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UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 1st December 2017 Organised by the United Kingdom Mathematics Trust SOLUTIONS 1. 132 Each circle immediately adjacent to the initial ‘4’ must contain the same integer, say, in order for the sum of those numbers at the end of each line segment to be the same. Those circles immediately adjacent to those with the integer must contain the integer 4 to preserve the sum of those numbers at the end of each line segment. Continuation of this pattern throughout the network eventually yields that the circle marked with a ‘7’ must contain the integer . Therefore . The completed network contains twelve ‘4’s and twelve ‘7’s with a total of .x x xx=7 12×4+12×7=132 2. 231 At the start of the race, the runners are in order 123. By the end of the race Primus and Tertius have exchanged places 11 times, so Tertius ends the race ahead of Primus. Also, since Secundus and Tertius have exchanged places 10 times, Secundus ends ahead of Tertius. So the result of the race is 231. 3. 221 In ascending order the list is 104, 113, 122, 131, 140, 203, 212, 221, 230, 302, 311, 320, 401, 410, 500. The median of this list is 221. 4. 216 Each of the squares has a side-length of 18 cm. Therefore each of the triangles has two sides of length 18 cm. Hence the triangles are isosceles. Let the angle contained by the two 18 cm sides of these triangle be . The interior angles of a square and a regular hexagon are and respectively. By considering angles at a point we have = 360. Therefore and the triangles are equilateral. All twelve outer edges of the figure are 18 cm in length. Therefore .x° 90°120° x+90+90+120x=60 P=12×18=216 5. 625 A BC F D E Draw a line through that is parallel to . Let be the intersection of that line with extended, as shown in the diagram.DABF BC NowEDC+CDF=EDC+ADE=90° ThereforeCDF=ADE=90°ŠECD AlsoDF C=DEA=90° Therefore triangles and are similar because they have the same set of angles. Because they must also be congruent.ADE C DF AD=CD Therefore the area of quadrilateral is equal to the area of rectangle . By the congruent triangles and we know that . Therefore the required area is .ABC D FDEB ADE C DF DF=DE=25 25×25=625

6. 373 Winnie begins with 2017 integers. There are multiples of three which are erased. This leaves integers. There are multiples of six which are reinstated. This leaves integers. The only multiples of twenty-seven that then remain in the list are those that are multiples of six. Winnie therefore erases all the multiples of fifty-four. There are multiples of fifty-four which are erased. This leaves integers. Of the 2017 integers she began with are now missing.2016÷3=672 2017Š672=1345 2016÷6=336 1345+336=1681 1998÷54=37 1681Š37=1644 2017Š1644=373 7. 300 70110 80 40 70110 8040 70110 80 40 70110 8040 There are essentially four different configurations as shown in the diagram. The perimeters of the third rectangle in these configurations are 220, 240, 300 and 140 respectively. Therefore the maximum possible perimeter of the third rectangle is 300. 8. 276 Let the angle that each hand makes with the vertical be degrees and let the current time be seconds after midday.x s In one complete hour the hour hand will turn . There are seconds in an hour so it takes seconds for the hour hand to turn one degree. Therefore .30°60×60=3600 3600÷30=120 s=120x In one complete hour the minute hand will turn . There are 3600 seconds in an hour so it takes seconds for the minute hand to turn . But the minute hand has turned clockwise through an angle of °. Therefore . Equating the two expressions we have obtained for we obtain the equation . The solution to this equation is . Therefore the number of seconds elapsed since midday is . The number of whole seconds remaining is .360° 3600÷360=10 1° (360Šx ) s=10 (360Šx ) s 120x=10 (360Šx ) x= 360 13 120× 360 13= 43200 13=3323 1 13 3600Š3324=276 9. 013 Robin could score a total of zero either by missing the target with all three arrows or if any two of his arrows hit adjacent regions. Robin could score totals of 3, 9, 15, 21, 27 or 33 if all three of his arrows hit regions 1, 3, 5, 7, 9 or 11 respectively. The only scores Robin can obtain from his three arrows hitting non-adjacent regions are ,,,, , . 1+1+5=71+5+5=11 3+3+7=13 3+7+7=17 9+9+11=29 9+11+11=31 Robin’s set of possible score is therefore {0, 3, 7, 9, 11, 13, 15, 17, 21, 27, 29, 31, 33}. Hence Robin can obtain 13 different possible scores. 10. 004 Suppose that there is a truther at . There must be two liars and one truther adjacent to . Let us suppose, without loss of generality, that is a truther and and are liars. Since is a truther and is adjacent to , then and are liars. This shows that there cannot be more than 4 truthers. A A BDE BAC F If we now suppose that and are both truthers, then each of the Bunchkins' statements fits the conditions. So 4 is the maximum possible number of Bunchkins.GH A B CD E FGH

