Monthly. Maths. I s s u e 2 2

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1 Monthly Planning ahead Maths I s s u e 2 2 Will the world end in 2012? D e c What have the Mayans done for us? 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! The Mayan long count calendar finishes one of its great cycles in December This has fuelled countless doomsday theories about the world ending at 11:11 on December 21, 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. Click here for the MEI Maths Item of the Month 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. 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

2 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 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. 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. 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? 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."

3 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. 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... 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.

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5 What have the Mayans done for us?

6 The Mayan civilization spread all over south-eastern Mexico, Guatemala, Belize, and Honduras between 2500 B.C. and A.D They are well known for their mathematical and astronomical systems.

7 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.

8 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.

9 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?

10 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

11 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.

12 Writing Bigger Numbers In base 10, the headings are 10 0, 10 1, 10 2, 10 3, 10 4 etc. In base 20 the headings are 20 0, 20 1, 20 2, 20 3, 20 4 etc. What are these values?

13 Number Bases Our base 10: The column headings are: Mayan base 20: The row headings are: 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.

14 Writing Mayan Numbers What numbers are shown? Remember: is 0 is 1 and is 5

15 Write this Write the following using Mayan numbers. Use the row headings to help you if you need to

16 Writing Mayan Numbers: answers 21 would be: 63 would be: 40 would be: 97 would be: 100 would be: 372 would be:

17 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 is too big, so start with as many 400s as possible, then work down with what s left for the 20s and 1s = 12 x 400 = 4800 (324 left) 16 x 20 = 320 (4 left) 4 x 1 = 4

18 Problem 1 How would you write 1377 in Mayan numbers? Now try writing 2012.

19 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.

20 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.

21 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.

22 Problem 2 How many (Long Count) years in a baktun? There are 13 baktuns in a great cycle. How many years is this?

23 Long Count Calendar Starting at year zero the very beginning of a Long Count period - the read-out of the calendar was set at: 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 , the end of the 13th baktun.

24 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: ? 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

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26 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.

27 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, This date is the end of the Mayan long count period but then another long count period begins for the Mayan calendar.

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29 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.

30 Teachers Notes: short answers from slides Slide 9: values for base 10 are values for base 20 are Slide 11: numbers shown are

31 Answers Problem = 2000 = 160 = 000 = 17 = 12 = 1377_ Problem years; 5200 years Problem 3 12 baktun, 19 katun, 19 tun, 17 winal, and 19 kin, or December 20, 2012.

32 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.

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34 Seasonal Problems A selection of short (and not so short) problems

35 Calendars This year, December 1 st is on a Saturday, in which year does this happen next? How many times this century will December 1 st fall on a Saturday?

36 A perpetual calendar consists of 2 numbered cubes and 3 cuboids with the months on. How should the cubes be numbered? Calendars

37 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 1 st 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 days in a year)

38 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?

39 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?

40 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)

41 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 25 th pupil changes the multiples of 25. When the teacher arrives which windows on the calendar are open? Can you explain why?

42 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?

43 Answers & notes Calendar problems December 1 st is next on a Saturday in 2018 How many Saturday, December 1 st 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 1 st this year: 2012 to 2039 (4), 2040 to 2067 (4), 2068 to 2095 (4) is the start of the next 28 year cycle (another Saturday) and then think about 2000 to (2001 and 2007 had a Saturday Dec 1 st ). 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 days which is 72 of our years so 2085

44 Calendar problems Mayan Calendar Answers & notes 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 = (days) then = 72 (years) When accounting for leap years, using the fact that we actually have 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) and 360 multiplied by 4 become 1461 and 1440 The LCM of 1461 and 1440 is Divide this by = (days) Then = 480 (years)

45 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 = 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

46 Old Favourites Adults shaking hands. Answers & notes 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!

47 Handshakes for a certain number of people People Handshakes People Handshakes

48 Answers & notes Super Santa The first possible issue with this problem is knowing how to write a billion: 1 x 10 9 or Rounding answers to different degrees of accuracy will give slightly different answers: He has to deliver to = households He has 31 x 60 x 60 = seconds to do this So he has = seconds to deliver to each household Or = 2560 households per second Or 2560 x 60 = 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.

49 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.

50 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

51 Cake Dilemma Answers & notes 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.