Monthly
Maths
I s s u e
Planning ahead
Assuming it’s safe
to make plans for
the end of
December, we
have included
links to festive
teaching
resources, as well
as including an
MEI ‘selection box’
of seasonal
teaching
resources at the
end of this edition.
There is also an
MEI resource
about the Mayan
number system
and calendar, with
some classroom
activities and links
to further activities
produced by
others.
All of us at MEI
would like to wish
you, your
colleagues and
your students
Season’s
Greetings and a
Happy New Year.
In 2013 we will
be celebrating
MEI’s 50th
birthday!
2 2
Will the world end in 2012?
The Mayan long count calendar finishes
one of its great cycles in December 2012.
This has fuelled countless doomsday
theories about the world ending at 11:11
on December 21, 2012. The last day of
the Mayan calendar corresponds with the
December Solstice (or Winter Solstice as
it is known in the Northern Hemisphere),
which has played a significant role in
many cultures all over the world.
www.mei.org.uk
D e c
2 0 1 2
What have the Mayans done for us?
In one of the two teaching resources that
accompanies this edition, we look the
mathematical contributions of the Maya
civilisation to future generations. These
include the value and use of zero and the
place system - they were the first
civilisation to use a placeholder for zero.
The Mayan number system dates back to
the fourth century and was approximately
1,000 years more advanced than the
European systems of that time. We
currently use a decimal system, which
has a base 10, but the Mayans used a
vigesimal system, which had a base 20.
The Mayan system used a
combination of two
One theory is that
symbols. A dot was used to represent the
on that date the
Sun will align with units (one to four) and a line was used to
the centre of the represent five. It has been suggested that
Milky Way, which counters may have been used, such as
pebbles, to represent the units, short
only happens
once every 25,772 years. Proponents of sticks to represent the fives, and a shell to
represent zero.
this theory claim that the Maya knew
about this alignment and set their Long
The Mayans also developed a complex
Count calendar to end on this day
calendar system, one using base 20 and
because the alignment will cause
base 18. Each month contained 20 days,
something to happen.
with 18 months to a year. This left five
NASA has a web page to explain away
the planetary alignment theory and
various others surrounding the supposed
apocalypse. In November NASA
scientists reiterated the facts in a
question and answer format to allay 2012
doomsday fears: Beyond 2012: Why the
World Won’t End.
Click here for the MEI
Maths Item of the Month
days at the end of the year. In this way,
the Mayans had invented the 365 day
calendar. A pyramid was used as a
calendar: four
stairways, each with
91 steps and a
platform at the top,
making a total of 365.
Two new MEI teaching resources
are at the end of this bulletin. Click here
to download them from our website.
Disclaimer: This newsletter provides links to other Internet sites for the convenience of users.
MEI is not responsible for the availability or content of these external sites, nor does MEI endorse or guarantee the products, services, or information
Counting down
to Christmas
The
Mathematics of
Santa Claus’
Present
Delivery
System
A fun blog by
William M Briggs,
Adjunct Professor
of Statistical
Science, Cornell
University, who
claims to be one
of the group of
consultants hired
by Santa to help
with the
complicated
computer code
that is necessary
to bring about the
massive toy
movement on
Christmas Eve.
Briggs outlines the
modern
mathematical
ideas that Santa
Claus now
employs, such as
the “Santa Claus
Gift Momentum
Equation” and the
“Gift Probability
Equation” here.
Maths Advent Calendars
We’re delighted to report that both
NRICH and Plus Magazine have
published 2012 versions of their
wonderful advent calendars:
NRICH Advent Calendar
2012 - Secondary has a
The PowerPoint includes links to outside
sites and introductory music, to add to the
fun and excitement! Included in the
downloads are the instructions in word
format for how to work the PPT and the
answers (recommended reading before
using in class, to get the most out of it).
poster with an activity to try
each day in December until Christmas Eve.
The 2012 Plus advent
calendar features Plus
magazine’s favourite moments from the
last year.
