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Senior Kangaroo 2018 Organised by the United Kingdom Mathematics Trust a member of the Association Kangourou sans Frontières SolutionsThese 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. ©UK Mathematics Trust Enquiries about the Senior Kangaroo should be sent to: Senior Kangaroo, UK Mathematics Trust, School of Mathematics, University of Leeds, Leeds LS2 9JT T 0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk

Senior Kangaroo 2018 Solutions1.My age is a two-digit number that is a power of 5. My cousin’s age is a two-digit number that is a power of 2. The sum of the digits of our ages is an odd number. What is the product of the digits of our ages? Solution 240 My age must be 25. My cousin’s age is a two-digit power of two, so is 16, 32 or 64. The sums of the digits of our ages would then be 14, 12 and 17 respectively. Since this sum must be odd my cousin’s age must be 64. Therefore the product of the digits of our ages is 2× 5× 6× 4= 240 . 2. Let Kbe the largest integer for which n200

Senior Kangaroo 2018 Solutions5.Rachel and Steven play games of chess. If either wins two consecutive games s/he is declared the champion. The probability that Rachel will win any given game is 0.6. The probability that Steven will win any given game is 0.3. There is a 0.1 probability that any given game is drawn. The probability that neither is the champion after at most three games is P. Find the value of 1000P. Solution 343 We use R to denote the event ‘Rachel wins a game’; Sto denote ‘Steven wins a game’; and D to denote ‘a game is drawn’. We use R to denote the event ‘Rachel does not win a game’, with S defined similarly. •Event R,D has probability 0.6 × 0.1 = 0.06 . • Event R,S , S has probability 0.6 × 0.3 × 0.7 = 0.126 . • Event D,R , R has probability 0.1 × 0.6 × 0.4 = 0.024 . • Event D,S , S has probability 0.1 × 0.3 × 0.7 = 0.021 . • Event S,D has probability 0.3 × 0.1 = 0.03 . • Event S,R , R has probability 0.3 × 0.6 × 0.4 = 0.072 . • Event D,D has probability 0.1 × 0.1 = 0.01 . Therefore P= 0.06 +0.126 +0.024 +0.021 +0.03 +0.072 +0.01 = 0.343 ; hence 1000 P= 343 . ©UK Mathematics Trust www.ukmt.org.uk 3

Senior Kangaroo 2018 Solutions6. The line segments PQ RSandW XY S intersect circle C 1 at points P,Q ,W and X.The line segments intersect circle C 2 at points Q,R,X and Y. The lengths Q R ,RS and XY are7, 9 and 18respectively. The length W Xis six times the length Y S. What is the sum of the lengths of PSand W S? Solution 150 Use the intersecting chords theorem on each circle. Let abe the length of SY . Then, in circle Q XY R ,a (a + 18 )= 9(9+ 7)giving solutions of a = 6,− 24 . Since a > 0we conclude a= 6. In circle PW X Q ,24 (24 +6× 6) = 16 (16 +z). The solution is z= 74 . Therefore PS+W S =74 +7+ 9+ 36 +18 +6= 150 . 7. The volume of a cube in cubic metres and its surface area in square metres is numerically equal to four-thirds of the sum of the lengths of its edges in metres. What is the total volume in cubic metres of twenty-seven such cubes? Solution 216 Let each of the twelve edges of the cube have length xmetres. Then x 3 + 6x2 = 4 3 × 12 x. This simplifies to x3 + 6x2 − 16 x = 0or x(x − 2)( x+ 8)which has solutions x = − 8,0,2. However x must be positive and so x= 2. Then 27such cubes have a volume of 27×23 = 27 ×8= 216 . © UK Mathematics Trust www.ukmt.org.uk 4

