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Andrew Jobbings Senior Kangaroo ©2021 UK Mathematics Trust a member of the Association Kangourou sans Frontières supported by SolutionsThese are polished solutions and do not illustrate the process of failed ideas and rough work by which candidates may arrive at their own solutions. It is not intended that these solutions should be thought of as the ‘best’ possible solutions and the ideas of readers may be equally meritorious. Enquiries about the Andrew Jobbings Senior Kangaroo should be sent to: UK Mathematics Trust, School of Mathematics, University of Leeds, Leeds LS2 9JT T0113 365 1121 challenges@ukmt.org.uk www.ukmt.org.uk

Andrew Jobbings Senior Kangaroo 2021 Solutions1.Adil was born in 2015 . His younger sister Bav was born in 2018 . What is the minimum number of days by which Adil is older than Bav? Solution 732 The closest the birth dates can be are 31st December 2015 and 1st January 2018. Since 2016 was a leap year, the number of days these are apart is 1+ 366 +365 =732 . 2. The total � is obtained as the sum of the integers from 2006 to 2036 inclusive. What is the sum of all the prime factors of �? Solution 121 We can write the sum as (2021 − 15 ) + ( 2021 − 14 ) + . . .+ ( 2021 − 1) + 2021 + ( 2021 + 1) + . . . + ( 2021 +14 ) + ( 2021 +15 )= 31 × 2021 = 31 × 43 × 47 . The sum of all the prime factors of �is 31 +43 +47 =121 . 3. How many of the figures shown can be drawn with one continuous line without drawing a segment twice? Solution 003 All are possible except for the second figure. Each of the small lines must either start or end the continuous line. But then only one of the semicircles in the second figure can be included. 4. On each side of a right-angled triangle, a semicircle is drawn with that side as a diameter. The areas of the three semicircles are � 2 ,3�and 180 where � 2 and 3�are both less than 180. What is the area of the smallest semicircle ? Solution 036 The areas of a square of side 2� and its inscribed circle are in the ratio 4�2 :�� 2 = 4 : �. Therefore, the area of the square on each side of the triangle and the area of the corresponding semicircle are in the ratio 4 : � 2 = 8 : �. This is independent of side length and we therefore may use the Pythagoras’ equation on the areas of the semicircles. This gives �2 + 3� = 180 . The solutions to this equation are �= − 15 (which is impossible since this would give an area of 3× − 15=−45 ) and �= 12 , meaning the required area is 36. © 2021 UK Mathematics Trust www.ukmt.org.uk 2

Andrew Jobbings Senior Kangaroo 2021 Solutions5. �=p ( 2021 +2021 ) + (2021 −2021 ) + (2021 ×2021 ) + (2021 ÷2021 ). What is the largest prime factor of T? Solution 337 Let �= 2021 . Then �=√ 2 � + �2 + 1= �+ 1= 2022 . Since 2022=2× 3× 337 , the largest prime factor of �is 337 . 6.Into how many regions do the �-axis and the graphs of � = 2− �2 and � = �2 − 1split the plane? Solution 010 As can be seen in the diagram the two graphs and the �-axis split the plane into ten regions. 7. Five cards have the numbers 101,102 ,103 ,104 and105 on their fronts. On the reverse, each card has one of five different positive integers: �,�,�,� and � respectively. We know that �= � � ,� + � = �and �− �= �. Frankie picks up the card which has the largest integer on its reverse. What number is on the front of Frankie’s card? Solution 103 Since �− �= �we have �= �+ �. Therefore, � > �and� > � . Since �= � � we have � > � and � > � , for if either of �,� equalled 1then �would equal the other. Therefore, � > � > �,� > � > � and� > � . The largest integer is �and the correct answer is 103. © 2021 UK Mathematics Trust www.ukmt.org.uk 3������ ������ ������ = ������2 − 1 ������ = 2− ������2 101 102 103 104 105

