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Senior Kangaroo © 2019 UK Mathematics Trust a member of the Association Kangourou sans Frontières 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 Senior Kangaroo should be sent to: UK Mathematics Trust, School of Mathematics, University of Leeds, Leeds LS2 9JT T0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk

Senior Kangaroo 2019 Solutions1. What is the sum of all the factors of 144? Solution 403The factor pairs of 144 are 1,144 ;2,72 ;3,48 ;4,36 ;6,24 ;8,18 ;9,16 and 12 (squared). Their sum is 403. 2. When I noticed that 24 = 42, I tried to find other pairs of numbers with this property. Trying 2and 16, I realised that 216 is larger than 162 . How many times larger is 216 ? Solution 256 2 16 16 2 = 2 16 ( 2 4 ) 2 = 2 16 2 8 = 28 = 256 3. The two diagonals of a quadrilateral are perpendicular. The lengths of the diagonals are 14and 30. What is the area of the quadrilateral? Solution 210 Label the quadrilateral ABC Dand let AC=14 and B D=30 . Let Mbe the intersection of ACand B D. Let AM =a, B M =b,C M =cand D M =d. Then the sum of the areas is 1 2 × ( ab +ad +cb +cd ) = 1 2 × ( a + c) × ( b + d) = 1 2 × 14 ×30 = 210 . 4. The integer nsatisfies the inequality n+ (n + 1) + (n + 2) + · · · +(n + 20 )> 2019 . What is the minimum possible value of n? Solution 087 We solve the inequality n+ (n + 1) + (n + 2) + ... +(n + 20 )> 2019 Therefore, 21n+ 210 >2019 , i.e. 21n> 1809 and7n > 603 . Therefore, n> 603 7 = 86 .1 . . . , so nmust be at least 87. © 2019 UK Mathematics Trust www.ukmt.org.uk 2

Senior Kangaroo 2019 Solutions5.Identical regular pentagons are arranged in a ring. The partially completed ring is shown in the diagram. Each of the regular pentagons has a perimeter of 65 . The regular polygon formed as the inner boundary of the ring has a perimeter of P. What is the value of P? Solution 130 Let the regular N-gon at the centre of the figure have interior angles of size xdegrees. The interior angle of a pentagon is 108 ◦. By angles at a point we have x+ 2× 108 = 360 , so x = 144 . The exterior angle of the N-gon is 180 − 144 = 36 . Therefore, the N-gon has 360 36 = 10 sides. As each side has length 65 5 = 13 , the perimeter is 10×13 =130 . 6. For natural numbers a and b we are given that 2019 = a2 − b2 . It is known that a < 1000 . What is the value of a? Solution 338 We can write 2019 = ( a + b)( a − b) . The integers a + b and a− b must be a factor pair of 2019 . There are two such factor pairs: 2019 ,1 and 673 ,3. These yield ( a ,b ) = ( 1010 ,1009 ) and ( a ,b ) = ( 338 ,335 ) respectively. As the answer must be at most 999 , we conclude that a = 338 . 7. How many positive integers n exist such that both n + 1 3 and 3n + 1 are three-digit integers? Solution 012 For n + 1 3 to be a three-figure integer we require 99

Senior Kangaroo 2019 Solutions8. The function J(x ) is defined by: J (x ) =         4 + x forx≤ − 2, − x for−2 < x≤ 0, x forx> 0. How many distinct real solutions has the equation J(J (J (x ))) =0? Solution 004 The only solutions to J(x ) = 0are x= 0,− 4. Since J(0 ) = 0, both will also be solutions of J(J (J (x ))) =0.Any solution to J (x ) = − 4will also be a solution to J(J (x )) = 0. The only solution to J(x ) = − 4 is x= −8. Since J(x ) = 0, x = −8 is also a solution of J(J (J (x ))) =0. Any solution to J (x ) = − 8 will also be a solution to J(J (J (x ))) = 0. The only solution to J(x ) = −8 is x= −12 . Therefore, there are four distinct solutions, x= 0,− 4,− 8 and −12 . 9. What is the smallest three-digit number K which can be written as K =ab + ba , where both aand bare one-digit positive integers? Solution 100 As the problem is symmetrical in a,b we assume a≤ bwithout loss of generality. If a = 1then the maximum value of ab + ba is 19 + 91 = 1+ 9 = 10 . This is not a three-digit number, so cannot be a value of K. If a = 2then possible values for K include 29 + 92 = 512 + 81 = 593 ,28+ 82 = 256 + 64 = 320 , 27 + 72 = 128 +49 =177 and26 + 62 = 64 +36 =100 . As 100 can be attained then 100is the smallest three-digit number K. © 2019 UK Mathematics Trust www.ukmt.org.uk 4 = 4 = 2 24 4 xy

