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Intermediate Mathematical Challenge Organised by the United Kingdom Mathematics Trustsupported by Solutions and investigations 7 February 2019These solutions augment the printed solutions that we send to schools. For convenience, the solutions sent to schools are confined to two sides of A4 paper and therefore in many cases are rather short. The solutions given here are longer. In some cases we give alternative solutions, and we have included some exercises for further investigation. We welcome comments on these solutions. Please send them to enquiry@ukmt.org.uk. The Intermediate Mathematical Challenge (IMC) is a multiple-choice paper. For each question, you are presented with five options, of which just one is correct. It follows that often you can find the correct answers by working backwards from the given alternatives, or by showing that four of them are not correct. This can be a sensible thing to do in the context of the IMC. However, this does not provide a full mathematical explanation that would be acceptable if you were just given the question without any alternative answers. So for each question we have included a complete solution which does not use the fact that one of the given alternatives is correct. Thus we have aimed to give full solutions with all steps explained (or, occasionally, left as an exercise). We therefore hope that these solutions can be used as a model for the type of written solution that is expected when a complete solution to a mathematical problem is required (for example, in the Intermediate Mathematical Olympiad and similar competitions). These solutions may be used freely within your school or college. You may, without further permission, post these solutions on a website that is accessible only to staff and students of the school or college, print out and distribute copies within the school or college, and use them in the classroom. If you wish to use them in any other way, please consult us. ©UKMT February 2019 Enquiries about the Intermediate Mathematical Challenge should be sent to: IMC, UKMT, School of Mathematics Satellite, University of Leeds, Leeds LS2 9JT T 0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk1 B 2 E 3 D 4 B 5 D 6 B 7 B 8 E 9 C 10 B 11 C 12 C 13 A 14 C 15 A 16 A 17 D 18 B 19 E 20 B 21 D 22 E 23 A 24 C 25 DUKMT UKMTUKMT

Intermediate Mathematical Challenge 2019 Solutions and investigations1. What is the value of 2019 tenths? A 2019B 201.9C 20.19D 2.019E 0.2019 Solution B The value of 2019 tenths is 2019×1 10 = 201 .9 . For investigation 1.1 Write the following as decimals. (a)2019 hundredths, (b)2019 thousandths, (c)2019 millionths, (d)2019 billionths. 2. Each of the five shapes shown below is made from five unit cubes. Which has the smallest surface area? A B C D E Solution E Each cube has six faces, and so between them the five unit cubes have 30 faces.The surface area of a shape is determined by how many of the faces of each cube are on the outside of the shape. When two of the cubes touch, two faces are hidden and so don’t form part of the surface of the shape. Therefore the shape with the smallest surface area is the shape where the largest number of pairs of cubes touch each other. It can be seen that in each of the shapes A, B, C, and D the number of pairs of faces that touch is 4, but in the shape of option E, there are 5 pairs of cubes that touch. Hence the shape that has the smallest surface area is the shape of option E. For investigation 2.1 Find the surface area of each of the five shapes. 2.2 Is there a shape built from the five unit cubes that has a smaller surface area than the shape of option E? 2.3 What is the smallest surface area of a shape made from six unit cubes? © UKMT February 2019 www.ukmt.org.uk 2

Intermediate Mathematical Challenge 2019 Solutions and investigations3.There are 120 000 red squirrels living in Scotland. This represents 75% of their total UK population. How many more red squirrels live in Scotland than live in the remainder of the UK? A 30 000 B40 000 C60 000 D80 000 E90 000 Solution D Because 75% of the red squirrels live in Scotland, 25% live in the remainder of the UK. Now 25% is one third of 75%. Therefore the number of squirrels in the remainder of the UK is 1 3 × 120 000 =40 000 . Therefore there are 120 000 −40 000 , that is, 80 000 , more red squirrels living in Scotland than in the remainder of the UK. For investigation 3.1 Based on the numbers given in this question, how many red squirrels are there in the UK? 3.2 If the number of red squirrels in Scotland were 80% of the total UK population of 120 000 , how many more red squirrels would live in Scotland than in the rest of the UK? 3.3 It is estimated that only one in eight of the UK’s red squirrels live in England. (a)What is this as a percentage? (b) Based on this estimate, how many more red squirrels live in Scotland than in England? 3.4 It is estimated that there are 110 000 fewer red squirrels in Wales than in the rest of the UK. On the basis of this estimate, what percentage of the UK’s red squirrels live in Wales? 3.5 The number of humans living in the UK, to the nearest 1 million, is 67 million. To the nearest 10, how many humans are there in the UK for each red squirrel? 4. A 24-hour digital clock shows the time in hours and minutes. How many times in one day will it display all four digits 2, 0, 1 and 9 in some order? A 6B 10C 12D 18E 20 Solution B Commentary Because the number of arrangements of the digits which correspond to a time is quite small, the most straightforward method is just to list them all. This is the first method that we use. When answering a question of this type by listing all the possibilities, it is important to list them in a systematic way, so as to be sure that no case has been left out. Here, the systematic method is to list the cases in order of the times to which they correspond. © UKMT February 2019 www.ukmt.org.uk 3