Andrew Jobbings Senior Kangaroo 2021 Solutions8.The geometric mean of a set of �positive numbers is defined as the �-th root of the product of those numbers. Yasmeen writes down a set of four numbers which have a geometric mean of 2048. Zak writes down a set of four numbers which have a geometric mean of 8. What is the geometric mean of the combined set of the eight numbers written by Yasmeen and Zak? Solution 128 Let Yasmeen’s numbers be � 1, � 2, � 3 and � 4; and Zak’s be � 1, � 2, � 3 and � 4. We are given that 4 √ � 1� 2� 3� 4 = 2048 and4 √ � 1 � 2 � 3 � 4 = 8. Therefore 4 √ � 1� 2� 3� 4� 1 � 2 � 3 � 4 = 4 √ � 1� 2� 3� 4 × 4 √ � 1 � 2 � 3 � 4 = 2048 ×8= 211 ×23 = 214 . We now take the square-root of both sides getting 8 √ � 1� 2� 3� 4� 1� 2� 3� 4 = 27 = 128 . 9. In the figure shown there are three concentric circles and two perpen- dicular diameters. The three shaded regions have equal area. The radius of the small circle is 2. The product of the three radii is �. What is the value of �2 ? Solution 384 We note that the central circle, the inner hoop and the outer hoop are also equal in area. Let the radii of the middle circle and outer circle be �and �respectively. The area of the central circle is 4� . The area of the inner hoop is ��2 − 4� . Therefore, ��2 − 4� = 4� . This simplifies to �2 = 8. The area of the outer hoop is � �2 − �� 2 = � � 2 − 8� . Therefore, � �2 − 8� = 4� . This simplifies to �2 = 12 . The product of the radii, �, is 2× �× �. Therefore, �2 = 4× �2 × �2 = 4× 8× 12 =384 . 10. A dealer bought two cars. He sold the first one for 40% more than he paid for it and the second one for 60% more than he paid for it. The total sum he received for the two cars was 54% more that the total sum he paid for them. When written in its lowest terms, the ratio of the prices the dealer paid for the first and the second car was � :�. What is the value of 11�+ 13 �? Solution 124 Let the prices the dealer paid for the first and second cars be � and � respectively. He sells the cars for 1.4 � and 1.6 � . Since he received 54% more than he paid, we have 1.54 (� +�)= 1.4 � +1.6 � . Therefore, 0.14 �=0.06 �which gives us the ratio �:� =6 : 14 =3 : 7 . Therefore �= 3and �= 7and 11�+ 13 �= 11 ×3+ 13 ×7= 33 +91 =124 . © 2021 UK Mathematics Trust www.ukmt.org.uk 4

Andrew Jobbings Senior Kangaroo 2021 Solutions11. Billie has a die with the numbers 1, 2, 3, 4, 5 and 6 on its six faces. Niles has a die which has the numbers 4, 4, 4, 5, 5 and 5 on its six faces.When Billie and Niles roll their dice the one with the larger number wins. If the two numbers are equal it is a draw. The probability that Niles wins, when written as a fraction in its lowest terms, is � �. What is the value of 7� + 11 �? Solution 181 Niles rolls a 4with probability 1 2 , and then wins if Billie rolls a 1, 2 or 3, with probability 1 2 . Niles rolls a 5with probability 1 2, and then wins if Billie rolls a 1,2,3or 4, with probability 2 3. Therefore, the probability of Niles winning is 1 2 × 1 2 + 1 2 × 2 3 = 1 4 + 1 3 = 3 12 + 4 12 = 7 12 . Therefore �= 7, � = 12 and 7� + 11 �= 7× 7+ 11 ×12 =181 . 12. There are 2021 balls in a crate. The balls are numbered from 1 to 2021 . Erica works out the digit sum for each ball. For example, the digit sum of 2021 is 5, since 2 + 0+ 2+ 1= 5. Erica notes that balls with equal digit sums have the same colour and balls with different digit sums have different colours. How many different colours of balls are there in the crate? Solution 028 The largest possible digit sum is that of 1999 which is 28 . The smallest is that of 1, which is 1. Each of the 9s in 1999 can be replaced by any of 0,1,. . . ,8. So all digit sums between 1and 28 can be achieved. Therefore there are 28different digit sums and colours. 13. A multiplication table of the numbers 1 to 10 is shown. What is the sum of all the odd products in the complete table? Solution 625 The odd numbers in the table are all the products of two of the odd numbers 1,3 ,5, 7,9. So their sum is (1 + 3+ 5+ 7+ 9) ( 1+ 3+ 5+ 7+ 9) = 25 2 = 625 . © 2021 UK Mathematics Trust www.ukmt.org.uk 5 10 10 20 30 · · · 100 . . . . . . . . . 2 2 4 6 · · · 20 1 1 2 3 · · · 10 × 1 2 3 · · · 10

