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UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 2nd December 2016 Organised by the United Kingdom Mathematics Trust SOLUTIONS 1. 121 The sum has eleven terms. Therefore the value of the required sum is .1+3+5+7+9+11+13+15+17+19+21 11×11=121 2. 950 123456789101050 11 12 13 14 15 16 17 18 19 20 * We observe that in all but the rightmost column the value in the second row is ten larger than the value in the first row. There are 10 such columns. Therefore the sum of the leftmost ten elements of the second row is 100 more than the corresponding sum in the first row. To achieve the same total in each row, * will need to be 100 less than the value above it. Therefore * = 950. 3. 27 We first observe that any pair of cubes are mathematically similar. These cubes' surface areas are in the ratio 1:9, so that their lengths are in ratio 1:3 and that their volumes are in ratio 1:27. Therefore Andrew may fill the larger cube in 27 visits, provided the smaller cube is completely filled on each occasion. 4. 125 The number will be of the form where , and are any odd digits and . Hence there are 5, 5, 5 and 1 possibilities for and respectively. Therefore there are such numbers.‘abcd’ab c d=5 a,b,cd 5×5×5×1=125 5. 9 The enclosed area is a concave quadrilateral with vertices at , , and . Considering this as two conjoined congruent triangles we find the area as .(Š3, 3)(0, 0)(3, 3) (0,Š3) 2× 1 2 ×3×3=9 6. 160 The tangent-radius property gives . From the angle sum of quadrilateral we may conclude that and therefore that . By considering the angle sum of triangle we conclude that the required total is .PSO=PT O=90° PT OSTPS=30°TPY=20° PT Y160°

7. 315 Suppose that the cross-section of the prism is an -gon with edges. The prism will have edges in each of its ‘end’ faces and a further edges connecting corresponding vertices of the end faces. Therefore the number of edges is and hence is a multiple of 3. The only multiples of 3 in the given range are 312, 315 and 318. Since we know the total is odd, the prism has 315 edges.NN NN 3N 8. 7 Since squares of real numbers are non-negative, the sum can only be 0 if each expression in brackets is zero. Therefore the solutions of the equation are , and . We observe that the maximum and minimum values for are and , and that since is the sum of one even and two odd numbers, that itself will be even.x=±3y=±2 z=±1x+y+z6 Š6x+y+z x+y+z It suffices to show that each even total between and can be attained.+6Š6 (+3)+(+2)+(+1)=+6(+3)+(+2)+(−1)=+4 (+3)+(−2)+(+1)=+2(+3)+(−2)+(−1)=0 (−3)+(+2)+(−1)=−2(−3)+(−2)+(+1)=−4 (Š3)+(Š2)+(− 1)=−6 Hence there are seven possible values for .x+y+z 9. 256 Let the radii of the larger and smaller circles be and respectively. Draw radius of the larger circle and drop the perpendicular from to . By the tangent-radius property this perpendicular will be a radius of the smaller circle.Rr OA OAB Now the area of the shaded region = area of larger circle − area of smaller circle. The area of the shaded region .=R 2Šr 2= (R2Šr 2) But (by Pythagoras' theorem), hence the area of the shaded region and therefore .R 2Šr 2=16 2=256 =256k=256 A B R rO 10. 103 Note that . We observe that if any multiplication sign, other than the first, is replaced by an addition sign then each remaining product is at most 360. Therefore we retain each multiplication sign except the first which may be replaced by an addition sign to obtain a maximal value of 721.6!=1×2×3×4×5×6=720 The prime factors of 721 are 7 and 103, of which 103 is the largest. 11. 183 We first observe that exactly one odd year and exactly one even year are under consideration. In an odd year we need only consider odd months. January, March, May and July each has 16 odd days while September and November has 15. Therefore the number of days Stephanie will swim in the odd year is .4×16+2×15=94 In an even year we need only consider even months. April, June, August, October and December has 15 even days and February has 14 (regardless of whether or not it is a leap year). Therefore the number of days Stephanie will swim in the even year is . Hence she will swim for days over the two years. 5×15+14=89 94+89=183