Intermediate Mathematical Challenge 2019 Solutions and investigationsOur second method avoids listing all the cases by counting them in a systematic way using combinatorial methods.Method 1 The times when the four digits 2, 0, 1 and 9 are displayed in some order are 01:29 02:19 09:12 09:21 10:29 12:09 19:02 19:20 20:19 21:09. We see that there are 10 times in the day when the clock displays the four given digits in some order. Method 2 The four digits 2, 0, 1 and 9 can be arranged in order in 24 ways, but not all these arrangements correspond to actual times. The digit 9 can occur only in the second and fourth positions because there are only 24 hours in a day and only 60 minutes in an hour. There is no hour 29, so if the digit 9 is in the second position, the first digit can only be 0 or 1. In each of these two cases, the remaining two digits can be arranged in the remaining two places in either order. So there are 2× 2= 4times when the digit 9 occurs in the second place. If the digit 9 is in the fourth position, the remaining three digits can be arranged in the remaining three places in any order. Three numbers may be arranged in order in 3×2×1= 6different ways. So there are 6 times when the digit 9 occurs in the fourth place. This makes a total of 4+ 6= 10 times in one day when the digits 2, 0, 1 and 9 are displayed in some order. For investigation 4.1 Another 24-hour digital clock shows the time in hours, minutes and seconds. (a) How many times a day does the clock display all the six digits 0, 1, 2, 3, 4 and 5 in some order? (b) How many times a day does the clock display all the six digits 1, 2, 3, 4, 5 and 6 in some order? (c) How many times a day does the clock display all the six digits 2, 3, 4, 5, 6 and 7 in some order? (d) How many times a day does the clock display all the six digits 3, 4, 5, 6, 7 and 8 in some order? © UKMT February 2019 www.ukmt.org.uk 4

Intermediate Mathematical Challenge 2019 Solutions and investigations5. The answers to the three calculations below are to be written in descending order. X0.6 × 0.5 + 0.4 Y 0.6 × 0.5 ÷ 0.4 Z 0.6 × 0.5 × 0.4 What is the correct order? A Y Z X BXY Z CX ZY DY X Z EZY X Solution D We have X0.6 × 0.5 + 0.4 = 0.3 + 0.4 = 0.7 , Y 0.6 × 0.5 ÷ 0.4 = 0.3 ÷ 0.4 = 0.75 , Z 0.6 × 0.5 × 0.4 = 0.3 × 0.4 = 0.12 .We therefore see that expression Y has the largest value and expression Z has the smallest value. Hence the correct descending order is Y X Z. For investigation 5.1 Arrange the following in increasing numerical order. (a) 0.6 × 0.5 + 0.4 × 0.3 (b) 0.6 + 0.5 × 0.4 × 0.3 (c) 0.6 ÷ 0.5 + 0.4 × 0.3 (d) 0.6 × 0.5 × 0.4 × 0.3 (e) 0.6 + 0.5 + 0.4 + 0.3 5.2 Each of the symbols □,^ and ♥ may be replaced by one of the operations + ,− ,× and ÷. Different symbols may be replaced by the same operation. Which choices of the operations make the value of the expression 0.6 □ 0.5 ^ 0.4 ♥ 0.3 as large as possible? © UKMT February 2019 www.ukmt.org.uk 5

Intermediate Mathematical Challenge 2019 Solutions and investigations6.The diagram shows part of a tessellation of the plane by a quadrilateral. Khelen wants to colour each quadrilateral in the pattern so that no two quadrilaterals that meet (even at a point) have the same colour. What is the smallest number of colours he needs? A 3B 4C 5D 6E 7 Solution B The quadrilaterals that meet at each vertex need to be given different colours. Because four quadrilaterals meet at each vertex, Khelen needs at least 4 colours. The figure shows that it is possible to colour the quadrilaterals using just 4 colours, so that no two quadrilaterals that meet at a vertex have the same colour. Hence the smallest number of colours that Khelen needs is 4. For investigation 6.1 Suppose that quadrilaterals that meet along an edge must be coloured differently, but quadrilaterals that meet just at a point may have the same colour. What is the smallest number of colours that Khelen would need in this case? 6.2 What is the smallest number of colours needed to colour the edges so that edges that meet at a vertex are coloured differently? 7. How many positive cubes less than 5000 end in the digit 5? A 1B 2C 3D 4E 5 Solution B The cube of a positive integer ends in the digit 5 only when the positive integer itself ends in the digit 5. So we check the cubes of the positive integers that have units digit 5, until we reach a cube which is greater than 5000. We have 53 = 125