We’ve found two maths Advent calendars
on the TES Resources site:
Advent Calendar full of fun, engaging
starters (by dannytheref )
An impressively interactive Advent
Calendar in PowerPoint format, behind
which are 11 different fun & engaging
starter/plenary activities covering the full
maths spectrum of topics & abilities.
Designed to be done as a running
competition between 2 teams, the games
include: Pointless, Catchphrase, Cookie
Monsters, Hangman, Countdown, 1-on-1
battles, Key Word Challenge, and more.
Christmas Countdown ( by blue117)
This KS3 resource provides an Excel file
with clues and answers, all to do with
factors, multiplies primes, triangular and
Fibonacci numbers. Each pupil has a 5x5
grid onto which they will stick their clues.
A new clue is given out every day (to get
through the 25 during December, you will
need to give out more than one a day!).
This clue could be shown to the whole
class as a starter activity or left in
prominent place for the students to look
at in their own time. Add the digits of a big
number, keep adding.
They stick their
If they add up to this
answer onto the
number, then you can
divide exactly by this
relevant section of
number. Which
their grid.
number am I?
A teacher who has tried this activity
comments on the TES resources site:
"Some of the clues are easier than
others, so group work encouraged the
students to work together to complete the
calendar. The clues which were not
immediately solved were set as a
homework to be discussed next lesson."
Seasonal selection
Santa’s
Christmas Eve
Workload,
calculated
Another
mathematical look
at the practicalities
of Santa delivering
presents to all the
children worldwide
aged under the
age of 14.
“The equation is
this: compare
population of
young people with
density of
Christianity and
plot it on the
globe. From that,
you've got total
population and
the times at which
Santa should hit
them.”
Philip Bump
considers time
zones, the
Christian
population and
population
distribution,
different dates for
celebrating
Christmas, hours
of darkness and
other factors in
The Atlantic
article.
PNC Christmas Price Index 2012
Some more fabulous Christmas themed
resources found on TES Resources. You
do need to be registered to download the
resources, but it’s free to register.
Christmas Tree Diagrams
(by alutwyche)
A KS3/4 resource that has been selected
to feature in the TES secondary maths
newsletter. The activity is based on
calculating the probability of two children
receiving combinations of presents from
their Christmas lists.
Maths Christmas Activities Booklet
(by Ryan Smailes)
A collection of Christmas
themed Maths puzzles and
problems for KS3/4 that is ideal
for the end of term, and which
has also been selected to
feature in the TES secondary
maths newsletter.
Word problems
Two sets of resources that you might like
to pick and choose from, here and here.
They are both from
American sites, so dollars
and yards are used in some
questions. They can act as
a starting point for you to
devise some word questions
of your own.
Don’t forget that at the end of this issue
there is an MEI seasonal resource with
some more word problems and more...
The PNC Christmas Price
index® for 2012 has been
published, showing the
current cost for one set of
each of the gifts given in
the song "The 12 Days of Christmas."
The True Cost of Christmas is the
cumulative cost of all the gifts when you
count each repetition in the song – so it
reflects the cost of 364 gifts. For 2012,
PNC has created a website with a global
journey to help deliver the information.
The updated site includes several pages
of animated gifts, and interactive chart
and an explanation of how the PNC
Christmas Price Index® was determined.
The PNC Christmas Price Index® is
similar to the Consumer Price Index,
which measures changes in price of
goods and services like housing, food,
clothing, transportation and more that
reflect the spending habits of the average
American. The total Christmas Price
Index® can be found here.
Although the prices are listed in US
dollars, the interactive Price Index is very
entertaining and clearly gets across the
message of price increases and
percentages. The change in the prices of
Christmas gifts, year over year, provides
an excellent lesson on inflation and other
economic trends. Plus, similarities to the
U.S. Consumer Price Index make it a fun
and easy way to study economic
indicators.
The PNC 2012 Press Release gives
more information on these economic
trends.
What have the
Mayans done
for us?
