advertisement

12 After all the arguments last year, Santa has sorted out the sleeping quarters for the reindeer. Rudolph, Donner, Blitzen, Dasher, Comet, Cupid, Vixen, Dancer & Prancer are all eager to know which pen will be theirs 1 2 3 for the festive season. 4 5 6 However, instead of simply telling them which pens to go to, Santa has 7 8 9 left a set of clues. 12 • Rudolph will be in pen number 1 (of course) • No reindeer will be in a pen that has the same number as the number of letters in his or her name. • Dancer and Prancer will be in pens which are numbered with square numbers • Cupid is not next to Prancer • In each row and column there will be a reindeer with a 5 letter name, a 6 letter name and a 7 letter name • Cupid’s number + Vixen’s number = Dasher’s number 11 • At a festive gathering, Uncle Joe is again demonstrating his amazing mathematical abilities. • He claims to easily be able to square numbers ending in 5 in his head. • Here is an example of how he does it for 752: • • • • ‘Split’ the number Work out 7x8 and 5x5 separately Put the answers together 752 = 5625 Does it work for these numbers: 952 1052 1952? 10 Aunt Dorothy has 17 young nieces and nephews and she likes to put together a Christmas stocking filled with small presents for each one. However, she likes each child to have a different set of presents to all the other nieces and nephews. She needs 6 presents to fill each stocking. She could try to find 102 different presents so that they all have completely different sets of presents, but this would take a very long time. What’s the minimum number of different items she needs to find? 9 Santa Claus is making 9 Christmas puddings. He’s made them all of equal size and weight, and intended to put a silver coin in each one. Unfortunately he has one coin left over, which means that one of the puddings is without a coin… He needs to work quickly (whilst Mrs Claus is feeding the reindeer) to identify the pudding, but he only has a simple set of scales to hand. How many times does he need to use the scales to be sure which pudding is without its coin? 8 Frosty started life 2 metres tall. Each time there is a sunny day, he loses 20% of his current height. How many sunny days will it be before he’s reduced to a small pile of snow of less than 10cm? 7 Three of the Seven Dwarves are arguing about a restaurant bill. They’ve worked out that Grumpy (yellow hat) owes half the bill, Bashful (orange hat) owes a third and Doc (red hat) owes an eighth of it. The bill is £23, but they each only have pound coins, The manager arrives and suggests they round it up to £24 instead as that will be easier to split. They are all happy with this and the manager gets £23… how can this be? 6 Place the numbers 1 to 12 in the circles (once each) so that each row of 4 numbers adds up to the same total. 5 5 machines are making sweets. They are all supposed to be making them exactly the same size, shape and mass. Unfortunately, one of the machines is making them 1 gram too heavy. How can the operator identify which machine it is by weighing just one batch of sweets? 4 Two identical large equilateral triangles form a ‘regular’ star. What fraction of the star is shaded? 3 All is not well in the Claus household. Santa wants to put up some lights… …but Rudolph, who is obsessed with numbers and sequences, has tinkered with them so that each one will only light up if the number that Santa enters appears in a certain sequence (a different sequence for each light). It should look like the set below. Please help Santa! 2 In each of the word sums on the next slide replace each letter by a digit from 0 to 9. The same letter is replaced by the same digit within a sum, and the sum must ‘add up’ correctly. Each digit can only be used once in each word sum (they’re not all the same solutions). Notice that Example: ‘O’ is ON E 4 3 2 replaced by + O N E could be + 4 3 2 ‘4’ in all places T W O 8 6 4 2 E A T V I X E N + V I X E N C O M E T + E A T F U L L S M I L E + S M I L E H A P P Y S N O W + M A N C O M E T + C O M E T C O L D D A S H E R 1 The numbers 1 to 15 have all been assigned a letter. Solve the clues on the next slide to crack the code below (dots mark the end of each word): 9 5 12 3 . 5 . 7 11 3 5 6 . 9 10 1 2 4 5 8 . 3 12 3 11 8 10 13 3 ; 14 3 3 . 8 10 15 . 5 7 5 2 13 . 2 13 . A x N x (O x E + L) 1a 2T = 12 G>I A2 = 25 AE = U G+I = 9 V = 3D L2 = L Y+T = S N+G = 20 2G = S 3E = 9 H+1 = O R-E = Y 1b 2N+I = 28 N+2I = 17 A2 = 25 AE = U G= 49½ (V-D)2 = Y2 L2 = L 2G = S R2-O2=21 (x + E)(x + Y) = x2 +11x +24 Y>E 2H-T = 12 H+T = 15 Teacher notes: Twelve Days of Christmas This is a selection of short activities and puzzles – and some a little longer – with a slightly festive twist to count down to the end of term. They could be used as starter activities for a wide range of students in KS3 and KS4 or as a bit of fun for Post-16 students. Some have natural extensions that would make them more challenging. In the teacher notes are answers for all of the problems and suggestions for additional questions for some of the problems. Teacher notes: Twelve Days of Christmas 12: You could print out slide 3 for students to work from 1 Rudolph 2 Cupid 3 Donner 4 Dancer 5 Blitzen 6 Vixen 7 Comet 8 Dasher 9 Prancer Teacher notes: Twelve Days of Christmas 11: Yes, this method always works. 