Download [PDF] 2016 UKMT UK SMC Solutions Leaflet Senior Mathematical Challenge United Kingdom Mathematics Trust

File Information

Filename:	[PDF] 2016 UKMT UK SMC Solutions Leaflet Senior Mathematical Challenge United Kingdom Mathematics Trust.pdf
Filesize:	119.43 KB
Uploaded:	09/08/2021 13:23:55
Keywords:	2016 ukmt uk smc solutions leaflet senior mathematical challenge united kingdom mathematics trust pdf
Description:	Download file or read online UKMT past exam paper senior mathematical challenge 2016 UK SMC solutions leaflet - United Kingdom Mathematics Trust
Downloads:	6

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/3ft

Full Page Link

https://www.edufilestorage.com/3ft/PDF_2016_UKMT_UK_SMC_Solutions_Leaflet_Senior_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf

HTML Code

<a href="https://www.edufilestorage.com/3ft/PDF_2016_UKMT_UK_SMC_Solutions_Leaflet_Senior_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/2491/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/3ft/PDF_2016_UKMT_UK_SMC_Solutions_Leaflet_Senior_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/2491/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2016 UKMT UK SMC Solutions Leaflet Senior Mathematical Challenge United Kingdom Mathematics Trust.pdf | Plain Text

UKMT UKMT UKMT UK SENIOR MATHEMATICAL CHALLENGE Tuesday 8 November 2016 Organised by the United Kingdom Mathematics Trust from the School of Mathematics, University of Leeds SOLUTIONS LEAFLET This solutions leaflet for the SMC is sent in the hope that it might provide all concerned with some alternative solutions to the ones they have obtained. It is not intended to be definitive. The organisers would be very pleased to receive alternatives created by candidates. For reasons of space, these solutions are necessarily brief. There are more in-depth, extended solutions available on the UKMT website, which include some exercises for further investigation: http://www.ukmt.org.uk/ The UKMT is a registered charity 1. B Since the answer to is , the digit 9 appears once.987654321?98 888 888 889 2. DJames bought a book for £5.50, so the normal price would be £5. On Friday, another copy of this book would therefore cost £4.50. 3. BAs the circle rolls, the centre of the circle moves along four straight lines shown as dashed lines. Each dashed line has length so the total distance travelled is which is 12.5?(1H1F4?3 1 1 5 4. COne angle in Alex's triangle is . Let be the smaller of the other two angles so is the third angle. The difference between these angles is then . Considering each option:80?? E100?)? E100?2)? A: gives both and to be 50. This triangle is therefore isosceles and not scalene.100?2=0100? B: gives to be 20 and to be 80. This is again isosceles.100?2=60100? Option D gives angles of 80, 0 and 100. Option E gives angles of and 110. Neither of these cases forms a triangle.80,?10 C: gives to be 10 and to be 90. All three angles are different so this is the correct option.100?2=80100? 5. AFrom the available digits the squares could be 256, 324 or 625. Since the middle digits must be the same, the centre digit must be 2.

6. D Let be the point of intersection of the lines and . Let the length of be . Then, using similar triangles, , so giving . The shaded region is a trapezium, so has area which is . FAE CD CF h CF CE= BA BE h 1= 3 4h= 3 4 ABC F 1 2 E3H 3 4 F ?3= 45 85 5 8 A B CD E F h 3 31 7. AThe number has value 1. As , . The values of their reciprocals, and are then in the opposite order. So the five options given are in numerical order, with , or , being the smallest. 2016 0 2016>1 2016 1/2

