D e c 2 0 1 2 I s s u e 2 2

Monthly Maths

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

symbols. A dot was used to represent the

units (one to four) and a line was used to

represent five. It has been suggested that

counters may have been used, such as

pebbles, to represent the units, short

sticks to represent the fives, and a shell to

represent zero.

The Mayans also developed a complex

calendar system, one using base 20 and

base 18. Each month contained 20 days,

with 18 months to a year. This left five

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.

Click here for the MEI

Maths Item of the Month

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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!

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.

One theory is that

on that date the

Sun will align with

the centre of the

Milky Way, which

only happens

once every 25,772 years. Proponents of

this theory claim that the Maya knew

about this alignment and set their Long

Count calendar to end on this day

because the alignment will cause

something to happen.

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.

www.mei.org.uk

Two new MEI teaching resources

are at the end of this bulletin. Click here

to download them from our website.

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).

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.

They stick their

answer onto the

relevant section of

their grid.

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."

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

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.

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.

Add the digits of a big

number, keep adding.

If they add up to this

number, then you can

divide exactly by this

number. Which

number am I?

PNC Christmas Price Index 2012

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.

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.

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...

Seasonal selection

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:

Mayan base 20:

The row headings are:

1000 100 10 1

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

8000

400

20

1

8000

400

20

1

8000

400

20

1

8000

400

20

1

What numbers are shown?

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

• 40

• 100

• 63

• 97

• 372

Writing Mayan Numbers: answers

• 21 would be:

• 40 would be:

• 100 would be:

• 63 would be:

• 97 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 20 = 320

(4 left) 4 x 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 = 1 uinal = 20 days 18 uinals = 1 tun = 360 days 20 tuns = 1 katun = 7,200 days 20 katuns = 1 baktun = 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 = 2000 =

160 = 000 =

17 = 12 =

1377_

• 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.

A selection of short

(and not so short) problems

Seasonal 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.