Senior Kangaroo 2018 Solutions8.An integer xsatisfies the inequality x 2 ≤ 729 ≤ − x3 .P and Q are possible values of x. What is the maximum possible value of 10(P −Q)? Solution 180 We observe that 729 = 36. First consider x 2 ≤ 729 . This has solution −27 ≤ x≤ 27 . Now consider 729≤ −x3 . This may be rearranged to give x3 ≤ − 729 with solution x≤ − 9. These inequalities are simultaneously satisfied when −27 ≤ x≤ − 9. The maximum value of P −Q is therefore −9 − (− 27)= 18 . So the answer to the question is 10×18 =180 . 9. The two science classes 7A and 7B each consist of a number of boys and a number of girls. Each class has exactly 30 students. The girls in 7A have a mean score of 48. The overall mean across both classes is 60. The mean score across all the girls of both classes is also 60. The 5 girls in 7B have a mean score that is double that of the 15 boys in 7A. The mean score of the boys in 7B is µ. What is the value of 10µ? Solution 672 In 7A there are 15boys and 15girls; in 7B there are 25boys and 5girls. All the girls have a total score of 20×60 =1200 . Girls in 7A have a total score of 15×48 =720 . Hence girls in 7B have a total score of 1200−720 =480 and a mean of 96. All pupils have a total score of 60×60 =3600 . Hence the 40boys have a total score of 2400and a mean of 60. The 15boys in 7A have a mean of 96 2 = 48 and total score of 720. Hence the 25boys in 7B have a total score of 2400−720 =1680 and a mean of 1680 25 . Hence the required number is 1680 25 × 10 =672 . © UK Mathematics Trust www.ukmt.org.uk 5

Senior Kangaroo 2018 Solutions10.The function SPF (n ) denotes the sum of the prime factors of n, where the prime factors are not necessarily distinct. For example, 120 = 23 × 3×5, so SPF ( 120 )= 2+2+2+3+5= 14 . Find the value of SPF(2 22 −4). Solution 100 Write 222 −4as a product of primes: 222 −4= 4× ( 220 −1) = 22 × ( 210 −1)( 210 +1) = 22 × ( 25 − 1)( 25 + 1)( 1024 +1) = 22 × ( 32 −1)( 32 +1)( 1025 ) = 22 × 31 ×33 ×5× 205 = 22 × 31 ×3× 11 ×5× 5× 41 = 22 × 3× 52 × 11 ×31 ×41 . Therefore S P F(2 22 −4) = 2+ 2+ 3+ 5+ 5+ 11 +31 +41 =100 . 11. A sequence U 1, U 2, U 3, . . . is defined as follows: • U 1= 2; • ifU nis prime then U n+ 1 is the smallest positive integer not yet in the sequence; • ifU nis not prime then U n+ 1 is the smallest prime not yet in the sequence. The integer kis the smallest such that U k+ 1 − U k> 10 . What is the value of k× U k? Solution 270 The sequence is 2,1,3,4,5,6,7,8,11 ,9,13 ,10 ,17 ,12 ,19 ,14 ,23 ,15 ,29 , . . . so that U 18 = 15 and U 19 = 29 . These are the first two consecutive terms with a difference greater than 10. Therefore k= 18 and k× U k= 18 ×15 =270 . © UK Mathematics Trust www.ukmt.org.uk 6

Senior Kangaroo 2018 Solutions12.The diagram shows a 16 metre by 16 metre wall. Three grey squares are painted on the wall as shown. The two smaller grey squares are equal in size and each makes an angle of 45 °with the edge of the wall. The grey squares cover a total area of Bmetres squared. What is the value of B? Solution 128 The wall has a width of 16 metres so the diagonal of each smaller grey square is 8 metres. Let the side-length of each smaller grey square be xmetres. Then, by Pythagoras’ Theorem, x 2 + x2 = 82, giving x= √ 32 . Therefore each smaller grey square has an area of 32 m2 . The side-length of the larger grey square is equal to the length of the diagonal of one of the smaller grey squares. Therefore the larger grey square has area 82 = 64 m 2 . Hence the total area covered by the grey squares, B, is 32+32 +64 =128 m 2 . 13. A nine-digit number is odd. The sum of its digits is 10. The product of the digits of the number is non-zero. The number is divisible by seven. When rounded to three significant figures, how many millions is the number equal to? Solution 112 None of the digits in the number may be zero since we know that their product is non-zero. As they sum to 10, we know the digits must be eight 1s and one 2, in some order. It remains to check the eight possible odd integers for divisibility by seven. Of those, only 112 111 111 has zero remainder when divided by seven, and to three significant figure this number is 112 million. Therefore the answer is 112. © UK Mathematics Trust www.ukmt.org.uk 7