14. DThe five options give the values of to be considered. In option B, 11 is prime so that can be discounted. The options A, C and E are 6, 27 and 51 which are not prime and subtracting 2 from each of these gives 4, 25 and 49 which are also not prime. However in D, which is not prime but is prime.n n=33n?2=31 15. BA non-regular hexagon can be drawn on the diagram as shown. Three of the exterior angles of the hexagon are then , and . Since corresponding angles on parallel lines are equal, the other three exterior angles are , and . The total of the exterior angles of any polygon is . Hence and so . 55?60?65? p?q?r? 360? pHqHrH55H60H65=360pHqHr=180 55? 60?65? r?p? q? 16. D The expression can be written as which is . This simplifies to which is and, using index notation, this can be written as . Hence . 2016H 56 25 ?3 2 ?7H 23 ?7 42 ?3 2 ?2?7H 22 ?2?712 14H2 14 14 14 14 3/2 k= 3 2 17. DOne way to count the possible codes is in descending numerical order of the three-digit codes. The list begins: 789; 689, 679, 678; 589, 579, 578, 569, 568, 567; ... . Each initial digit produces part of the list with the triangular number of possible codes, where . The total number of possible codes is then the sum of these triangular numbers including 1 code starting with the digit 7, all the way to 28 codes starting with the digit 1. The total number of codes that Aaron can choose is 84.nE8?nFth n7 1H3H6H10H15H21H28 18. E The four arcs are of equal length and their end-points lie on a circle, so the four end-points can be joined to make a square. As two of the arcs are ‘turned over’, the two unshaded regions inside the square have areas equal to the two shaded regions outside the square. 1 The total shaded area is therefore equal to the area of the square. The radius of the circle is given as 1 so, by Pythagoras' Theorem, the side-length of the square is . So the area of the shaded region is . 12 H1 2 = 2 2? 2=2 19. DLet consist of in ascending order of size. We want to be as small as possible. Given three side-lengths, there is a quadrilateral with non- zero area with a specified fourth side-length if and only if the fourth side-length is less than the sum of the other three side-lengths. To ensure that are not the side-lengths of such a quadrilateral, we must have . Likewise, considering , we must have . Since the smallest possible values of and are 1, 2 and 3 respectively then so 6 is the smallest value of . Also so 11 is the smallest value of .ShIjIkIlImm jIkIlIm m jHkHl hIjIkIll hHjHk hIjk l 1H2H3lm 2H3H6 m 20. ELet the point be directly below , so that is a rectangle. As the length of is 2 cm, the distance from to the nearest corner of the square is 3 cm. The area of is which is . The diagonals and split into quarters and each has area . The central grey octagon is formed from a square with side of length 2 cm together with four triangles, each of area . The ZXXYZZ XY Y XY ZZ 2cm? 3cm 6cm 2 XZ Y Z XY ZZ1 1 2 cm 2 ZZ 1 1 2 cm 2 XY Z Z 3 2 3 total area of the shaded octagon is which is .2?2H4?1 1 2 10 cm 2

21. C As there are 10 discs, the adjacent lines drawn from the centre of the inner circle to the centre of each disc are separated by an angle of . The line is a tangent to both the disc with centre and the disc with centre . So the points , and lie on a straight line as angles and are both .36? OB A CAB COBAOBC 90? In the second diagram, from triangle we have which rearranges toOAB sin 18?= r 1Hr sin 18? 1−sin18?=rK The radius of the outer circle is 1H2r=1H 2sin18? 1−sin18?= 1Hsin 18? 1−sin18?K r A B O 18? 1r r A B C O 18? 18? 1Hr 22. CSuppose first that Cam is telling the truth, so Ben and Dan are both lying. Then Ben's statement is actually correct, as is Dan's. There is a clear contradiction. So we know that Cam is in fact lying. Therefore at least one of Ben and Dan is telling the truth. If Ben is telling the truth, then we learn that Dan is telling the truth. If Dan is telling the truth then we learn that Ben is telling the truth. So both Ben and Dan are telling the truth. This means that only Cam is lying. 23. EFor the cuboid to be contained within a sphere of smallest possible radius, all eight vertices of the cuboid must lie on the sphere. The radius of the smallest sphere is then half of the length of the body diagonal of the cuboid, so . If the largest cube which will fit inside this sphere has side-length , then . Thus , so and so . The side-length of the largest cube is 14.r r= 12H5 2H11 2= 147 2xr= x2Hx 2Hx 2 3x 2=147x 2=49 x=7 24. BLet the square have side-length 2, and let be the point of contact between and the circle. Two tangents to a circle which meet at a point are of equal length. So as so does . Similarly . Applying Pythagoras' Theorem to triangle gives so and RT=hA TU QU=1AU T A=TS=2?h URT 1 2 Hh 2 =(1H2?hF 2 1Hh 2 =9?6hHh 2 P Q RS T A U therefore which gives . The required ratio is then .8?6h=0h= 4 3 4:3 25. D For to be the smallest integer for which has 2016 digits, must start with 1, be followed by 2014 zeros and end with a digit . When this number is divided by 7, the answer is formed from the repeating sequence of 6 digits 142857. The remainders also form a repeating sequence 3, 2, 6, 4, 5, 1. These sequences are repeated 335 times as is 2010. The last 4 zeros (to make 2014 zeros in total) and the final create the last section of the division as shown: n7n7n a 6?335 a 1 4 2 8 ? 103020604aK .......... Finally, must be divisible by 7 and be as small as possible. So and as the units digit of is 6.40Haa=2 42?7=6n