Intermediate Mathematical Challenge 2019 Solutions and investigations8.Three consecutive positive integers less than 20 are, in ascending order, prime, even and triangular. What is the product of these three integers? A 6B 60C 990D 1786E 2730 Solution E The triangular numbers that are less than 20 are 1, 3, 6, 10 and 15. The largest of the three consecutive positive integers is one of these numbers. It is odd as it immediately follows an even number. It is not 1 as it is the largest of three positive integers. Hence it is either 3 or 15. Therefore the three consecutive integers are either 1, 2, 3 or 13, 14, 15. The first number in the sequence is a prime number. This rules out the sequence 1, 2, 3 because 1 is not a prime number. However, 13 is a prime number. So the three consecutive positive integers are 13, 14, 15. Their product is given by 13×14 ×15 =13 ×210 =2730 . It is important to remember that 1 is not regarded as a prime number. This is only a convention, but it is a standard convention. As Vicky Neale says in her recent book, Closing the Gap: the quest to understand prime numbers , Oxford University Press, 2017, “In case you’re wondering, I’ll mention that 1 is not prime. That’s not because of some interesting philosophical point, or because it’s only got one number that divides into it rather than two, or anything like that. Definitions (such as that of a prime number) don’t get handed to mathematicians on stone tablets. Rather, part of the job of mathematicians is to make good definitions, ones that lead to interesting mathematics, and it turns out to be better to define 1 not to be prime.” For investigation 8.1 Is it possible to have three consecutive positive integers that are all primes? 8.2 Is it possible to have four consecutive integers, all greater than 2, such that three of them are prime? 8.3 Is it possible to have four consecutive positive integers none of which is prime? 9. What is the value of ( 7 − 6× (− 5)
− 4× (− 3) ÷ (− 2)? A −17 B−11 C 31D 37E 43 Solution C We have 7− 6× (− 5) = 7+ 30 =37 and 4× (− 3) ÷ (− 2) = −12 ÷ − 2= 6. Hence ( 7 − 6× (− 5))
− 4× (− 3) ÷ (− 2) = 37 −6= 31 . © UKMT February 2019 www.ukmt.org.uk 7

Intermediate Mathematical Challenge 2019 Solutions and investigations10.A recent report about the amount of plastic created in the last 65 years stated that the 8.3 billion tonnes produced is as heavy as 25 000 Empire State Buildings in New York or a billion elephants. On that basis, how many elephants have the same total weight as the Empire State Building? A 4000B 40 000C400 000 D4 000 000 E 40 000 000 Solution B A billion is 1 000 000 000 . So we are told that 1 000 000 000 elephants weigh the same as 25 000 Empire State Buildings. Therefore the number of elephants that have the same weight as one Empire State Building is given by 1 000 000 000 25 000 = 1 000 000 25 = 40 000 . For investigation 10.1 Based on the information given in the question, what is the weight of the Empire State Building, and what is the weight of a typical elephant? 11. Which of the following is equal to 3 9 9 3 ? A 3B 9C 27D 81E 243 Solution C Because 9= 32 , we have 93 = (3 2 )3 = 36 . Therefore, 3 9 9 3 = 3 9 3 6 = 39 − 6 = 33 = 27 . For investigation 11.1 Our solution uses the facts that (32 )3 = 36and 39 3 6 = 39 − 6 . These are special cases of the general results that, for all positive integers mand n, and all real numbers a, we have (a) (a m )n = amn , and (b) for m>n, a m a n = am −n . Explain why these are true. 11.2 Find the integer nsuch that 27 81 81 27 = 3n . 11.3 Find the integer nsuch that 16 256 256 16= 2n . © UKMT February 2019 www.ukmt.org.uk 8

Intermediate Mathematical Challenge 2019 Solutions and investigations12.The game of Rorrim2 is played on a 4× 4board, starting with a counter in one corner, as shown. At each turn, the player moves the counter to a cell that is the reflection of its current cell in one of the six dashed lines. How many cells could the counter occupy after precisely three turns? A 4B 6C 8D 12E 16 Solution C In the figures below we have shaded alternate squares of the board to make it easier to see what is going on. In Figure 1 we have shown all the cells that the counter could occupy after one turn. These positions are found by considering all the possible moves that the player can make when the counter starts off in the bottom left hand cell of the board. Figure 2 shows all the cells that the counter could occupy after two turns. These positions are found by considering, in turn, all the player’s possible moves when the counter is in one of the positions marked ‘1’ in Figure 1. In a similar way Figure 3 shows all the cells that the counter could occupy after three turns. From Figure 3 we see that there are 8 cells which the counter could occupy after precisely three turns. For investigation 12.1 How many cells could the counter occupy after at most three turns? 12.2 How many cells could the counter occupy after precisely four turns. 12.3 (a) Now consider the same game played on a 6× 6board. For k= 1,2, ... determine how many cells the counter could occupy after kturns. (b) Which is the least ksuch that each of the cells could be occupied after at most kturns? © UKMT February 2019 www.ukmt.org.uk 9

