Senior Challenge ‘20 the Solutions

Senior Challenge ‘20 The Solutions University of Southampton Entries 485 students 31 schools 41 prizes Top Schools School Number of prizes Wilson’s School 9 Brookfield Community School 5 King Edward VI Camp Hill School for Girls 3 King’s School Winchester 2 (including 1st and a 2nd prize) Well done to all schools! 1. Flicks for Four Families Four families go to the cinema. The Reid family take 2 adults and 2 children for £18; the McGhee family take 1 senior, 2 adults and 1 child for £18.50; and the Griffiths family take 1 senior and 3 adults for £19.50. How much will it cost for the Linton family to take 1 senior, 2 adults and 3 children? 1. Flicks for Four Families ① Reid 2a + 2c = 18 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 ④= ③ – ② (s+3a) – (s+2a+c) = a – c 19.50 – 18.50 = 1 So ④ is a – c = 1 1. Flicks for Four Families ① Reid 2a + 2c = 18 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 ④ a – c = 1 ① +2 × ④ (2a + 2c) + 2 × (a – c) = (2a + 2c) + (2a – 2c) = 4a 18 + 2 × 1 = 20 so 4a = 20, so a = 5 1. Flicks for Four Families ① Reid 2a + 2c = 18 a = £5.00 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 Substitute a=5 into ① gives 2 × 5 + 2c = 18, so 10 + 2c = 18, so 2c = 8, so c = 4 1. Flicks for Four Families ① Reid 2a + 2c = 18 a = £5.00 ② McGhee s + 2a + c = 18.50 c = £4.00 ③ Griffiths s + 3a = 19.50 Substitute a=5 into ③ gives s + 3 × 5 = 19.50, so s + 15 = 19.50, so s = 4.50 So a=£5, c=£4, s=£4.50 1. Flicks for Four Families a=£5, c=£4, s=£4.50 Lintons are s + 2a + 3c = 4.50 + 2 × 5 + 3 × 4 = 4.50 + 10 + 12 =£26.50 2. I Can See Here From My House Sarah claimed she could see Russia from her house in Wasilla (Alaska) although the nearest point in Alaska to Russia (called Wales, oddly enough) is 635 miles away from Wasilla. If you square 635, and subtract 1 from the result, you get a number divisible by four (try it). Show that the same is true of 7. Can you prove that the same is true of ANY odd number? 2. I Can See Here From My House 635² – 1 = 403,225 – 1 = 403,224 403,224=4×100,806 7² – 1 = 49 – 1 = 48 48=4×12 A number of entries lost marks here because they omitted the first part of the question or tried to divide 403,224 by 7. We also had far too many people who said ‘0 is not divisible by 4’! 2. I Can See Here From My House An even number is 2n for some integer n. An odd number is one more than an even number, so can be written as 2n+1 The square of an odd number is then (2n+1)×(2n+1)= 4n²+4n+1 Subtract 1 gives 4n²+4n This can be factorised to give 4(n²+n) which shows that 4 is a factor, so 4n²+4n is divisible by 4 3. Spectacular! Six students, Oleg, Bonnie, Pete, Georgia, Rachel and Simon, all need new glasses. The opticians have 6 pairs of glasses, each with a different coloured frame: red, blue, green, silver, pink and orange. Each of the boys has 2 colours that he’d prefer and each girl has 1 colour she’d prefer. Oleg prefers silver and pink, Simon prefers green and red, Pete prefers orange and blue, Bonnie prefers orange, Rachel prefers pink, and Georgia prefers blue. One of the girls (who has an r in her name) realises that not everyone can have their preference. How do they know? She then offers to choose a different colour so that everyone else can have their preference. Who was it and what colour did they choose? Which colour does each student ultimately select? 3. Spectacular! Colours Students Simon is the only person who likes Red and Green Blue Bonnie Green Georgia If Oleg has Silver, then Orange Oleg Rachel has Pink, so her changing wouldn’t help. Pink Pete Bonnie doesn’t have an r Red Rachel So Georgia changes her mind. Silver Simon 3. Spectacular! Colours Students This gives us two possible answers, with these 4 Blue Bonnie matches fixed Green Georgia If Simon has red, Georgia gets green and vice versa. Orange Oleg Pink Pete Red Rachel Silver Simon 3. Spectacular! Without the ‘r’ condition, Colours Students Bonnie could change, this gives these 4 as the fixed Blue Bonnie matches Green Georgia