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UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 29th November 2013 Organised by the United Kingdom Mathematics Trust SOLUTIONS

2013 Senior Kangaroo Solutions 1. 9 There are 30 sweets in total so, since the boys all finish with the same number of sweets, they must then have sweets. Carl gains 5 sweets from Bill and gives 4 sweets to Adam so has a net gain of 1 sweet. Since Carl finishes with 10 sweets, he must start with sweets.30÷3=10 10−1=9 2. 110 The perimeter of each -rectangle is 22 cm. Therefore, the sum of the length and the width is 11 cm. All the sides of the -rectangle are whole numbers so the possible - rectangles are with area , with area , with area , with area and with area . Hence the sum of the areas of all possible -rectangles is . Therefore the value of is 110.i ii 1×10 10 cm 2 2×918cm 2 3×8 24 cm 2 4×728cm 2 5×630cm 2 i10+18+24+28+30=110 cm 2 A 3. 25 Let the distance of the second knot from the other end of the rope be m.d This part of the rope will become the hypotenuse of the right-angled triangle so, on applying Pythagoras' Theorem, we have the equation . Now expand the brackets to get . This simplifies to , which has solution .d 2 =15 2+(45−d) 2 d2 =225+2025−90d+d 2 90d=2250d=25 Hence the second knot is 25 m from the other end of the rope. 4. 12 Let each side of the original cube have length so that the cube has surface area . Then the cuboid has side-lengths and , so has surface area .x6x 2 2x, 3x6x 2× (2x×3x+2x×6x+3x×6x )=72x 2 Hence the value of is .N72x 2÷6x 2 =12 5. 25 Dean has answered 5 questions incorrectly so 5 questions must represent 20% of the questions. 20% is equivalent to so the total number of questions is . 1 5 5×5=25 6. 2 Let the length of be . Therefore, the lengths of and are and respectively. The length of the upper path is . The length of the lower path is .AE4xADDE3xx 1 2×π×4x=2πx1 2×π×3x+ 1 2 ×π×x=2πx Therefore the ratio of the length of the upper path to the length of the lower path is 1 : 1. Hence the value of is 2.a+b 7. 13 In any triangle, the length of the longest side is less than the sum of the lengths of the other two sides. Apply this result (known as the triangle inequality) to the triangle to obtain or . In the same way, apply this result to triangle to obtain or . But we are told that is an integer and so . BC D x

8. 7 In addition to the original square, four squares can be drawn that share two adjacent vertices of the original square and a further four squares can be drawn that share two opposite vertices of the original square. The union of these squares creates the octagon as shown. The octagon is made up of five squares of side 1 unit and four halves of a square of side 1 unit. Hence the area of the polygon is equal to .7×1×1=7 A BCD 9. 60 Let the sizes of angles , and be , and respectively.AB C a°b°c° From the question we have . Therefore .b=0.75×cc= 4 3b Similarly we have . Therefore .b=1.5×aa= 2 3b Angles in a triangle add up to , so that we have , which means that . It follows that .180°180= 2 3b+b+ 4 3b 180=3bb=60 10. 3 Factorise both sides of the equation to get . Thus we have which is equivalent to . Since 2 and 3 have no factors in common other than 1, a power of 2 cannot equal a power of 3 unless both powers are zero when both sides of the equation equal 1. Therefore we have and . Hence the value of is 3 (and the value of is 1).2 m(21+1 ) =3 n(32−1 ) 2m×3=3 n×82 m−3 =3 n−1 m−3=0n−1=0 mn 11. 128 Let the radii of the larger and smaller semicircles be and respectively. Then the shaded area as .Rr 1 2 ×πR 2− 1 2 ×πr 2 = 1 2π(R2−r 2) Let be the centre of the larger semicircle and let be the midpoint of . Since is parallel to then and . Apply Pythagoras' Theorem to triangle to give or .XY CD CD AB XY=r∠XY D=90° XY D R 2 =r 2+16 2 R2−r 2=256 Therefore the shaded area is . Hence the value of is 128. 1 2π(R2−r 2)=128π k ABCD X Y 12. 11 Let the smallest number be . From the information in the question, we obtain the equationn n+n+1+n+2+n+3+n+4=n+5+n+6+n+7. Therefore we get , which has solution .5n+10=3n+18n=4 Hence the largest number is 4 + 7 = 11. 13. 8 Let Zoe be years old. Therefore, her mother's age is years old. Now divides if and only if divides 24. The positive factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 and so Zoe's age is a factor of her mother's age on 8 birthdays, when her mother's age will be 25, 26, 27, 28, 30, 32, 36 and 48.xx+24x x+24x 14. 992 Since is an integer, is a square number. The square numbers near 1000 are , and . Clearly, if then is greater than 1000, so this is not possible. However, if , then , which is less than 1000. Hence the largest three-digit integer than can be written in the given form is 992.n+ nn 30 2 =900 31 2 =961 32 2 =1024n=32 2 n+ n n=31 2 n+ n=961+31=992 15. 8 If the equation has integer solutions, then it can be written in the form for integers and . This means that .x 2+ax+2013=0 (x+b)(x+c)=0bc bc=2013 As the prime factorisation of 2013 is , so the possible factor pairs of 2013 are 1 and 2013, 3 and 671, 11 and 183 and 33 and 61. However, these only take into account the cases when both and are positive and four further pairs are possible if both and are negative. Thus there are 8 distinct values of , namely , , and .3×11×61 bc bc a±2014±674 ±194±94