Intermediate Mathematical Challenge 2019 Solutions and investigations13.Megan writes down a list of five numbers. The mean of her first three numbers is −3. The mean of her first four numbers is 4. The mean of her first five numbers is −5. What is the difference between her fourth number and her fifth number? A 66B 55C 44D 33E 22 Solution A Because the mean of Megan’s first three numbers is −3, the sum of her first three numbers is 3 × − 3= −9. Similarly, the sum of her first four numbers is 4× 4= 16 , and the sum of her first five numbers is 5 × − 5= −25 . It follows that Megan’s fourth number is 16 − (− 9)= 25 and her fifth number is −25 − 16 = − 41 . Therefore the difference between her fourth number and her fifth number is 25− (− 41)= 66 . For investigation 13.1 Find the sequence of four numbers for which the first is 1, the mean of the first two is 2, the mean of the first three is 3, and the mean of the first four is 4. 13.2 An infinite sequence of numbers has the property that, for each positive integer n, the mean of the first nnumbers in the sequence is n. Find a formula for the nth number in the sequence. 13.3 An infinite sequence of numbers has the property that, for each positive integer n, the mean of the first nnumbers in the sequence is n2 . Find a formula for the nth number in the sequence. © UKMT February 2019 www.ukmt.org.uk 10

Intermediate Mathematical Challenge 2019 Solutions and investigations14. There are four people, some of whom always tell the truth. The others always lie. The first person said, “An odd number of us always tell the truth”. The second person said, “An even number of us always tell the truth”. The third person said, “A prime number of us always tell the truth”. The fourth person said, “A square number of us always tell the truth”. How many of these four people were telling the truth? A 0B 1C 2D 3E 4 Solution C Commentary Our first method involves thinking about what the different people say logically. The second method indicates a more computational approach.The first method leads to a shorter solution. The second method might be of more use in a more complicated problem of this type. Method 1 We note first that if a person is telling the truth, then they always tell the truth. Similarly, if they are telling a lie, then they always tell lies. An integer is either even or odd, but not both. Therefore one of the first two people always tells the truth and one always lies. A square is not a prime, so at most one of the last two people always tells the truth. We deduce from this that the number of the people who always tell the truth is either one or two. However, if just one person always tells the truth and the other three always lie, both the first and fourth people are telling the truth, as 1 is both odd and a square. Hence two of the people always tell the truth. This is contradiction. So this case doesn’t occur. If just two people always tell the truth, then the second and third people are telling the truth, because 2 is even and a prime, and the other two are not telling the truth. So in this case there is no contradiction. We deduce that 2 is the number of people who were telling the truth. © UKMT February 2019 www.ukmt.org.uk 11

Intermediate Mathematical Challenge 2019 Solutions and investigationsMethod 2In columns two to five of the table below we have systematically listed all 16 different combinations of the people who always tell the truth, indicated by T, and who always lie, indicated by L. We use P1, P2, P3, P4 to indicate the people. In the sixth column we give the number of truth tellers according to the information in the earlier columns. The last 4 columns indicate whether or not the four people are telling the truth based on what they say and the number of truth tellers as given in the sixth column. For example, in row 16, because 0 is even and a square, in this row it is just P2 and P4 who are truth tellers. row no. P1 P2 P3 P4 no. of truth tellers P1 P2 P3 P4 1 T T T T 4 L T L T 2 T T T L 3 T L T L 3 T T L T 3 T L T L 4 T T L L 2 L T T L 5 T L T T 3 T L T L 6 T L T L 2 L T T L 7 T L L T 2 L T T L 8 T L L L 1 T L L T 9 L T T T 3 T L T L 10 L T T L 2 L T T L 11 L T L T 2 L T T L 12 L T L L 1 T L L T 13 L L T T 2 L T T L 14 L L T L 1 T L L T 15 L L L T 1 T L L T 16 L L L L 0 L T L T If, in a given row, the information in the last four columns is not the same as that in columns two to five, the situation in that row cannot occur. From the table we see that the only situation that can occur is that described in row 10. We deduce that 2 of the people were telling the truth. For investigation 14.1 For each person there are 2 possibilities. They are either a truth teller or a liar. Why does it follow that we needed to have 16 rows in the above table? How many rows would be needed if there were 5 people? 14.2 Columns 2 to 5 in the above table have been set out systematically to make sure that none of the combinations has been excluded. Can you describe the system that has been used? 14.3 Suppose that in the first four columns you replace T by 1 and L by 0. What do you then notice? © UKMT February 2019 www.ukmt.org.uk 12