If Simon has red, Bonnie gets Orange Oleg green and vice versa. Pink Pete We marked this question so that students got the more Red Rachel generous mark from the two versions. Silver Simon 4. Whatever the Weather In a network of weather stations, each station is required to know how many hours of daylight there were at each of the others. Unfortunately, their email system has been infected by a virus and they can only communicate via phone calls. For three stations, show that three separate calls are required so that all three sets of results are known at all three stations. For five stations, show that at least six separate calls are required. What is the minimum number of calls if there are seven stations? Extend your reasoning for n stations. 4. Whatever the Weather Calls Stations 0 1 2 3 A A AB AB ABC B B AB ABC ABC C C C ABC ABC A C B 4. Whatever the Weather Calls Stations 0 1 2 3 4 5 6 A A A AB AB AB AB AB ABCDE B B AB ABC ABC ABCDE ABCDE ABCDE B C C C C ABC ABC ABC ABCDE ABCDE D D D D DE ABCDE ABCDE ABCDE D E E E E E DE DE ABCDE ABCDE 4. Whatever the Weather Calls Statio 0 1 2 3 4 5 6 7 8 9 10 ns F ABCD ABCD ABCD A A A AFG ABFG ABFG ABFG ABFG ABFG EFG EFG EFG ABCF ABCF ABCD ABCD ABCD ABCD ABCD G B B B B ABFG G G EFG EFG EFG EFG EFG ABCF ABCF ABCF ABCD ABCD ABCD ABCD C C C C C G G G EFG EFG EFG EFG A ABCD ABCD ABCD ABCD ABCD D D D D D D DE EFG EFG EFG EFG EFG ABCD ABCD ABCD ABCD E E E E E E DE DE B C EFG EFG EFG EFG ABCD F F FG FG FG FG FG FG FG FG FG EFG D E ABCD ABCD G G FG AFG AFG AFG AFG AFG AFG AFG EFG EFG 4. Whatever the Weather For 4 stations, the information can be B C shared with 4 calls as shown here. Each additional station needs 2 calls: one D E to put their information into the loop, and one to take everyone else’s information back out. So for n ≥ 4, the number of calls is 2n - 4 5. Staying Out of the Spotlight Andrew is standing in the centre of a circle. Above the edge of the circle there are 3 spotlights, pointed straight down. These are equally spaced around the circle and can be moved around it in either direction by Nicole whilst remaining the same distance from each other. The spotlights can move twice as fast as Andrew. Andrew gets caught if a spotlight hits him. Determine whether Andrew can get out of the circle without being caught. 5. Staying Out of the Spotlight Andrew can’t make it out by simply running for the edge. He moves r, the light 2휋푟 has to move 6 The light is moving at twice his speed, so catches him 5. Staying Out of the Spotlight TheThis dottedmeans circlehe can is halfwayreach the to halfway the edge. Withinpoint, whilst this circle, keeping Andrewone of the can lights move behind aroundhim. faster than the lightsHe then as hemakes has aless dash thanfor the half edge the anddistance can toescape. cover for a given angle. 6. Say What You See Lilly, Shaun and Ellen are stood in a line, one behind the other, looking in the same direction along the line. Lilly can see both Shaun and Ellen; Shaun can only see Ellen, who cannot see either of the other two. They are blindfolded and a hat is placed on each of their heads. They are told that the hats are selected at random from a box containing two red hats and three green hats. The blindfolds are removed, and they are asked to state, without turning around or conferring, what colour hat they themselves are wearing. After a pause, during which no-one has spoken, Ellen announces that she knows the colour of her hat. What is the colour of her hat and how does she know this? 6. Say What You See Lilly Shaun Ellen Looking this way Ellen does not turn round She doesn’t take her hat off There is no mirror There is no box with the other hats in! 6. Say What You See Looking this way Lilly Shaun Ellen Lilly can see both red hats, so knows hers is green If Lilly is silent, she must be able to see at least one green hat. If Shaun can see a red hat, he knows his must be green If Shaun is silent, he cannot see a red hat, so Ellen’s hat must be green.

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