The Mayan civilization spread all over south-eastern
Mexico, Guatemala, Belize, and Honduras between
2500 B.C. and A.D. 250. They are well known for
their mathematical and astronomical systems.
Mayan Mathematics
• Inscriptions show the Mayans
were working with calculations
up to hundreds of millions.
• They produced extremely
accurate astronomical data
from naked eye observations.
• The Mayans measured the
length of the year to a high
degree of accuracy, but
approximated it to 360 days.
Mayan Number System
• The Mayans used a number system based on
1s, 5s and 20s.
• They were one of the only ancient civilizations to
use a zero.
• The Mayans devised a counting system that was
able to represent very large numbers by using
only 3 symbols: a dot, a bar, and a glyph for
zero, usually a shell.
Mayan Number symbols 0 -19
• Zero is represented by a
shell; 1 to 4 are
represented by dots.
• Multiples of five are
represented by lines, with
extra dots being added to
complete the numbers as
shown.
• Can you work out the
missing diagrams?
Writing Bigger Numbers
• Our own number system is base 10, which
means that we have 9 ‘symbols’
(1,2,3,4,5,6,7,8,9) plus a zero.
• The Mayan number system had 19 symbols (as
on the previous slide) plus a shell for zero.
• When writing numbers, once we get to ‘9’ we
then have to move across to the next column.
We write a ‘one’ followed by a ‘zero’ to show that
we have moved across – zero is a ‘place-holder’
Writing Bigger Numbers
• The Mayans used a similar system using their
19 symbols and then moving to the next section
and putting a zero (represented by the shell) as
a placeholder.
• Another difference is that they used rows instead
of columns, starting from the bottom and working
upwards.
Writing Bigger Numbers
• In base 10, the headings are 100, 101, 102, 103,
104 etc.
• In base 20 the headings are 200, 201, 202, 203,
204 etc.
• What are these values?
Number Bases
Our base 10:
The column headings
are:
1000
100
10
1
Mayan base 20:
The row headings are:
8000
400
20
1
It will help initially to see the row headings, but they would not
normally be shown… just as our young children use 1, 10 and
100 as column headings when they begin writing numbers.
Writing Mayan Numbers
What numbers are shown?
8000
8000
8000
8000
400
400
400
400
20
20
20
20
1
1
1
1
Remember:
is 0
is 1
and
is 5
Write this
Write the following using Mayan numbers.
Use the row headings to help you if you need to.
• 21
• 63
• 40
• 97
• 100
• 372
Writing Mayan Numbers: answers
• 21 would be:
• 63 would be:
• 40 would be:
• 97 would be:
• 100 would be:
• 372 would be:
Larger Mayan Numbers
To write larger numbers start with the highest row that
can be subtracted from the number you are trying to
write.
Example: Writing 5124.
‘8000’ is too big, so start with as many 400s as
possible, then work down with what’s left for the 20s
and 1s.
5124 =
12 x 400 = 4800
(324 left)
16 x
(4 left)
4 x
20 = 320
1 =
4
Problem 1
• How would you write 1377 in Mayan numbers?
• Now try writing 2012.
Other Mayan Number Systems
• The Mayan had a second
Number System, used for
dating buildings and on
Calendars, etc.
• This would be a more
formal system, rather
than a number system
used for calculation.
Mayan Calendar
• Maya dates combined at least two calendars one, the ‘Calendar Round’, covering 365 days
and the other 260 days, such that every day had
two names, which reset every 52 years.
• The Maya also used a “Long Count" system of
187,2000 days that added a numeral at the end
of a cycle to keep a constant count of years.
• It is important to note that the Long Count's
version of a year, the tun, is only 360 days, not
the solar count of 365.
Mayan Calendar
• The basic unit for the Mayan calendar is the kin.
20 kins = 1 uinal
= 20 days
18 uinals = 1 tun
= 360 days
20 tuns = 1 katun
= 7,200 days
20 katuns = 1 baktun = 144,000 days
• Every date expressed in long count terms
contained five numerals, that is, the number of
baktuns, katuns, tuns, uinals (or winals) and
kins elapsed from the "beginning of time",
according to the Maya system.