Senior Kangaroo 2019 Solutions10. What is the value of q 13 +p 28 +√ 281 ×q 13 −p 28 +√ 281 ×p 141 +√ 281 ? Solution 140 q 13 +p 28 +√ 281 ×q 13 −p 28 +√ 281 ×p 141 +√ 281 = r 13 2 −  p 28 +√ 281  2 × p 141 +√ 281 = r 169 − 28 +√ 281  × p 141 +√ 281 = p 141 −√ 281 ×p 141 +√ 281 =r  141 −√ 281  ×  141 +√ 281  = √ 141 2 − 281 =√ 141 2 − 282 +1= q ( 141 −1)2 = √ 140 2 = 140 11.In the triangle ABC the points M and N lie on the side ABsuch that AN=AC and B M =BC . We know that ∠M C N =43 °. Find the size in degrees of ∠AC B . Solution 094 Let ∠AC M =x° and ∠BC N =y°. Using the base angles property of isosceles triangles AC N and BC M , we have ∠ AN C = 43 +x and ∠B M C = 43 +y . In triangle C M N,43 +( 43 +x) + ( 43 +y) = 180 . Therefore, ∠AC B =x+ 43 +y= 94 . © 2019 UK Mathematics Trust www.ukmt.org.uk 5 A BC M N43 ◦ A BC M N43 ◦ x ◦ y◦

Senior Kangaroo 2019 Solutions12. What is the value of A2 + B3 + C5 , given that: A = 3 p 16 √ 2 B = p 9 3 √ 9 C =[( 5 √ 2 )2 ]2 Solution 105 A 2 =  3 p 16 ×√ 2  2 =  2 4 × 21 2  2 3 =  2 9 2  2 3 = 218 6 = 23 = 8 B 3 = p 9 × 3 √ 9 3 =  9 × 91 3  3 2 =  9 4 3  3 2 = 912 6 = 92 = 81 C 5 =   5 √ 2 2  2 5 =  2 1 5  2× 2× 5 = 220 5 = 24 = 16 A 2 + B3 + C5 = 8+ 81 +16 =105 13.The real numbers aand b, where a > b , are solutions to the equation 32x − 10 ×3x+ 1 + 81 = 0 . What is the value of 20a2 + 18 b2 ? Solution 198 In the equation 32 x − 10 × 3x+ 1 + 81 = 0, replace 3x with y. The equation becomes y2 − 10 × 3× y+ 81 = 0. This factorises as ( y − 3)( y− 27 ) = 0with solutions y = 3,27 . This means 3x = 3or 3x = 27 . The x-values are 1,3respectively, so a = 3and b= 1. The value of 20 a2 + 18 b2 = 20 ×9+ 18 ×1= 198 . 14. A number N is the product of three distinct primes. How many distinct factors does N 5 have? Solution 216 Let the three distinct prime factors of N be p,q and r. Therefore, N 5 = p5 × q5 × r5 . Each factor of N 5 may be written as pa × qb × rc , where a ,b, c ∈ { 0,1,2,3,4,5}. Since there are 6 choices for the value of each of a,b, c there are 6× 6× 6= 216 distinct factors of N5 . © 2019 UK Mathematics Trust www.ukmt.org.uk 6