Andrew Jobbings Senior Kangaroo 2021 Solutions14.The graph of ( � 2 + �2 − 2�)2 = 2(� 2 + �2 )2 meets the �-axis in � different places and meets the �-axis in �different places. What is the value of 100�+ 100 �? Solution 400 The graph of (� 2 + �2 − 2�)2 = 2(� 2 + �2 )2 meets the �-axis when �= 0. i.e. (� 2 − 2�)2 = 2(� 2 )2 i.e. �4 − 4�3 + 4�2 = 2�4 i.e. 0= �4 + 4�3 − 4�2 i.e. �= 0or 0= �2 + 4� − 4 i.e. �= 0or �= −2 ± 2√ 2 . Therefore there are three points where the graph meets the �-axis, so �= 3. The graph of (� 2 + �2 − 2�)2 = 2(� 2 + �2 )2 meets the �-axis when �= 0, i.e. (� 2 )2 = 2(� 2 )2 i.e. �4 = 2�4 , with only one solution of �= 0. Therefore there is only one point where the graph meets the �-axis, so �= 1. The value of 100�+ 100 �is 100 ×3+ 100 ×1= 400 . 15. Which is the lowest numbered statement which is true? Statement 201: "Statement 203 is true". Statement 202: "Statement 201 is true". Statement 203: "Statement 206 is false". Statement 204: "Statement 202 is false". Statement 205: "None of the statements 201, 202, 203 or 204 are true". Statement 206: "1 + 1 = 2". Solution 204 Statement 206 is clearly true, so Statement 203 must be false. In turn Statement 201 must be false and therefore Statement 202must also be false. Statement 204 is therefore true. Therefore, Statement 204 is the lowest numbered true statement. ©2021 UK Mathematics Trust www.ukmt.org.uk 6

Andrew Jobbings Senior Kangaroo 2021 Solutions16.A polygon is said to be friendly if it is regular and it also has angles that when measured in degrees are either integers or half-integers (i.e. have a decimal part of exactly 0.5). How many different friendly polygons are there? Solution 028 The interior angle will be an integer or half-integer precisely when the exterior angle is an integer or a half-integer respectively. A regular polygon of �sides has exterior angle 360 � . This is either an integer or a half-integer if, and only if, 720 � is an integer. Therefore, � must be a factor of 720 . The factors of 720 are 720 ,360 ,240 ,180 ,144 ,120 ,90 ,80 ,72 ,60 ,48 ,45 ,40 , 36 ,30 ,24 ,20 ,18 ,16 ,15 ,12 ,10 ,9,8,6,5,4 and 3 (ignoring 2 and 1 neither of which is a valid number of sides for a polygon). There are 28 numbers in this list, so there are 28 different friendly polygons. 17. Find the value of �, given that the numbers �and �are defined as: � =202 1 + 20 21 +2021 ; � is the remainder when �is divided by 1000. Solution 354 The term 20 21 is divisible by 10 21 and therefore is divisible by 1000 , meaning it will not contribute to the remainder on division by 1000. The term 2021 =221 =210 ×210 ×2= 1024 ×1024 ×2= (1000 +24 ) × ( 1000 +24 ) × 2. When multiplied out, all terms involving a 1000can be ignored. So this contributes 24×24 ×2= 1152 , or equivalently 152, to the remainder. Hence the total remainder is �=202 1 + 152 =354 . © 2021 UK Mathematics Trust www.ukmt.org.uk 7