Problem 2
• How many (Long Count) years
in a baktun?
• There are 13 baktuns in a
“great cycle”.
• How many years is this?
Long Count Calendar
• Starting at ‘year zero’ – the very beginning of a
Long Count period - the read-out of the calendar
was set at: 0.0.0.0.0.
• When each value was numerically accomplished to
its maximum, it would then reset to ‘0’ and the total
would be carried forward into the next time cycle to
its left.
• The beginning of the current cycle corresponds to
August 13, 3114 B.C. on the Gregorian calendar.
• This cycle is due to end on 13.0.0.0.0, the end of
the 13th baktun.
Problem 3
• These are typically recorded by archaeologists
translating the ancient Maya script, like this:
baktun.katun.tun.winal.kin
• Can you work out what date is represented by:
12.19.19.17.19?
Remember:
20 kins
18 uinals
20 tuns
20 katuns
=
=
=
=
1 uinal
1 tun
1 katun
1 baktun
=
=
=
=
20 days
360 days
7,200 days
144,000 days
Mayan Apocalypse?
• The Mayan calendar finishes
one of its great cycles in
December 2012, which has
fuelled countless theories
about the end of the world at
11:11 on December 21, 2012.
Not the End of the World
• Just as the calendar you have on your wall does
not cease to exist after December 31, the Mayan
calendar does not cease to exist on December
21, 2012.
• This date is the end of the Mayan long count
period but then another long count period begins
for the Mayan calendar.
Teachers Notes: structuring the work
The work on Mayan number systems and calendars could be split into
two sections according to need and time, using sections 1 to 15 and
then 16 to 23 at a later time. The information from the first section is
not required in order to complete the second section.
Both sections reinforce working with number, and although working in
other number bases is currently not part of most GCSE syllabuses,
working on this type of activity often helps pupils to better understand
and appreciate the structure of base 10.
The activities are best suited to a combination of short teacher-lead
information sessions to help pupils to understand the systems, followed
by paired working on the problems.
Teachers Notes: short answers from slides
Slide 9: values for base 10 are 1 10 100 1000 10000
values for base 20 are 1 20 400 8000 160000
Slide 11: numbers shown are 20 410 900 551
Answers
• Problem 1
1200 =
160 =
17 =
1377_
2000 =
000 =
12 =
• Problem 2
400 years; 5200 years
• Problem 3
12 baktun, 19 katun, 19 tun, 17 winal, and 19 kin, or
December 20, 2012.
External Resources
• Cracking the Maya Code
Students see how scientists began to unravel the meaning of Maya
glyphs and then determine their own birth date using the Maya Long
Count calendar system.
• How to Calculate with Mayan Numbers
Workbook for students to practice addition, subtraction, multiplication,
division, and square roots using Mayan numbers.
• The Exploratorium’s Mayan Calendar
In this 1-2 hour activity, students will learn about the two calendars
the Maya used, and solve the problem of how often the two cycles
coincided, by making and rotating gears, and by using prime
numbers and smallest common multiples.
Seasonal Problems
A selection of short
(and not so short) problems
Calendars
• This year, December 1st is on a Saturday, in
which year does this happen next?
• How many times this century will December 1st
fall on a Saturday?
Calendars
• A perpetual calendar
consists of 2
numbered cubes and
3 cuboids with the
months on. How
should the cubes be
numbered?
Calendars
• The ancient Mayans thought there were 360 days in
a year.
• We have 365 days a year (366 in a leap year)
• If the first day of the calendar year (January 1st for
us) were to coincide on both calendars in 2013, how
many years until it will coincide again?
– Ignoring leap years and assuming there are 365
days in our year
– Accounting for the fact that we have leap years?
(assume 365.25 days in a year)
Some old favourites…
• In a class there are 30 pupils. Each pupil gives
a Christmas card to each of the other pupils in
the class. How many cards are sent in total?