11. 714 The difference between any two terms is either 4 or is a multiple of 4. So the term-to- term difference in the progression must be a divisor of 4. Since all the terms of the progression are integers the only feasible differences are 1, 2 and 4. If the term-to-term difference is 1 then the one-hundredth term will be .7+99×1=106 If the term-to-term difference is 2 then the one-hundredth term will be 7+99×2=205. If the term-to-term difference is 4 then the one-hundredth term will be 7+99×4=403. The sum of these numbers is 714. 12. 121 Take each shaded semi-annulus that is below the line and reflect it in the line then move it one centimetre to the left. A shaded semicircle of diameter 11cm is obtained. Therefore the whole shaded area is . Hence the value of is 121. 121 8 k 13. 002 The expression may be simplified to . k·a·n·r·o·o m·e The smallest possible numerator is .5×4×3×2×1×1=120 The largest possible denominator is .9×8=72 The smallest possible value of the expression, whether integer or not, is therefore . 120 72=1 2 3 We may obtain a value of 2 for the expression via . 6×4×3×2×1 9×8 Therefore the smallest possible integer value of the expression is 2. 14. 027 There are six subsets consisting of one element: {1}, {2}, {3}, {4}, {5}, {6}. There are eleven subsets consisting of two elements: {1, 2}, {1,3}, {1,4}, {1, 5}, {1, 6}; {2, 3}, {2, 5}; {3, 4}, {3, 5}: {4, 5}; {5, 6}. There are eight subsets consisting of three elements: {1, 2, 3}, {1, 2, 5}; {1, 3, 4}, {1, 3, 5}; {1, 4, 5}; {1, 5, 6}; {2, 3, 5}; {3, 4, 5}. There are two subsets consisting of four elements: {1, 2, 3, 5}; {1, 3, 4, 5}. Hence the answer is .6+11+8+2=27 15. 829 The possible answers for 1 across are 16, 25, 36, 49, 64 and 81. The possible answers for 2 down are 124, 215, 342, 511, 728, 999. By considering the last digit of 1 across (which must be the same as the first digit of 2 down) we see that the only possible pairs of answers for 1 across and 2 down are (25, 511), (49, 999) and (81, 124). The pair (49, 999) would leave no possible answer for 5 across, so may be disregarded. The pair (25, 511) gives an answer of 16 for 5 across and thence 36 for 4 down and 25 for 1 down. However these answers contradict the clue for 4 down so this case may be disregarded. (81, 124) gives an answer of 49 for 5 across and hence 99 for 4 down and 88 for 1 down. These answers satisfy all the conditions in the clues and therefore the answer to this Kangaroo question is 829. 16. 074 The curve is a circle of radius 5 centred at the origin.x 2+y 2 =25 The polygon has vertices at coordinates (0,5), (3,4), (4,3), (5,0), (4,–3), (3,–4), (0,–5), (–3, –4), (–4, –3), , (–4, 3), (–3, 4).P (Š5, 0) We may find the area of that part of in the upper-right quadrant by splitting it into two trapezia and a triangle as shown in the diagram. This area is P 5 5 4 4 3 3y Ox . (1 2(5+4)×3 )+ (1 2(4+3)×1 )+ (1 2 ×3×1 ) = 37 2 Therefore the area of is .P4× 37 2 =74

17. 447 Three-digit squares beginning with 2, 4, 5, 6 or 8 may be disregarded since on reversal these will be divisible by two or five. The residual three-digit squares are 100, 121, 144, 169, 196, 324, 361, 729, 784, 900 and 961. We may disregard 144, 324, 729 and 961 since these are divisible by three and will remain so on reversal. We may also disregard 169, 961, 100 and 121 since these each form a square on reversal. This leaves 196, 361 and 784 which on reversal form 163, 487 and 691. Each of these numbers is prime and they have a mean of . 1 3(163+487+691)=447 18. 603 Let be the centre of the semicircle and let and be the feet of the perpendiculars drawn from to and respectively. Let be the intersection of the diagonals of the rhombus.OMN OAB AD G and . So . PO=10OPC=30°OC=10 tan 30°= 10 3 and . So . MO=10OAM=60°AO= 10 sin 60°= 20 3 A B C MN G OD P Q Therefore .AC= 10 3+ 20 3= 30 3=10 3 Hence and . So .AG=5 3GBA=30°BG= 5 3 tan 30°=15 Therefore the area of triangle is .BGA 1 2 ×15×5 3= 75 2 3 So the area of the rhombus is .4× 75 2 3=150 3 Therefore and , so a=150b=3ab+a+b=450+150+3=603. 19. 343 F1(x)=x; F 2(x)= 1 1Šx; F 3(x)= 1 1Š 1 1Šx =1Š 1 x; F 4(x)= 1 1Š (1Š 1 x)=x. Hence we have, .F 1(x)=F 4(x)=F 7(x)= ƒ=F 3kŠ2 (x) Therefore we are required to solve where is a three- digit cube (given). The cubes are 125, 216, 343, 512, 729. The only one of the format for some positive integer is 343.F 3kŠ2 (3kŠ2)=3kŠ23kŠ2 3kŠ2k 20. 077 It is clear that each of , and must have prime factors including 2 and 3 and, since we are seeking a minimal number of factors of , these must be the only prime factors. ab c abc Let , and where and are positive integers or zero. Since we have .a=2 p3q b=2 r3s c=2 v3w p,q,r,s,vw a 2 =2b 3 =3c 5 22p32q =2 3r+1 33s =2 5v35w+1 Considering indices of 2 we have .2p=3r+1=5v The smallest values of which satisfy this equation are (5, 3, 2).(p,r,v) Considering indices of 3 we have .2q=3s=5w+1 The smallest values of which satisfy this equation are (3, 2, 1).(q,s,w) Using these values, .abc= (2533)× (2332)× (2231) =2 1036 Any factors of will be of the form where and .abc2 y3z y{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} z{0, 1, 2, 3, 4, 5, 6} There are 11 possibilities for and 7 possibilities for . Hence has factors.yzabc11×7=77