Intermediate Mathematical Challenge 2019 Solutions and investigations15.The diagram shows six congruent equilateral triangles, of side-length 2, placed together to form a parallelogram. What is the length of P R? A 2√ 13 B 7C 6√ 3 D 9E 7√ 3 Solution A The point Q is a vertex of the parallelogram, as shown. We extend the line PQto the point Uso that QUhas length 2. It can be seen that RQU is an equilateral triangle with side length 2. We let Tbe the foot of the perpendicular from Rto QU . It can be seen that the triangles RT QandRT U are congruent. It follows that QT=T U =1. By Pythagoras’ Theorem, applied to the right-angled triangle RT Q , we have RQ 2 = RT 2 + T Q 2 . It follows that RT=p RQ 2 − T Q 2 = √ 2 2 − 12 = √ 3 . We see that PT=7. By Pythagoras’ Theorem, applied to the right-angled triangle RT P , we have P R 2 = PT 2 + RT 2 = 7 2 + √ 3 2 = 49 +3= 52 . Therefore P R=√ 52 =2√ 13 . For investigation 15.1 Prove that the triangle RQUis equilateral. 15.2 Prove that the triangles RT QandRT U are congruent. 15.3 The diagram shows 2ncongruent equilateral triangles, of side length 2, placed together to form a parallelogram, where nis some positive integer. (a)Find the length of PS. (b)Prove that there is no positive integer nfor which the length of PSis an integer. © UKMT February 2019 www.ukmt.org.uk 132 P R

Intermediate Mathematical Challenge 2019 Solutions and investigations16. Two numbers xand yhave a product which is equal to their sum. Which of these expressions gives xin terms of y? A y y − 1 B y y + 1 C y + 1 y D y − 1 y E y 2 y + 1 Solution A Because the product of xand yis the same as their sum, xy = x+ y. By subtracting xfrom both sides, we see that this equation is equivalent to xy − x= y. This last equation is equivalent to x(y − 1) = y.We should now like to divide both sides of this last equation by y − 1, but we need first to check that we are not dividing by an expression that could have the value 0. We see that if ywere equal to 1, the equation xy = x+ y would become x= x+ 1, which is impossible. We deduce that ycannot have the value 1 and therefore y− 1cannot be 0. It therefore follows that x= y y − 1. 17. Which of these is equal to 0.? 8 + 0.0 ? 7 ? A 0.8 ? 7 B 0.88C 0.9 ? 5 D0.9 ? 6 E0.9 ? 8 Solution D Commentary The difficulty with this question is that 0. ? 8 and 0.0? 7are recurring decimals, so that if we could write them in full they would each have an infinite number of digits. When we add numbers with only a finite number of places, the method is to start at the right, adding the corresponding digits and keeping track of the need to carry a 1 to the next place. For example, when adding 1.234 and 5.678 there are two carries (from the units column to the tens column, and from the tens column to the hundreds column): 1.234 + 5.678 6 .912 © UKMT February 2019 www.ukmt.org.uk 14

Intermediate Mathematical Challenge 2019 Solutions and investigationsHowever, with infinite decimal expansions, there is no last place on the right where we can begin the addition: 0.8888888 ..... + 0.0777777 ..... ? .???????? ..... We can get over this difficulty in two ways. We can convert the recurring decimals into fractions, add the fractions, and then convert the answer back into a decimal. Alternatively, we can replace the infinite decimal expansions by finite approximations. Method 1 We have 0.? 8 = 8 9 and 0.0 ? 7 = 7 90 . Therefore 0.? 8 + 0.0 ? 7 = 8 9 + 7 90 = 80 90 + 7 90 = 87 90 = 81 90 + 6 90 = 9 10 + 2 30 = 0.9 + 0.0 ? 6 = 0.9 ? 6 . Method 2 The standard method for adding decimals with a finite number of decimal places gives us 0.88 +0.07 =0.95 0 .888 +0.077 =0.965 0 .8888 +0.0777 =0.9665 0 .88888 +0.07777 =0.96665 and so on. We therefore see that, ‘in the limit’, 0.? 8 + 0.0 ? 7 = 0.9 ? 6 . For investigation 17.1 Express as a single decimal (a) 0.? 6 + 0.? 7 , (b) 0.? 7 + 0.0 ? 6 and (c) 0.? 6 ? 7 + 0.0 ? 7 ? 6 . © UKMT February 2019 www.ukmt.org.uk 15

Intermediate Mathematical Challenge 2019 Solutions and investigations18. Two numbers xand yare such that x+ y= 2 3 and x y = 2 3 . What is the value of x− y? A −2 3 B −2 15 C 2 25 D 2 5 E 2 3 Solution BWe avoid fractions by multiplying both sides of the first equation by 3, and both sides of the second equation by 3y. This gives 3x + 3y = 2 and 3x = 2y. Substituting from the second equation into the first, 2y + 3y = 2, that is, 5y = 2. Therefore y= 2 5 . Hence, as x+ y= 2 3 , x= 2 3 − 2 5 = 10 −6 15 = 4 15 Therefore x− y= 4 15 − 2 5 = 4 − 6 15 = − 2 15 . For investigation 18.1 The two equations given in the question may be written as 3x + 3y = 2and 3x − 2y = 0. Show that x− y= − 1 15 ( 3 x + 3y) + 2 5 ( 3 x − 2y). Use this to calculate the value of x− ywithout the need to find xand yseparately. © UKMT February 2019 www.ukmt.org.uk 16