• At a New Year’s Eve party there are 12 young
people who all shake hands with each other.
How many hand shakes are there in total?
• At a much larger gathering there are 50 people,
how many handshakes now?
Can you find a formula for this?
Some old favourites…
At a New Year’s Eve party there are some adults
who all shake hands with each other… except
there are a small (unfriendly) group who all refuse
to shake hands with each other, but will shake
hands with everyone else.
There are 135 handshakes in total. How many
people are at the party? How many are there in
the ‘unfriendly’ group?
How many different answers can you find for this?
Super Santa
This year Santa has to deliver to all
the young people of the world, roughly
2 billion under 18’s in total.
Assuming, on average, that there are 3.5 young people per
household, how long does he have for each household to
get down the chimney, eat the mince pie, grab the carrot
for the reindeer and fill the stockings with goodies?
(Because of the earth’s rotation he probably has about 31
hours available).
…and how many households is that per second or per
minute? (whichever seems appropriate)
Class Calendar
A class of 25 pupils have an advent calendar. The first
pupil decides to open all the windows on the calendar. The
second pupil goes and closes all of the windows that are a
multiple of 2. The third pupil changes all the multiples of 3
– if they are open then she closes them, if they are closed
then she opens them. The fourth pupil changes all the
multiples of 4, if they are open then he closes them, if they
are closed then he opens them. The fifth pupil changes the
multiples of 5 and so on until the 25th pupil changes the
multiples of 25.
When the teacher arrives – which windows on the calendar
are open?
Can you explain why?
Cake Dilemma
Mrs Claus was delighted to find an unusual shaped
Christmas cake in her local supermarket; a regular
hexagonal one instead of the usual circular or square
ones. She bought one immediately and took it home
to show to Santa.
Santa also loved the new cake shape but pointed out
one small problem… “There are 5 of us eating and
we’ll need to share the cake equally, how are you
going to cut it up to give 5 equal portions?”
(the same volume)
“Easy”, said Mrs Claus…
How did she do it?
Answers & notes
Calendar problems
• December 1st is next on a Saturday in 2018
•
•
•
•
How many Saturday, December 1st are there this century? Because of leap years, the
answer is not ‘simply divide by 7’.
The day of the week that any particular date falls on runs in a 28 year cycle, where
each day of the week will occur 4 times. Hence beginning with there being a Saturday
December 1st this year: 2012 to 2039 (4), 2040 to 2067 (4), 2068 to 2095 (4). 2096 is
the start of the next 28 year cycle (another Saturday) and then think about 2000 to
2012. (2001 and 2007 had a Saturday Dec 1st).
Perpetual calendar: 1, 2, 3, 4, 5, 6 and 0, 1, 2, 7, 8, 9 works. Both cubes must
have 1 and 2 on, the 3 and 0 must be on different cubes, the others can be on either.
Mayan Calendar: If we assume there are 365 days in our year then we are looking for
the lowest common multiple of 360 and 365, which will tell us how many days it will
be until they coincide. This is 26280 days… which is 72 of our years… so 2085
Answers & notes
Calendar problems
Mayan Calendar
• If we have 365 days a year and the Mayans have 360, then one way to solve the
problem is to find the Lowest Common Multiple (LCM) of 365 and 360. This can be
done using factor trees to identify the prime factors for each number and then finding
the Highest Common Factor.