Andrew Jobbings Senior Kangaroo 2021 Solutions18.Next year, 2022 , has the property that it may be written using at most two different digits, namely 2and 0. How many such years will there be between 1and 9999 inclusive? Solution 927 All of the 99years 1to 99 may be written using at most two different digits. Three-digit years (i.e. those between 100 and 999 inclusive) can be written using only one different digit if they are of the form ‘aaa’, for example 111or222 . There are 9such years. Three-digit years can be written using exactly two different digits � and � if they have any of the three following forms: ‘aab’, ‘aba’ or ‘abb’, where a,b are distinct digits and � ≠ 0. There are 9 possibilities for � and for each there are then 9 possibilities for �. Therefore there are 9× 9× 3= 243 such years. Four-digit years (i.e. those between 1000 and 9999 inclusive) can be written using only one different digit if they are of the form ‘aaaa’, for example 3333 or 4444 . There are 9such years. Four-digit years can be written using exactly two different digits � and � if they have any of the seven following forms: ‘aaab’, ‘aaba’, ‘abaa’, ‘aabb’, ‘abab’, ‘abba’ or ‘abbb’, where a,b are distinct digits and �≠ 0. There are 9 possibilities for � and for each there are then 9 possibilities for �. Therefore there are 9× 9× 7= 567 such years. Therefore, the total number of such years is 99+9+ 243 +9+ 567 =927 . 19. The function �(� ) is defined as �(� ) = � − 1 � + 1 . The equation �(� 2 ) × �(� ) = 0.72 has two solutions �and �, where � > �. What is the value of 19�+ 7� ? Solution 134 We are required to solve � 2 − 1 � 2 + 1 × � − 1 � + 1 = 18 25 , which simplifies to ( � − 1) ( �− 1) � 2 + 1 = 18 25 . This leads to the quadratic equation 7�2 − 50 � + 7= 0which factorises as (7� − 1) ( �− 7)= 0. Therefore, �= 7and �= 1 7 and 19�+ 7� = 19 ×7+ 7× 1 7 = 133 +1= 134 . © 2021 UK Mathematics Trust www.ukmt.org.uk 8

Andrew Jobbings Senior Kangaroo 2021 Solutions20.Each cell in this cross-number can be filled with a non-zero digit such that all of the conditions in the clues are satisfied. The digits used are not necessarily distinct. What is the answer to 2 DOWN? Solution 397 The only two-digit � � with � = �+ 4are 77 = 7× 11 and 21 = 3× 7. The latter of these is not possible, since 3 ACROSS cannot begin with a 1as it is twice a three-digit number. So the answer to 1 DOWN is 77. We now know the first digit of 1 ACROSS is 7, meaning that as primes 71 ,73 and 79 are possible. Of these, the only one that can be expressed as the sum of two squares is 73 = 64 +9. The clue for 4 DOWN tells us that 5× 4 DOWN = 3× 5 ACROSS. Hence 5 ACROSS is a multiple of 5and 4 DOWN is a multiple of 3. The first of these shows that the bottom right entry is 5(since 0is not allowed); and the second then shows that 4 DOWN is a multiple of 15 . By the clue for 3 ACROSS, the first digit of 4 DOWN is even. It cannot be 0,2,6, or 8because 4 DOWN is a multiple of 3. Therefore 4 DOWN = 45and 5 ACROSS = 75. Finally, given the first digits of 2 DOWN and 3 ACROSS and the clue for 3 ACROSS, we see that the middle digit must be at least 5and we then can check quickly that 9is the only possible value. Hence 397is the solution. © 2021 UK Mathematics Trust www.ukmt.org.uk 91 2 3 4 5 ACROSS 1. A prime which is the sum of two squares 3. Twice the answer to 2 DOWN DOWN 1.������× ������, where ������,������ are prime and������= ������+ 4 4. 60% of 5 ACROSS