Intermediate Mathematical Challenge 2019 Solutions and investigations19. Which of these expressions has the largest value? A 1 2 B 1 3 + 1 4 C 1 4 + 1 5 + 1 6 D 1 5 + 1 6 + 1 7 + 1 8 E 1 6 + 1 7 + 1 8 + 1 9 + 1 10 Solution E Because 1 3 > 1 4 , we have 1 3 + 1 4 > 1 4 + 1 4 = 1 2 . Because 1 5 > 1 6 , we have 1 4 + 1 5 + 1 6 > 1 4 + 1 6 + 1 6 = 1 4 + 1 3 = 1 3 + 1 4 . Because 1 7 > 1 8 , we have 1 5 + 1 6 + 1 7 + 1 8 > 1 5 + 1 6 + 1 8 + 1 8 = 1 5 + 1 6 + 1 4 = 1 4 + 1 5 + 1 6 . Finally, because 1 9 > 1 10 , we have 1 6 + 1 7 + 1 8 + 1 9 + 1 10 > 1 6 + 1 7 + 1 8 + 1 10 + 1 10 = 1 6 + 1 7 + 1 8 + 1 5 = 1 5 + 1 6 + 1 7 + 1 8 .We therefore see that the values of the expressions given in the question increase as we go from option A to option E. Hence the expression with the greatest value is that given by option E. For investigation 19.1 Show that (a) 1 2 + 1 3 + 1 4 > 1. (b) 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 > 3 2 . (c) 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + 1 9 + 1 10 + 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16 > 2. 19.2 Prove that for each positive integer n, 1 + 1 2 + 1 3 + · · · + 1 2 2 n − 1+ 1 2 2 n > 1+ n 2 . Note : It follows that the infinite series (called the harmonic series) 1 + 1 2 + 1 3 + · · · +1 n + . . . does not converge to a finite limit. © UKMT February 2019 www.ukmt.org.uk 17

Intermediate Mathematical Challenge 2019 Solutions and investigations20.Three equilateral triangles with sides of length 1 are shown shaded in a larger equilateral triangle. The total shaded area is half the area of the larger triangle. What is the side-length of the larger equilateral triangle? A √ 5 B√ 6 C5 2 D 3√ 3 2 E 1+ √ 3 Solution B We let sbe the side length of the larger equilateral triangle. The ratio of the area of similar triangles (or, indeed, other pairs of similar figures) is the same as the ratio of the squares of the lengths of their corresponding sides. Therefore the area of the larger equilateral triangle is s2 times the area of each of the three equilateral triangles with side length 1. Since the total shaded area is half the area of the larger triangle, the area of the larger equilateral triangle is twice the total area of the three smaller equilateral triangles, and hence is six times the area of any one of these triangles. It follows that s2 = 6, and therefore s= √ 6 . For investigation 20.1 Five regular pentagons, each with side length 1, are shown shaded in a larger regular pentagon. The total shaded area is half the area of the larger regular pentagon. What is the side-length of the larger regular pentagon? 20.2 Six regular hexagons, each with side length 1, are shown shaded in a larger regular hexagon. The total shaded area is half the area of the larger regular hexagon. What is the side-length of the larger regular hexagon? 20.3 Find, in terms of n, the side length of a regular polygon with nsides which has twice the total area of nregular polygons, each of which has nsides of length 1. 20.4 The above solution uses the fact that the ratio of the areas of similar triangles is the same as the ratio of the squares of the lengths of their corresponding sides. Prove that this is correct. © UKMT February 2019 www.ukmt.org.uk 18

Intermediate Mathematical Challenge 2019 Solutions and investigations21.The diagram shows a right-angled triangle PQ R . The point S is the midpoint of the side Q Rand tan∠Q P R =3 2 . What is the value of sin∠Q PS ? A 1 √ 3 B 1 √ 2 C 1 2 D 3 5 E 4 5 Solution D We let QShave length s. Then, as Sis the midpoint of Q R, the length of Q Ris2s. Now tan∠Q P R =Q R Q P . Therefore, by the information given in the question, we have 2s Q P = Q R Q P = tan ∠Q P R =3 2 . and therefore Q P=2 3 × 2s = 4 3 s . Hence, by Pythagoras’ Theorem, applied to the right-angled triangle SQ P, S P 2 = QS 2 + Q P 2 = s2 +  4 3 s 2 = s2 + 16 9 s 2 =  1 + 16 9  s2 = 25 9 s 2 . If follows that S P=5 3 s . Therefore sin∠Q PS =QS PS = s 5 3 s = 3 5 . Note You will see that we could have avoided almost all of the fractions in this solution by letting the length of QS be 3sat the start of the question. But this is being wise after the event, and given the limited time that you have for the IMC, you could be forgiven for not looking ahead in this way. For investigation 21.1 Suppose that QShas length 3s. What in this case would be the length of PS? 21.2 What are the values of sin∠SQ P andcos∠SQ P ? 21.3 Evaluate cos2 ∠SQ P +sin 2 ∠SQ P . Note: cos2 t and sin2 t are the standard ways of writing (cos t)2 and (sin t)2 . 21.4 Does your answer to 21.3 generalize? If so, can you prove this? © UKMT February 2019 www.ukmt.org.uk 19P Q R S