Senior Kangaroo 2018 Solutions14.A square ABC D has side 40 units. Point F is the midpoint of side AD . Point G lies on C F such that 3CG =2G F . What is the area of triangle BCG? Solution 320 Start by drawing a diagram. By Pythagoras’ Theorem, FC = 20 √ 5 and hence CG = 8 √ 5 . Now, sin ∠BCG =sin ∠C F D = 40 20 √ 5 = 2 √ 5 . Hence the area of triangle BCGis1 2 × 40 ×8√ 5 × 2 √ 5 = 320 . 15. In the sequence 20 ,18 ,2,20 , − 18 , . . . the first two terms a 1 and a 2 are 20 and 18 respectively. The third term is found by subtracting the second from the first, a3 = a 1 − a 2 . The fourth is the sum of the two preceding elements, a 4 = a 2 + a 3 . Then a 5 = a 3 − a 4 , a 6 = a 4 + a 5, and so on. What is the sum of the first 2018 terms of this sequence? Solution 038 The sequence is 20 ,18 ,2,20 , − 18 ,2,− 20 ,− 18 ,− 2,− 20 ,18 ,− 2,20 ,18 , . . . . This is periodic and will repeat every twelve terms. The sum of the first twelve terms is 0. Note also that 2018 = 12 ×168 +2. Therefore the first 2018 terms will consist of 168 cycles of the first twelve terms with zero sum, followed by a 2017 = 20 and a 2018 = 18 . Therefore the sum of the first 2018 terms is 168 ×0+ 20 +18 =38 . © UK Mathematics Trust www.ukmt.org.uk 8F G A BC D 20 40

Senior Kangaroo 2018 Solutions16.A right-angled triangle has sides of integer length. One of its sides has length 20. Toni writes down a list of all the different possible hypotenuses of such triangles. What is the sum of all the numbers in Toni’s list? Solution 227 Consider a hypotenuse of length 20 . Let the shortest side be of length a. For integer side-lengths we require 20 2 − a2 to be a square. We observe that of 20 2 − 12,20 2 − 22,20 2 − 32, . . . , 20 2 − 19 2 only 20 2 − 12 2 = 256 and 20 2− 16 2 = 144 are squares. Therefore a hypotenuse of length 20 is possible. Now we consider that one of the shorter two sides has length 20 . Let the hypotenuse and the other shorter side be of lengths hand brespectively. By Pythagoras’ Theorem 20 2= h2 − b2 , yielding ( h − b)( h + b) = 400 . We now consider factor pairs (m ,n ), m ≤nof 400 to find (h − b, h + b). • (m ,n ) = (1 ,400 )gives (h ,b ) = (200 .5 ,199 .5 ) which are non-integers. • (m ,n ) = (2 ,200 )gives (h ,b ) = (101 ,99 )with hypotenuse 101. • (m ,n ) = (4 ,100 )gives (h ,b ) = (52 ,48 )with hypotenuse 52. • (m ,n ) = (5 ,80 )gives (h ,b ) = (42 .5 ,37 .5 ) which are non-integers. • (m ,n ) = (8 ,50 )gives (h ,b ) = (29 ,21 )with hypotenuse 29. • (m ,n ) = (10 ,40 )gives (h ,b ) = (25 ,15 )with hypotenuse 25. • (m ,n ) = (16 ,25 )gives (h ,b ) = (20 .5 ,4 .5 ) which are non-integers. • (m ,n ) = ( 20 ,20 )gives (h ,b ) = ( 40 ,0)which is a degenerate case, that is it includes a side of zero length. Therefore the possible hypotenuses are 101,52 ,29 ,25 and 20with sum 227. 17. Sarah chooses two numbers aand bfrom the set {1,2,3, . . . , 26 }. The product ab is equal to the sum of the remaining 24 numbers. What is the difference between aand b? Solution 006 The sum of the numbers in the set is 1 2 × 26 × 27 = 351 . Numbers a and bwill satisfy the equation 351−a− b= ab . Therefore 352 =ab +a+ b+ 1= (a + 1)(b+ 1). We now search for a factor pair (a + 1,b + 1)of 352 with a,b ≤ 26 and a ≤ b . The only such pair (a + 1,b + 1)is (16 ,22 ). Therefore aand bare 15 and 21respectively and their difference is 6. © UK Mathematics Trust www.ukmt.org.uk 9