LCM = a x b ÷ HCF
• The answer is 365 x 360 ÷ 5 = 26280 (days) then 26280 ÷ 365 = 72 (years)
•
•
•
•
•
When accounting for leap years, using the fact that we actually have 365.25 days a
year, a similar method can be used, but both values need to be multiplied by 4 to
obtain integer values initially to be able to use factor trees (whilst maintaining the
correct ratio), and then dividing the answer by 4 again at the end. (To understand this
it may help to consider what happens with smaller values such as finding the LCM of
1.25 and 3)
365.25 and 360 multiplied by 4 become 1461 and 1440
The LCM of 1461 and 1440 is 701280
Divide this by 4 701280÷ 4 = 175320 (days)
Then 175320 ÷365.25 = 480 (years)
Answers & notes
Old Favourites
• Class cards: Each person sends to 29 people, so 30 x 29 = 870
•
Handshakes: this is a way into triangle numbers and follows on from the question
above. Each person must shake every other person’s hand. 12 x 11 = 132,
however, the difference here is that if A has shaken B’s hand and then B has shaken
A’s hand, they would have shaken hands twice, so the answer needs to be divided by
2. 132 ÷ 2 = 66. pupils might arrive at this answer through experimentation, or by
thinking about a smaller group of people and searching for a pattern in the numbers
or they may be able to go straight to the calculation.
•
Pupils might be encouraged towards finding a formula for triangle numbers by
thinking about a very large group of pupils, the example given is 50 people. For this
the calculation is 50 x 49 ÷ 2 = 1225
The formula for the number of handshakes for n people (triangle numbers) is
•
ℎ𝑎𝑛𝑑𝑠ℎ𝑎𝑘𝑒𝑠 =
𝑛(𝑛−1)
2
Answers & notes
Old Favourites
Adults shaking hands.
This problem continues the triangle numbers theme, but with a less obvious method of
solution.
One way to solve it is to consider the first 20 triangle numbers which are given on the
next slide. This could be displayed to pupils, but it would spoil the opportunity for pupils
to think if it is shown too early on.
To find an answer for the numbers of people at the party and in the ‘unfriendly’ group
simply search for a pair of triangle numbers with a difference of 135.
There are 3 possible answers for this:
• 17 at the party, with 2 who refuse to shake each others’ hand
• 19 at the party, with 9 who all refuse to shake each others’ hands
• 20 at the party, with 11 who all refuse to shake each others’ hands … which doesn’t
sound like much of a fun party to be at!
Handshakes for a certain
number of people
People
Handshakes
People
Handshakes
2
1
12
66
3
3
13
78
4
6
14
91
5
10
15
105
6
15
16
120
7
21
17
136
8
28
18
153
9
36
19
171
10
45
20
190
11
55
21
210
Answers & notes
Super Santa
The first possible issue with this problem is knowing how to write a billion: 1 x 109
or 1 000 000 000
Rounding answers to different degrees of accuracy will give slightly different answers:
•
•
He has to deliver to 2 000 000 000 ÷ 3.5 = 285 714 286 households
He has 31 x 60 x 60 = 111 600 seconds to do this
•
So he has 111 600 ÷ 285 714 286 = 0.0004 seconds to deliver to each household
•
•
Or 285 714 286 ÷ 111 600 = 2560 households per second
Or 2560 x 60 = 153 600 households per minute
You might like to pose some ‘localised’ questions, (or ask members of the class to pose
some) such as how long Santa will take to deliver to all the pupils in school, or the local
village, town or city, or the UK.
Answers & notes
Class Calendar
This problem is adapted from a ‘prison door’ problem.
Each door is initially opened by child 1 and each is then subsequently opened or closed
by any child whose ‘number’ is a factor of it.
The cycle clearly alternates:
open close open close etc.
Any number which has an even number of factors will end up closed; any number with an
odd number of factors will end up open.
Hence, calendar doors with square numbers end up as the only open ones.
Answers & notes
Cake Dilemma
For the problem given of dividing a hexagon into five equal sections,
Firstly divide
each side into
5 equal
lengths.
Find the
centre
Draw a
segment from
the centre to
any marked
point
Count round
the hexagon
marking every
sixth segment
Answers & notes
Cake Dilemma
Can you prove that the 5 segments are equal?
Consider each of the small triangles… what is the base and height of each?
Would this idea work for other
polygons and/or number of
required equal sections?
Dynamic geometry would be
helpful to demonstrate that the
sections have an equal area, but
also to show that if equal angles
are taken at the centre, this does
not result in sections of equal
area.