18 hours
It will take about 18 hours to fill the 8 gig storage device.
8,388, 608 KB 
3. Complete the following table showing the length of time needed to download a given number of songs
(assuming each is 4 megabytes in size).
Number of Songs
Time Needed to
Download
0
0
5
160 sec = 2.7 min
10
320 sec = 5.3 min
15
480 sec = 8 min
20
640 sec = 10.7 min
25
800 sec = 13.3 min
30
960 sec = 16 min
4. Create a graph showing the relationship between the number of songs downloaded and the length of time needed
to download the songs. Write an explanation for why you chose to design the graph in the way that you did.
Specifically explain the choice of which quantity to place on which axis.
NOTE: Students may plot Number of Songs as a function of Download Time instead. Their explanation will be
key in judging.
5. Create a formula that could be used to determine the length of time (L) needed to download a given number of
songs (S).
S  4 1024
 32S
128
S  4 1024 8
L (minutes) 
 S  0.53 S
128  60
15
L (seconds) 
6. How long it would take to download enough songs to fill a 1 terabyte storage device.
1 TB  1024 GB  1024 1024 MB
1024 1024 MB
 262,144 songs
4 MB per song
1024 1024 MB  1, 048,576 MB  1, 048,576 1024 KB
1, 048,576 1024 KB
 8,388, 608 seconds
128 KB per second
=139,810.13 minutes
=2,330.168 hours
=97.090370 days
It would take just over 97 days to fill a 1 TB storage device.
Part 4 – Making Sense of the Petabyte
The United States Geological Survey office in Sioux Falls, SD has a special department called the Earth Resources
Observation and Science (EROS) Center. They “provide science and imagery to better understand our Earth.” For
example, EROS takes satellite images showing the effects of wildfires in the western United States.
On the EROS website (eros.usgs.gov), they share high resolution photos that allow for interactivity. EROS has
many, many high quality images and need to store them on high-capacity storage devices. Their storage devices
are at the petabyte level of capacity and they currently have around 4 petabytes worth of digital images. This part
of the project is designed to help us to understand how large this really is!
1. Suppose that we represent a single byte of data using a cube that is 1 foot by 1 foot
by 1 foot. That is, each of these cubes contains one byte of data. How many times
could we fill the Grand Canyon with a 1 petabyte quantity of these cubes (each
cube representing 1 byte of data)? According to the National Park Service, the
Grand Canyon has a volume of 5.45 trillion cubic yards.
27 cubic feet
 147,150, 000, 000, 000 cubic feet
1 cubic yard
The volume of the Grand Canyon is 147,150,000,000,000 cubic feet.
5, 450, 000, 000, 000 cubic yards 
1 Grand Canyon
1, 000, 000, 000, 000, 000 cu. ft. 21.21 Grand Canyon


147,150, 000, 000, 000 cu. ft.
1 petabyte
1 petabyte
The Grand Canyon could be filled over 21 times with 1 petabyte worth of data (where each byte is represented
by 1 cubic foot).
2. The EROS group has approximately 4 petabytes worth of digitial image data. How many times could the Grand
Canyon be filled with 4 petabytes worth of data given the same assumptions as in #1 above?
21.21 Grand Canyons
 4 petabytes  84.84 Grand Canyons
1 petabyte
The Grand Canyon could be filled over 84 times with 4 petabyte worth of data (where each byte is represented
by 1 cubic foot).
Part 5 – A Very Large Sugar Bowl?
Assume that a sugar cube is a perfect cube with
dimensions 1.25 cm 1.25 cm 1.25 cm
(www.webstaurantstore.com/dixie-crystals-fine-sugar-cubes-1-poundbox/99929001.html) . Also, assume that a sugar cube contains 725,000
sugar crystals (www.chsugar.com/familyfun/sugart.html).
1. What is the volume of a single sugar crystal?
1.95 cu. cm.
 0.00000269 cu. cm. per sugar crystal
725, 000 crystals
2. On January 2, 2013, the Louisville Cardinals played the
Florida Gators in the 79th Sugar Bowl football game held in
the Superdome in New Orleans, LA. This stadium has a
volume of 3,500,000 m3
(en.structurae.de/structures/data/index.cfm?id=s0000384).
Suppose a byte of data is represented by a single sugar crystal.
How many times could the Superdome be filled with sugar
crystals if we had 4 petabytes worth of sugar crystals (now
that’s a big sugar bowl!)?
3,500, 000 cu. m. 1, 000, 000 cu. cm. 3,500, 000, 000, 000 cu. cm.


1 Superdome
cu. m.
1 Superdome
If each sugar crystal represents 1 byte, then
1, 000, 000, 000, 000, 000 sugar crystals 0.00000269 cu. cm.
cu. cm.

 2, 700, 000, 000
1 petabyte
1 sugar crystal
petabyte
1 Superdome
2, 700, 000, 000 cu. cm.
Superdome

 0.00077
3,500, 000, 000, 000 cu. cm.
1 petabyte
petabyte
Only 4*0.00077 = 0.003 copies of the volume of the Superdome will be filled with sugar with 4 petabytes of
data (if each petabyte is represented by a single sugar crystal).
That is, if we could pour 4 petabytes worth of data into the Superdome, only 0.3% of the Superdome would be
filled.
Part 6 – “640 kilobytes ought to be enough for anyone!”
There is a rumor that in 1981, Bill gates said “640 kilobytes ought to be
enough for anyone!” when referring to the newly released IBM personal
computer that had 640 kilobytes of RAM (random access memory). Nobody
can confirm that Bill Gates said this and he denied ever saying it! In fact,
some say that he actually said that “640 kilobytes should keep people happy
for 10 years.” After 6 years, we surpassed the 640 kilobyte limit.
1. Suppose that there was a way to connect these 1981 computers so that
together they would have enough RAM to store one 4 megabyte photo.
How many of these 640 kilobyte computers would it take to provide
storage for one photo?
4 MB  4096 KB
4096 KB
 6.4 computers
640 KB per computer
We would need 7 computers linked together to have enough memory to store one photo.
2. Suppose that 640 kilobytes of storage space is represented by a 1 foot by 1 foot square tiles. Now suppose that
we created a number of these tiles to represent 1 gigabyte of storage space. If these tiles (1 gigabyte worth) were
lined up side by side, how far would they stretch? Express this distance in the most appropriate unit (feet?
yards? miles?).
1 GB = 1,048,576 KB
1, 048,576 KB
 1638.4 sq. ft.
640 KB per sq.ft.
If we lined up the tiles side-by-side, they would stretch out to over 1638.4 feet which is about 0.3 miles.
Part 7 – Googol the Number not the Web Site
One googol is the number 1 with one hundred zeros after it:
10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.
Let’s assume that the standard memory needs for
a personal laptop computer today is 8 gigabytes
and that the amount of standard memory is
doubling every 3 years. How long will it be in
years before a personal laptop will need 1 googol
bytes of memory?
1 googol bytes = 11091 gigabytes.
Since the amount of memory is doubling every 3
years, we write the equation:
8 2
t
3
 1091
While the junior high students may not do this, I
show the solution just for our own assistance in
judging the work of the teams.
8 2
t
3
 1091
1091
t  3  log 2
8
It will take a little almost 898 years for the standard memory to hit 1 googol bytes of memory.