Senior Kangaroo 2018 Solutions18. How many zeros are there at the end of 2018! 30! ×11! ? Solution 493Each zero at the end of N!must be generated by a ( 2× 5) in its prime factorisation. In general N!= 2a × 5b× K . For all values of N,awill be greater than b. We therefore determine the value of bin each of 2018!,30! and 11!. In 2018! =1× 2× . . . ×2018 there are • 403 multiples of five; • 80 multiples of 52 = 25 , each contributing one additional five; • 16 multiples of 53 = 125 , each contributing one further additional five; and • 3multiples of 54 = 625 , each contributing one yet further additional five. Therefore, for 2018!,b = 403 +80 +16 +3= 502 . In 30! there are 6multiples of five and 1multiple of 25. So, for 30!,b = 6+ 1= 7. In 11! there are 2multiples of five. So, for 11!,b = 2. Therefore the power of five in the fraction 2018! 30! ×11! is 502 − 7− 2= 493 and so there are 493 zeros at the end of the number. 19. Shan solves the simultaneous equations xy = 15 and (2 x − y)4 = 1 where xand yare real numbers. She calculates z, the sum of the squares of all the y-values in her solutions. What is the value of z? Solution 122 From (2x − y)4 = 1we know (2x − y)2 = ± 1. Since any squared quantity must be non-negative, we know (2 x − y)2 = 1from which 2x − y= ±1. Consider the case 2x − y= 1. Multiplying by ygives 2xy − y2 = y , but xy = 15 and so 30 −y2 = y. Therefore y= −6,5 . Consider the case 2x − y= − 1. Multiplying by ygives 2xy − y2 = −y , but xy = 15 and so 30 −y2 = −y. Therefore y= −5,6 . Hence z= (− 6)2 + 52 + (− 5)2 + 62 = 122 . © UK Mathematics Trust www.ukmt.org.uk 10

Senior Kangaroo 2018 Solutions20. Determine the value of the integer ygiven that y= 3x2 and 2 x 5 = 1 1 − 2 3 + 1 4 − 5 6 − x Solution 147 Note first that 4− 5 6 − x= 24 −4x − 5 6 − x = 19 −4x 6 − x = F, say. Then, 3+ 1 F = 3+ 6 − x 19 −4x = 57 −12 x+ 6− x 19 −4x = 63 −13 x 19 −4x = G,say. Then, 1− 2 G = 1− 2 × ( 19 −4x) 63 −13 x = 63 −13 x− ( 38 −8x) 63 −13 x = 25 −5x 63 −13 x= H,say.So we get the equation 2x 5 = 1 H = 63 −13 x 25 −5x which simplifies to 2x2 − 23 x + 63 = 0. This has solutions x = 4.5and x = 7; the corresponding values of yare 60 .75 and 147 respectively. Therefore y= 147 , since we are told yis an integer. © UK Mathematics Trust www.ukmt.org.uk 11