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UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 2nd December 2011 Organised by the United Kingdom Mathematics Trust SOLUTIONS 1. 71 Since the entries are different, no entry can be 1, and the smallest total will come from using the integers 2, 3 and 4 on the top line of the diagram. Treating an ordering and its reverse as the same, since they give the same total, we can arrange these in the orders 2, 3, 4 or 2, 4, 3 or 3, 2, 4. Of these, 3, 2, 4 gives the smallest overall total of 71. 2. 390 Over the years 2007 to 2010, the school accepted students. The mean for 2007 to 2011 is 4% higher than 325, which is 338. This means that students were accepted over the years 2007 to 2011. Therefore 390 students were accepted in 2011.325×4=1300 338×5=1690 3. 16 On the first pass, all 200 people receive a pound coin. On the second pass, only people in even numbered positions receive a coin. On the th pass, people receive a coin if divides the number representing their position. Now 120 is divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120 so the 120th person receives 16 pound coins.nn [Note that, in general, if the prime decomposition of an integer,,is then the number of divisors of is.] Xp a11pa22pa33…p ann X (a1+1 )( a2+1 )( a3+1 )… (an+1 ) 4. 100 Using the labelling shown, we see that and are similar and have lengths in the ratio 2:1. Because the height of is 10 + the height of , the height of is 20 and its area is . The area of the smaller square is 100 so the shaded area is .AC EBC D AC EBC D AC E 1 2 ×20×20=200 200−100=100 A B C D E 5. 36 The square and cube of an integer end in the same digit if, and only if, the integer itself ends in 0, 1, 5 or 6. The two-digit numbers with this property can have any tens digit from 1 to 9 so there are such two-digit numbers.4×9=36

6. 84 The perpendicular height from any side of an acute- angled triangle is always less than the length of either of the other two sides so the height is 12. Thus we have the situation shown in the diagram alongside. Using Pythagoras' theorem in , we obtain and in we get . This means that and the area of triangle is .AC M AM=9CMB MB=5 AB=14ABC 1 2 ×14×12=84 M 1213 15 B A 7. 30 Let the centre of the circle be . Join to and to and , as shown in the diagram alongside.OACOA B In , since they are both radii, and we are given that has length equal to the radius so and is equilateral. Hence . Since the angle at the centre is twice the angle on the circumference, is .AO B AO=OB AB AB=AO=OBAO B ∠AOB=60° ∠AC B30° O BC A 8. 13 Let the original price, in pence, be .p The new price is 4% more than the original so, working in pence, 100n= 104 100×p, which may be rearranged to n= 13 1250×p. Now is an integer and 1250 is not divisible by 13, so is divisible by 1250. The smallest value of will be when , which means is 13.np np=1250n 9. 5 y x R Py x Let be and suppose that the -axis is a line of symmetry. Then is a vertex of the square since it is the reflection of the vertex in the -axis. Hence is either an edge or a diagonal of the square. In the first case there are two possible squares and in the second case there is one, as shown in the first figure.P(−1,−1)xQ(−1, 1) Px PQ Similarly, when the -axis is a line of symmetry there are three possible squares. However, one of these is the same as before, so in all there are exactly five squares possible.y

10. 4 By expanding the brackets, we obtain ( 8+2 7− 8−2 7) 2 =8+2 7−2 (8+2 7)(8−2 7)+8−2 7 =16−2 64−28=16−2 36=16−12=4. 11. 60 AreaABC=areaAC D+areaBC D+areaABD = 1 2 ×e×5+ 1 2×f×12+ 1 2×g×13= 1 2(5e+12f+13g ). But has sides 5, 12 and 13, hence it is a right-angled triangle and so has area . Therefore . ABC 1 2 ×5×12=30 5e+12f+13g=60 12. 40 From the information given about those who voted, we can conclude: Eaten broccoli? Voted Broccoli Party?YesYes y 0 NoNo x 9x where , and are the appropriate percentages of those who voted.x9xy We are given that and, since the table includes everyone, we also have . So and . Therefore and so the percentage that voted for the Broccoli Party is 40%.x+y=46 x+y+9x=100 9x=54x=6y=40 13. 65 Suppose represents the number of increments of €5 above (or below, if is negative) the selling price of €75. Then the number of sweaters sold is and the profit made, in Euros, is . So the profit is and is a maximum when . This gives a sale price of €65.nn 100−20n ((75+5n)−30 )(100−20n)=(45+5n)(100−20n)=100(5−n)(9+n) 100 (49−(n+2) 2) n=−2 14. 75 Using Pythagoras' theorem firstly in and then in we get and . It follows that is similar to as the corresponding sides are in the same ratio. Therefore, . Also , using alternate angles, so and is isosceles. Let be the mid-point of and join to . This gives two more right-angled triangles, and , also similar to . Thus which gives . Therefore the area of is .ABCAC E AC=20 AE=25ABCAC E ∠BAC=∠CAE∠BAC=∠AC F ∠CAF=∠AC FAF C M AC M FAM FCMFABC MF MA= BC BAMF= 15 2AC F 1 2× 15 2×20=75 15. 10 x y The diagram shows the dashed lines with equations , and . The solid lines form the graph of the function given in the question. We can see that the maximum value of occurs when crosses . y=3x+1y=2x+3y=−4x+24 f(x) y=2x+3y=−4x+24 At this point , therefore the maximum value of is 10.y=10 f(x)

16. 891 Since , we have where there are 98 nines and 98 zeroes. Therefore the sum of the digits is .m=10 99 −1m 2 = (10 99 −1 )2 =10 198 −2×10 99 +1 =999…9998000…001 98×9+8+1=891 17. 21 A B C D R QP M Dividing rectangle into 16 equal parts, as shown in the diagram above, demonstrates that the area of parts. Therefore the area of is of the area of rectangle so .ABC D APM=12− 1 2 ×3− 1 2 ×3− 1 2 ×8=5 APM 5 16 ABC D m+n=21 18. 4 There are six different numbers that can be formed with digits , and . The sum of these six numbers is ab c (100a+10b+c )+(100a+10c+b )+(100b+10a+c ) + (100b+10c+a )+(100c+10a+b )+(100c+10b+a ) =200 (a+b+c )+20 (a+b+c )+2 (a+b+c ) =222 (a+b+c )=1554 so . Thus the only possibility for , and is 1, 2 and 4 so .a+b+c=7ab c c=4 19. 3 From we have since . Therefore , which gives and so . This means that . (a+ 1 a ) 2 =6a+ 1 a= 6a>0 (a+ 1 a ) 3 = ( 6)3 a3+3a 2× 1 a+3a× 1 a 2 + 1 a 3 =6 6N 6+3 (a+ 1 a ) =6 6 N=3 20. 17 From we deduce that and hence that . This means , and . Therefore the value of is 17.f (x2+1 )≡x 4+4x 2≡ (x2+1 )2+2 (x2+1 )−3f(w)≡w 2+2w−3 f (x2−1 )≡ (x2−1 )2+2 (x2−1 )−3≡x 4−2x 2+1+2x 2−2−3≡x 4−4 a=1b=0c=−4a 2+b 2+c 2 An alternative solution is to realise that . So . This gives the same value for .f (x2+1 ) ≡ [(x2+1 )+1 ]2−4 f (x2−1 )≡ [(x2−1 )+1 ]2−4≡x 4−4a 2+b 2+c 2