A selection of short (and not so short) problems

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

3. 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?

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

5. 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)

6. 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?

7. 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?

8. 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)

9. 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?

10. 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?

11. Answers & notes Calendar problems • December 1stis 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

12. 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)

13. 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

14. 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!

15. Handshakes for a certain number of people

16. 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.

17. 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 who’s ‘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.

18. Answers & notes Cake Dilemma For the problem given of dividing a hexagon into five equal sections, Draw a segment from the centre to any marked point Firstly divide each side into 5 equal lengths. Find the centre Count round the hexagon marking every sixth segment

19. 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.