Intermediate Mathematical Challenge 2019 Solutions and investigations22.Four of the following six-digit integers are always divisible by 7, regardless of the values of the digits Pand Q. Which of the following is not neccesarily a multiple of 7? A ‘ PQQ PQQ ’B ‘ PQ P Q PQ’C ‘ Q PQ PQ P’D ‘ P P P P P P’ E ‘ P P P QQQ ’ Solution E Commentary The key facts that we use in this question are that both 1001 and 10 101 are multiples of 7. In fact, 1001 = 7× 11 × 13 , a factorization which it is well worth remembering, and 10 101 =3× 7× 13 ×37 , which is not quite so memorable. We also need to recall that the notation ‘PQ RST U ’ , where P,Q,R,S,T and U are digits, represents the number 100 000P+10 000 Q+1000 R+ 100 S+ 10 T+U. We see that ‘PQQ PQQ ’= 1001 בPQQ ’, ‘ PQ PQ PQ ’= 10101 בPQ ’, ‘ Q PQ PQ P ’= 10101 בQ P ’, and ‘P P P P P P ’= 1001 בP P P ’. Now 1001 = 7× 11 × 13 and 10 101 = 3× 7× 13 × 37 . It follows that all of ‘PQQ PQQ ’ , ‘PQ PQ PQ ’, ‘Q PQ PQ P ’ and ‘P P P P P P ’ are necessarily multiples of 7. In the context of the IMC we can now deduce that the remaining option, ‘P P PQQQ ’ is not necessarily a multiple of 7. [Exercise 22.3 below asks you to check this.] For investigation 22.1 Check that 1001=7× 11 ×13 and that 10 101=3× 7× 13 ×37 . 22.2 Check that (a ) ‘PQQ PQQ ’= 1001 בPQQ ’ ( b ) ‘PQ PQ PQ ’= 10101 בPQ ’ ( c ) ‘Q PQ PQ P ’= 10101 בQ P ’ ( d ) ‘P P P P P P ’= 1001 בP P P ’ 22.3 Give an example of digits Pand Qsuch that ‘ P P PQQQ ’ is not a multiple of 7. 22.4 Is it possible for ‘ P P PQQQ’ to be a multiple of 7? 22.5 Which primes are factors of all integers of the form ‘ P P PQQQ’? © UKMT February 2019 www.ukmt.org.uk 20

Intermediate Mathematical Challenge 2019 Solutions and investigations23.The diagram shows a triangle with sides n 2 + n ,2n+ 12 and 3n + 3. What is the sum of all the values of nfor which the triangle is isosceles? A 7B 9C 12D 13E 16 Solution A Commentary In this solution we use the symbol ⇔to mean is equivalent to . For the triangle to be isosceles two of its sides need to be equal. We therefore equate, in turn, the three pairs of formulas for the side lengths. First we have n2 + n= 2n + 12 ⇔ n2 − n− 12 =0 ⇔ ( n− 4)( n + 3) = 0 ⇔ n= 4or n= −3. Now, when n = 4the sides of the triangle are 20, 20 and 15, which is possible for a triangle. If n were −3, the side lengths would be 6, 6 and −6, which is impossible, because the side length of a triangle is a positive number. So 4 is a possible value of n, but −3 is not. Second, n2 + n= 3n + 3⇔ n2 − 2n − 3= 0 ⇔ ( n− 3)( n + 1) = 0 ⇔ n= 3or n= −1 When n = 3the sides of the triangles are 12, 18 and 12, which is possible for a triangle. If nwere −1the sides lengths would be 0, 12 and 0, which is impossible. So 3 is also a possible value of n, but −1 is not. Finally, 3n + 3= 2n + 12 ⇔ n= 9 If nwere 9 the side lengths would be 90, 30 and 30. This is impossible for a triangle, because the length of one side of a triangle cannot be greater than the sum of the lengths of the other two sides and 90>30 +30 . Therefore, the values of nfor which the triangle is isosceles are 4 and 3. The sum of these values is4+ 3= 7. © UKMT February 2019 www.ukmt.org.uk 213 n + 32 n + 12 n 2 + n