16. 30 Let the coordinates of a relevant point on the sphere be . By the three- dimensional version of Pythagoras' Theorem, we have . The only solutions for which , and are positive integers are (3, 0, 0) and (1, 2, 2) in some order.(x,y,z) x 2+y 2+z 2=3 2 xy z There are solutions based on the values (3, 0, 0) as the 3 can go in any of the three positions and be either positive or negative. Similarly there are solutions based on the values (1, 2, 2) as the 1 can go in any of the three positions and all three of the values can independently be either positive or negative. This gives solutions in total. 3×2=6 3×2×2×2=24 6+24=30 17. 150 Let be the length of a side of the rhombus and let and be the lengths of the two diagonals. The area of the rhombus isxab areaQRS+areaPQS= 1 2x2sin∠SRQ+ 1 2x2sin∠SPQ. Opposite angles in a rhombus are equal so this simplifies to . However, the area of a rhombus can also be calculated in a similar way to a kite, i.e. half the product of the diagonals. This gives the equation . From the question, we know that so . Hence and so . Lines and are parallel and so, using co-interior angles, . This means .x 2sin∠SRQ x 2sin∠SRQ= 1 2ab x= ab x 2=absin∠SRQ= 1 2 ∠SRQ=30°SR PQ ∠PQR+∠SRQ=180°∠PQR=150° 18. 804 The triangular numbers are given by the formula . is a multiple of 5 if, and only if, one of or is also a multiple of 5. This means that two triangular numbers in every group of 5 consecutive triangular numbers will be a multiple of 5. None of the numbers 2011, 2012, 2013 or 2014 is a multiple of 5 and so none of , or is a multiple of 5 either. Hence the number of multiples of 5 in the first 2013 triangular numbers is .T n = 1 2n(n+1)T n nn+1 T 2011 T2012 T2013 2× 2010 5 =804 19. 741 Consider the numbers . Using at most one of each of these in a sum, the number of totals we can create is as each of the numbers can either be included or excluded from the sum but at least one number must be included so the choice of excluding all the numbers is discounted (and they are all distinct). For , this is 63 and so the 64th number in the sequence will be . Then the 70th term is equal to the 64th term + 6th term .n3 0, 3 1, 3 2,…, 3 n−1 2n−1n n=6 3 6 =729 =729+12=741 20. 50 Let the distance Rachel runs before they first meet be m. Let and be the respective speeds of Rachel and Nicky and let and be the times they take to get to their first and second passing points respectively (shown as and on the diagram below).xv R vN t1 t2 PQ 20 xP 10 Qx+10 As distance = speed × time, we have the following equations: and ; and x=v Rt1 20=v Nt1 2x+30=v Rt2 x+30=v Nt2 where the first and second pair give the distances travelled by Rachel and Nicky from the start to and respectively.PQ Divide each equation in the first set by the corresponding equation in the second set to eliminate and to obtain . Now multiply both sides by the common denominator to obtain . This simplifies to , i.e. . This factorises to with solutions and . As is measuring a distance, it must be positive so . Hence the length of the track is 30 + 20 = 50 m.v R vN x 2x+30= t1 t2 = 20 x+30 (x+30)(2x+30)x(x+30)=20(2x+30) x 2+30x=40x+600x 2−10x−600=0 (x−30)(x+20)=0x=30x=−20x x=30