952 = 8125 1052 = 11025 1952 = 38025 Try for other numbers. Additional question: why does this work? The number ‘P’5 can be written as 10P+5 (10P+5)2 = 100P2 +100P +25 = 100P(P+1) + 25 The 100 multiplier ‘shifts’ the answer to P(P+1) into the ‘hundreds’ column and beyond, leaving space for the ‘25’ Teacher notes: Twelve Days of Christmas 10: 8 different items needed It’s probably easiest thinking about this in terms of what she doesn’t put in. If she had 7 different items, she would have to leave 1 item out each time, so there are only 7 ways to do this. If she had 8 different items, she would need to leave 2 out of each stocking. There would be 8C2 possibilities, which is 28; she needs 17, so this is sufficient. Alternative approach to 8C2 Shaded cells indicate which pairs of presents can be left out: & 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 There are 28 pairs. Teacher notes: Twelve Days of Christmas 9: Just twice. We know that the pudding without the coin will be lighter. Group the puddings into 3 sets of 3 (A, B & C) Identify which set of 3 the lighter pudding is in by loading set A on one side and set B on the other. If one side rises, it’s within that set of 3, if they balance, it’s within set C. From the set of 3, put one pudding on one side and one on the other. If one side is lighter, it has been identified, if they balance, it’s the other pudding. Additional question: what if there were 27 puddings? (just 3 comparisons needed). Recommended to change the initial problem to 27 puddings for more challenge. Teacher notes: Twelve Days of Christmas 8: If Frosty is built on Day 1, then on Day 15 he drops below 10 cm. Teacher notes: Twelve Days of Christmas 7: The Dwarves were incorrect in their initial calculations: 1 2 + 1 + 1 = 23 3 8 24 Teacher notes: Twelve Days of Christmas 6: One solution is shown. Hint: What must the total for each line be? Each number is used twice, so the sum for the whole star is 2x(1+2+3…+11+12) = 156 156 ÷ 6 = 26 This forum has a good explanation about how to find solutions logically. Teacher notes: Twelve Days of Christmas 5: Sweet problem. Take: 1 sweet from machine 1 2 sweets from machine 2 3 sweets from machine 3 4 sweets from machine 4 5 sweets from machine 5. Weight them all together. The number of grams over the ‘proper’ weight will tell her which machine is malfunctioning. Teacher notes: Twelve Days of Christmas 4: 1/48th The middle hexagon is equal in area to the area of the 6 outside points. The interior hexagon has been divided into 6 similar shapes. One of the sections has been divided into 4 similar triangles. Can you show they are similar? Teacher notes: Twelve Days of Christmas 3: You will need the Excel file ‘Christmas Lights Sequence’; macros must be enabled for this file to work. The aim of this activity is to light all 4 lights simultaneously, where each light is linked to a different sequence. A number is entered into the yellow box and ‘checked’ if the number is in the sequence linked to the light, the light will appear, if not it remains hidden. Through entering a range of values, students work out what the 4 number sequences are and find the number that will light all 4 lights. Each sequence is linear. Eight sets of sequences are provided: for sequences 1,3,5,& 7 it is possible to light all 4 lights simultaneously, for sequences 2,4,6 & 8 it is not. This activity encourages problem solving and reasoning. Giving all students time to think and explain is important. Teacher notes: Twelve Days of Christmas 3: This is designed as a whole class activity; mini-whiteboards could be used to encourage participation by ‘voting’ for numbers or predicting what will happen. Possible structure and questioning. Explain that the intention is to make all 4 lights appear. Ask for a value to try and check it to see which lights appear. Ask for several more values and check each. Ask if anyone has any ideas about what numbers and/or sequences seem to be linked to which lights. Ask if students to give you a number to light a specific light. Give a number and ask students to predict which lights it will make appear. This is helpful if sequences don’t coincide as it encourages reasoning and explanation. Ask if anyone can give a number that will light all 4 lights. Ask if anyone can give a different number to light all 4 lights… is it just double the first solution? Why not? This is a very common misconception. For sequences 2, 4, 6 & 8 ask if anyone can explain why it is not possible to make all 4 appear.. Teacher notes: Twelve Days of Christmas 3: The file uses set 1 initially. On the ‘change values’ tab it is possible to select a different set of values (1 to 8) – you might choose to do a couple of sets that work and then choose one that doesn’t. Set 1: 23 Set 2: not possible to light 2 & 3 together Set 3: 60 Set 4: not possible to light 1 & 2 or 3 & 4 together Set 5: 61 Set 6: not possible to light 1 & 2 together Set 7: 86 Set 8: not possible to light 2 & 4 or 3 & 4 together Sets 9 and 10 are left blank for you to insert your own sequences or for students to devise their own sets to challenge their peers. Teacher notes: Twelve Days of Christmas 2: Word Sums The instructions for these are straightforward – and they simply rely on the rules of addition, however, some are quite tricky to solve, whilst some have several solutions. A solution for each one is given on the next slide. Teacher notes: Twelve Days of Christmas + E A T + E A T F U L L 6 7 2 + 6 7 2 1 3 4 4 S M I L E S M I L E H A P P Y S N O W M A N O L D + C + 3 2 0 5 9 3 2 0 5 9 6 4 1 1 8 2 + 3 C O M E T + C O M E T D A S H E R + 9 0 7 1 4 9 0 5 6 9 2 7 0 3 + 9 2 7 0 3 1 8 5 4 0 6 + V I X E N V I X E N C O M E T 4 7 8 0 1 4 7 8 0 1 9 5 6 0 2 Teacher notes: Twelve Days of Christmas 1: Coded message. 1a and 1b both have the same solutions, but 1b is more challenging, using quadratics and simultaneous equations. Answers: 1L 2I 3E 4D 5A 6T 7G 8Y 9H 10O 11R 12V 13N 14S 15U “Have a great holiday everyone; see you again in 2015” ‘Sticking points’ 1a: students should be able to find 4 letters quite easily, they then need to use G+I = 9 and realise that there is only one pair of numbers left which add up to 9, G>I enables them to move forward. Similarly, they will get to a point near the end when they will need to realise that only one pair of numbers remains which would work for V=3D