12. 144 The regular 18-gon has a circumcircle, that is, a circle passing through all of its vertices. This is also the circumcircle of each right-angled triangle formed. In order for one of these triangle's angles to be a right angle, the opposite side needs to be a diameter, There are 9 possible choices of diameter. For each choice of diameter, there are 8 vertices on each side for the right angle, making 16 choices overall. For each choice of diameter there are 16 choices for the third vertex of the right-angled triangle. 13. 143 For a year to be expressible as as the sum of two positive integers and where we require and . From the first of these, it follows that and hence . Ypq 2p=5qp+q=Y2p=5q 2p+2q=2Y5q+2q=2Y Therefore from which it follows that is also divisible by 7 (since 2 and 7 are coprime).7q=2YY We observe that will be an integer less than for all that are multiples of 7.q= 2Y 7 YY Then will also be an integer.p=YŠq We now must count all the multiples of 7 between 2000 and 3000 inclusive. Since and there are multiples of 7 between 2000 and 3000 and hence there are 143 such years.1995=285×7 2996=428×7428Š285=143 14. 35 Assume, without loss of generality, that . Since , are positive integers and , the possible values of are (1, 105), (3, 35), (5, 21), (7, 15).xyxy xy=105(x,y) Since we require for the triangle to exist, we may eliminate the first three of these possibilities, leaving only (7, 15) and conclude that the perimeter is .13+x>y 13+7+15=35 15. 110 For each small equilateral triangle, let the length of each side be and the perpendicular height be .xh We may trap the shaded triangle in a rectangle as shown, where one vertex is coincident with one of the vertices of the rectangle and the other two vertices lie on sides of the rectangle. The rectangle has width and height . Therefore the rectangle's area is .4x3h 12xh The three additional (unshaded) right-angled triangles in the rectangle have areas , and . Therefore their total area is . 1 2 ×4x×2h=4xh 1 2 × 1 2x×3h= 3 4xh 1 2 × 7 2x×h= 7 4xh 4xh+ 3 4xh+ 7 4xh= 13 2xh Therefore .K=12xhŠ 13 2xh= 11 2xh Each of the 36 smaller equilateral triangles has area so we know that and therefore that . 1 2xh 1 2xh=10xh=20 Therefore .K= 11 2 ×20=110

16. 87 The function has the property that . First observe that . Therefore and .f(x)3f(x)+7f ( 2016 x ) =2x 2016 8=252 3f(8)+7f (252 ) =16 3f(252)+7f(8)=2×252 Let and . Therefore and .f(8)=Vf(252)=W3V+7W=16 3W+7V=504 When these equations are solved simultaneously, we obtain and so that .V=87W=Š35 f(8)=87 17. 649 We observe that the total of all integers on the board at the start of the process is . 10(1+2+3+4+5+6+7+8+9+10)=550 On each turn this total is increased by 1. Since we start with one hundred integers on the board and at each turn this number of integers is decreased by one, then 99 turns will be required to complete the process. Therefore the total of all integers on the board will increase by 99 over the course of the process. Hence the remaining number will be . 550+99=649 18. 441 Let the smallest number be . Therefore and hence x x2+(x+1) 2+(x+2) 2+(x+3) 2=(x+4) 2+(x+5) 2+(x+6) 2 . This can be rewritten as or . Hence , which has solutions and . The question tells us that is positive and therefore . x 2+x 2+2x+1+x 2+4x+4+x 2+6x+9=x 2+8x+16+x 2+10x+25+x 2+12x+36 4x 2+12x+14=3x 2+30x+77x 2−18xŠ63=0 (xŠ21)(x+3)=0x=21x=Š3 xx=21 The square of the smallest of these integers is therefore .21 2 =441 19. 421 When an odd number is subtracted from an odd square, an even (and hence composite) number is obtained. Similarly, when a multiple of 3 (or 7) is subtracted from a square of a multiple of 3 (or 7), a multiple of 3 (or 7) is obtained which is also composite. Therefore we need only consider three-digit squares that are neither odd nor a multiple of 3 (or 7). Hence the only squares we need to consider are , , and which yield differences of 235, 379, 463 and 655 respectively. It is easy to see that 235 and 655 are multiples of 5 and hence composite. Therefore only 379 and 463 remain as possible primes satisfying the given condition. After checking divisibilty by 11, 13, 17 and 19 for both, both are indeed seen to be prime and their mean is 421.16 2 =256 20 2 =400 22 2 =484 26 2 =676 20. 116 Any code will start with a black strip and a white strip followed by a shorter barcode. Let be the number of distinct barcodes of width . C(m)m Those codes which start with BW will be followed by a code of width ; so there will be of these. Likewise, there will be codes starting BBW, the same number starting BWW, and starting BBWW; and that exhausts the possibilites. So it follows that .mŠ2 C(mŠ2)C(mŠ3) C(mŠ4) C(m)=C(mŠ2)+2C(mŠ3)+C(mŠ4) When , it is simple to list all possible barcodes; namely B, BB, BWB and BBWB, BWBB, BWWB. Therefore and . We can now calculate for .m4 C(1)=C(2)=C(3)=1C(4)=3 C(m)m>4 Thus , and continuing like this, we get ,,,, , , . C(5)=C(3)+2C(2)+C( 1)=1+2+1=4 C(6)=6C(7)=11C(8)=17C(9)=27C(10)=45C(11)=72 C(12)=116