Senior Kangaroo 2019 Solutions15.Five Bunchkins sit in a horizontal field. No three of the Bunchkins are sitting in a straight line. Each Bunchkin knows the four distances between her and each of the others. Each Bunchkin calculates and then announces the total of these distances. These totals are 17, 43, 56, 66 and 76. A straight line is painted joining each pair of Bunchkins. What is the total length of paint required? Solution 129 Each line’s length will be announced twice; once by each of the two Bunchkins at its ends. By adding up the total of the numbers announced we will therefore include the length of each line exactly twice. The total length of paint required is 1 2 × ( 17 +43 +56 +66 +76 ) = 258 2 = 129 . 16. The real numbers xand ysatisfy the equations: x y − x= 180 andy+ xy = 208 . Let the two solutions be (x 1, y 1) and (x 2, y 2) . What is the value of x 1 + 10 y 1 + x 2 + 10 y 2? Solution 317 Subtracting the two equations yields y + x= 28 . Substituting y= 28 − x into xy − x= 180 leads to the quadratic equation 0 = x2 − 27 x + 180 . This has solutions 12 ,15 . The solution sets are therefore ( 15 ,13 ) and ( 12 ,16 ) . The value of x 1 + 10 y 1 + x 2 + 10 y 2 is 15 +10 ×13 +12 +10 ×16 =317 . 17. In triangle ABC ,∠ B AC is 120 °. The length of AB is 123 . The point M is the midpoint of side BC. The line segments ABand AM are perpendicular. What is the length of side AC? Solution 246 Extend the line B A . Draw a line through C, parallel to M A , meeting the extended line B A at point N. By the intercept theorem, B A =AN = 123 , because B M =M C . In triangle N AC ,cos 60 =1 2 = 123 AC . Therefore, AC = 246 . © 2019 UK Mathematics Trust www.ukmt.org.uk 7A B C M N60◦ 30 ◦

Senior Kangaroo 2019 Solutions18.An integer is said to be chunky if it consists only of non-zero digits by which it is divisible when written in base 10. For example, the number 936 is Chunky since it is divisible by 9, 3 and 6. How many chunky integers are there between 13 and 113? Solution 014 For a two-digit number N=“ab ”we may write N=10 a+ b. If N is Chunky, then N will be divisible by aand therefore b = N − 10 awill also be divisible by a. It is therefore sufficient to check only those two-digit numbers which have a units digit divisible by their tens digit. Following checking, 15 ,22 ,24 ,33 ,36 ,44 ,48 ,55 ,66 ,77 ,88 and 99 are the only two-digit Chunky numbers (excluding 11 and 12 , which are not under consideration). Of those three-digit numbers under consideration, only 111 and 112 are Chunky. The answer is 14 . 19. The square ABC D has sides of length 105 . The point M is the midpoint of side BC . The point N is the midpoint of B M . The lines B D and AM meet at the point P. The lines B Dand ANmeet at the point Q. What is the area of triangle APQ? Solution 735 Let V be the centre of the square ABC D . Let W be the intersection between B Dand the line through Nparallel to AB. Triangles AP B and M PV are similar, with BP :PV =AB : MV =1 2 . Therefore, BP=2 3 × BV =2 3 × 1 2 × 105 √ 2 = 35 √ 2 . Similarly, triangles AQ B and N QW are similar, with BQ :QW = AB :N W =1 4 . Therefore, BQ =4 5 × BW =4 5 × 1 4 × 105 √ 2 = 21 √ 2 . The area of APQ is 1 2 × Q P ×V A =1 2 ×  35 √ 2 − 21 √ 2  × 1 2 × 105 √ 2 = 1 2 × 14 √ 2 × 1 2 × 105 √ 2 = 735 . © 2019 UK Mathematics Trust www.ukmt.org.uk 8 A B C D N M P Q V W

Senior Kangaroo 2019 Solutions20.Each square in this cross-number can be filled with a non-zero digit such that all of the conditions in the clues are fulfilled. The digits used are not necessarily distinct. What is the answer to 3 ACROSS? ACROSS DOWN 1. A composite factor of 1001 1. One more than a prime, one less than a prime 3. Not a palindrome 2. A multiple of 9 5. pq3 where p, q prime and p, q 4.p3 q using the same p, q as 5 ACROSS Solution 295 1 Across may be either 77 or 91 . The only possibility for 1 Down with 7or 9as its first digit is 72 . So 1 Across is 77and 1 Down is 72. In the clues for 5 Across and 4 Down we see that p, q must be 2,3in some order, since if any larger prime were used then pq 3 and q p 3 would not both be two-digit. Therefore, 5 Across and 4 Down are 3× 23 = 24 and 2× 33 = 54 in some order. We know that 3 Across is not a palindrome (so may not end in a 2). Therefore, 5 Across is 24and 4 Down is 54. The only three-digit multiples of 9beginning with a 7are 702 and 792 . As every digit in the completed crossnumber must be non-zero we have 2 Down is 792and 3 Across is 295. © 2019 UK Mathematics Trust www.ukmt.org.uk 912 34 5