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UKMT UKMT UKMT SENIOR ‘KANGAROO’ MATHEMATICAL CHALLENGE Friday 28th November 2014 Organised by the United Kingdom Mathematics Trust SOLUTIONS

2014 Senior Kangaroo Solutions 1. 5 First note that, on a standard die, the numbers on opposite faces add to 7. Let the number on the top of the tower be . The numbers on the touching faces going down the tower are then , , and respectively. The bottom number is . The numbers on a die are 1 to 6 so and hence . The question states that is even so . Hence the number on the bottom of the tower is . (It is easy to check that when all the numbers going down the tower are values that can appear on the face of a standard die.) n 7−n5−(7−n)=n−27−(n−2)=9−n5−(9−n)=n−4 7−(n−4)=11 −n11−n≤6 n≥5nn=6 11−n=5n=6 2. 1 All prime numbers greater than 2 are odd. For these numbers, is even and greater than 2 and so not prime. However, which is prime. Hence only one prime number, namely 2, has the desired property.pp 4+1 2 4+1=17 3. 11 A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Let the second and third digits of the combination be and respectively. Hence is divisible by 9. Since and we have or 27. This gives either , which has nine different solutions given by , , and so on up to or which has two different solutions, namely , and , . This means there are different combinations with the desired property.xy10+x+y 0≤x≤90≤y≤910+x+y=18 x+y=8x=0x=1 x=8x+y=17x=8 y=9x=9y=89+2=11 4. 108 Let . Triangle is isosceles with so . Triangle is also isosceles with so . ∠CAB=x°ABC C A=CB ∠CBA=x°BC D DB=DC∠BC D=x° The exterior angle of any triangle is equal to the sum of the interior opposite angles, so and hence, since triangle is isosceles, .∠CDA=2x° CAD∠AC D=2x° AB C D The angle sum of a triangle is , and applying this to triangle we have . Therefore and hence .180°CAD x+2x+2x=180x=36∠BC A=36°+2×36°=108° 5. 15 Let the number not used be . The sum of the seven numbers is 123 which is divisible by 3. The six numbers used are divided into three pairs with the same sum so is also divisible by 3. This means that is divisible by 3 and the only number in the list that is divisible by 3 is 15. The remaining six numbers can then be paired as 11 and 25, 20 and 16, 19 and 17 all with sum 36.x 123 −x x 6. 343 Multiply the two given equations together to obtain . Hence and so . Therefore the value of is .x 2yz 3×xy 2 =7 3×7 9 x3y3z3 =7 12 xyz=7 4 xyz 77 3 =343 7. 220 When the unwanted rows and columns are erased, 11 rows and 10 columns remain. The table then contains entries, all equal to 2. Hence the sum of the numbers remaining in the table is .11×10 11×10×2=220 8. 11 Let the numbers placed in the empty circles be , , and as shown and let be the number placed in the circle marked . Recall that the number placed in the circle marked is 9. The sum of the numbers in a closed loop of length 3 is 30 so and . Add these two equations to get . However, the sum of the numbers in a closed loop of length 4 is 40. Thus weabc dy Y X a+b+9=30 c+d+y=30 a+b+9+c+d+y=60 YXa b c d also have . This tells us that and hence that so Andrew should place number 11 in the circle marked .a+b+c+d=40 9+y=20y=11 Y

9. 153 Use the three-dimensional version of Pythagoras' Theorem to get . Hence so . AB 2=(3×3) 2+(2×3) 2+(2×3) 2=81+36+36AB 2 =153k=153 10. 35 Let the number of marks scored for each question be , , , and with . The number of marks scored for the two questions with the lowest number of marks is 10 and so . However, and so . Similarly and and hence . So and therefore , and . So the total number of marks Carl scored is .abcd e a

16. 225 Multiply each term of the second equation by to obtain . Square each side of the first equation to obtain . So and hence .xyz yz+xz+xy=0 (x+y+z) 2 =15 2 x2+2xy+2xz+y 2+2yz+z 2=225x 2+y 2+z 2 =225 17. 12 Let the length of the sides of the square be 2 units so its area is 4 units 2. Introduce points , and as shown on the diagram where and are parallel to and let the length of be units. The triangles , and are all similar and isosceles so and . Also triangles and are similar so which has solution . The shaded area is then . Hence the area of the square is times the shaded area and so .XY Z XM Y Z SP PZ x PZ Y PM X PQR Y Z=x XM=1SPM Y ZM 2 1= x 1−xx= 2 3 2× 1 2 ×1× (1− 2 3) = 1 3 4÷ 1 3 =12 k=12 P QR S M Z YX 18. 15 The total number of ‘man-days’ of work required for the project is . The number of ‘man-days’ completed is leaving 800 to be completed. To finish this in 20 days requires workmen and so an extra workmen are required.5×25×8=1000 25×8=200 800÷20=40 40 − 25=15 19. 16 The figures 2, 0 and 2 in the hundreds column lines 3, 4 and 5 of the calculation are not large enough to create any carry into the thousands column. Hence the first two missing figures in the third row of working must add with 2 and 9 to give 56 and so are 4 and 5. Note also that the final two digits in that row must be zeros from the structure of the sum so the third row of working is 45200. This means that one of the original 3-digit multiplicands is a 3-digit factor of 452 and so is 452, 226 or 113. The first row of the working is a 4-digit number starting 22 and so is 2260 as it is also a multiple of the same multiplicand. This means that this multiplicand is 452 and that the completed sum is as shown on the right. Hence the sum of the digits of the answer is . 5+6+5+0+0=16 452 125 565002260 9040 45200 20. 24 Draw in line as shown and let be the point where intersects . The corresponding sides of and are equal and so and are congruent (SSS). Hence . Note also that, because and are parallel, since they are alternate angles. This means that and so is isosceles and hence .PR X PR TSPQR PSRPQRPSR ∠PRQ=∠PRS TS QR∠PRQ=∠RX S ∠PRS=∠RX S XRS XS=RS=15 cm is parallel to and so and using corresponding angles. This means that and are similar and so which gives so . TX QR∠PT X=∠PQR ∠PX T=∠PRQ PT XPQR TX:QR=PT:PQ T X:15=15 : 25 TX=9 cm Hence .TS=TX+XS=9 cm+ 15 cm=24 cm P Q R XS T cm 15 cm 15 cm 15 cm 25