Intermediate Mathematical Challenge 2019 Solutions and investigations24.When 5655 is divided by a two-digit positive integer N, the remainder is 11. When 5879 is divided by the same positive integer N, the remainder is 14. What is the sum of the digits of N? A 6B 7C 8D 9E 10 Solution C Because 5655 has remainder 11 when divided by N, it follows that 5644 = 5655 −11 is a multiple of N. Similarly, because 5879 has remainder 14 when divided by N, it follows that 5879 − 14 = 5865 is also a multiple of N. Because 5644 and 5865 are both multiples of N, so also is their difference. Now 5865 − 5644 = 221 and therefore 221 is a multiple of N. In other words, N is a two-digit factor of 221. When we factorize 221 into primes, we obtain 221=13 ×17 . Therefore Nis either 13 or 17. It can be checked that 17 is a factor of both 5644 and 5865, but neither has 13 as a factor. Hence N=17 . Therefore the sum of the digits of Nis1+ 7= 8. For investigation 24.1 Check that (a) the remainder when 5655 is divided by 17 is 11, and the remainder when 5879 is divided by 17 is 14, (b)17 is a factor of both 5644 and 5865, (c)neither 5644 nor 5865 has 13 as a factor. 24.2 When 7000 is divided by a three-digit positive integer N, the remainder is 99. When 5000 is divided by the same positive integer N, the remainder is 56. What is the value of N? 24.3 When 5000 is divided by a positive integer N, the remainder is 9. When 1000 is divided by the same positive integer N, the remainder is 8. What is the value of N? 24.4 Prove that if the integers jare kare both divisible by the integer N, then both j + k and j− kare divisible by N. © UKMT February 2019 www.ukmt.org.uk 22

Intermediate Mathematical Challenge 2019 Solutions and investigations25.The diagram shows three touching semicircles with radius 1 inside an equilateral triangle, which each semicircle also touches. The diameter of each semicircle lies along a side of the triangle. What is the length of each side of the equilateral triangle? A 3B 13 4 C 3 2 √ 3 D2√ 3 E 4 Solution D Let P,Q and Rbe the vertices of the equilateral triangle. Let S,T be the points on Q R ,RP , respectively, that are the centres of two of the semicircles. Let K be the point where the semicircle with centre Stouches PQ , and let L be the point where the semicircle with centre T touches Q R . Because the tangent to a circle is perpendicular to the radius of the circle through the point where they touch, both ∠QK S = 90 °and ∠R LT =90 °. We leave it as an exercise to show that, because the semicircles with centres Sand T touch, the line ST joining their centres passes through the point where they touch. It follows that ST has length 2. We also leave it as an exercise to deduce that ∠ST R =90 °. Because they are the angles of an equilateral triangle ∠RQ P =∠P RQ =60 °. Because the triangles SK Q and ST R each have angles 60 °and 90 °, and the sum of the angles in a triangle is 180 °, they also each have a third angle of 30 °. Because these triangles have the same angles they are similar. Therefore, because the triangles SK Q and ST R are similar, and ST is twice the length of SK , the length of S Ris twice the length of QS. It follows that the length of Q Ris three times the length of QS. From the right-angled triangle QK S, we have K S QS = sin 60 °= √ 3 2 . Hence QS=2 √ 3 × K S =2 √ 3 × 1= 2 √ 3 = 2 3 √ 3 . It follows that Q R=3× 2 3 √ 3 = 2√ 3 . We conclude that the length of each side of the equilateral triangle is 2√ 3 . © UKMT February 2019 www.ukmt.org.uk 23

Intermediate Mathematical Challenge 2019 Solutions and investigationsFor investigation 25.1Show that, because the semicircles with centres Sand T touch, the line ST passes through the point where they touch. Deduce that SThas length 2. 25.2 Deduce, from the result of Exercise 25.1, that ∠ST R =90 °. 25.3 Let ABC be an equilateral triangle with sidelength 2. Let D be the point where the perpendicular from Ato BC meets BC. (a)Prove that Dis the midpoint of BC. (b) Use the right-angled triangle ADC to show that sin 60 ° = √ 3 2 and tan 60 °= √ 3 . (c)Find the values of cos 60°, sin 30 °, cos 30 °and tan 30 °. If you remember that a triangle with angles 30 °,60 °and 90 °forms half of an equilateral triangle, you can readily work out the values of cos 30 °,sin 30 °,tan 30 ,cos 60 °,sin 60 ° and tan 60 °, without the need to remember any of these values. 25.4 Use a right-angled triangle in which the other two angles are both 45 ° to find the values of (a) sin 45 °, (b) cos 45 °, (c) tan 45 °. Likewise, if you remember the triangle shown in this exercise, you don’t need to remember the values of cos 45°, sin 45 °and tan 45 °. 25.5 Find the ratio of the total area of the three semicircles to the area of the equilateral triangle. © UKMT February 2019 www.ukmt.org.uk 24