Download [PDF] 2020 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust

File Information

Filename:	[PDF] 2020 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf
Filesize:	272.16 KB
Uploaded:	09/08/2021 18:19:50
Keywords:	2020 ukmt senior kangaroo solutions mathematical challenge united kingdom mathematics trust pdf
Description:	Download file or read online UKMT past exam paper senior mathematical challenge SMC senior kangaroo 2020 solutions - United Kingdom Mathematics Trust
Downloads:	2

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/5Q1

Full Page Link

https://www.edufilestorage.com/5Q1/PDF_2020_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf

HTML Code

<a href="https://www.edufilestorage.com/5Q1/PDF_2020_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/2445/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/5Q1/PDF_2020_UKMT_Senior_Kangaroo_Solutions_Mathematical_Challenge_United_Kingdom_Mathematics_Trust.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/2445/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2020 UKMT Senior Kangaroo Solutions Mathematical Challenge United Kingdom Mathematics Trust.pdf | Plain Text

Senior Kangaroo © 2020 UK Mathematics Trust a member of the Association Kangourou sans Frontières supported by SolutionsThese are polished solutions and do not illustrate the process of failed ideas and rough work by which candidates may arrive at their own solutions. It is not intended that these solutions should be thought of as the ‘best’ possible solutions and the ideas of readers may be equally meritorious. Enquiries about the Senior Kangaroo should be sent to: UK Mathematics Trust, School of Mathematics, University of Leeds, Leeds LS2 9JT T0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk

Senior Kangaroo 2020 Solutions1.What is the difference between the greatest and the least of the following five quantities? 20 +20 20 ×20 202 +0 (2 0 ) ( 2 0 ) 20+2+ 0 Solution 399 The five quantities are 40,400 ,202 ,1 1 = 1and 22. The difference between the greatest and least of these is 400−1= 399 . 2. The positive integers �and �satisfy the equation �� 2 + � � 2 = 70 . What is the value of �4 + �4 ? Solution 641 We may assume that � ≤ � as the expressions ��2 + � � 2 and �4 + �4 are symmetrical in �and �. If �≥ 4(and so �≥ 4) then ��2 + � � 2 ≥ 2× 43 = 128 >70 . So �must be one of 1, 2,3. When �= 2we obtain the equation 4� + 2�2 = 70 which has the integer solution �= 5. The analogous equations in �obtained for �= 1,3 do not have integer solutions. Therefore, �= 2, � =5and �4 + �4 = 24 + 54 = 16 +625 =641 . 3. How many distinct integer solutions ( �, � ) are there to the equation 51 + 42+ 33+ 24 = �� ? Solution 006 The sum 51 + 42 + 33 + 24 is equal to 5+ 16 +27 +16 =64 =26 . The positive integer solutions (�, � ) to the equation �� = 64 are (64 ,1),(8,2),(4,3)and (2,6). There are also two further solutions involving negative integers, namely (−8,2 ) and (−2,6 ). Therefore there are six distinct integer solutions. 4. Two identical cylindrical sheets are cut open along the dotted lines and glued together to form one bigger cylindrical sheet, as shown. The smaller sheets each enclose a volume of 100 . What volume is enclosed by the larger sheet? Solution 400 Since the circumferences of the smaller cylinder and the larger cylinder are in the ratio 1 : 2 , the radii of their cross-sections are also in the ratio 1 : 2 . Therefore the areas of their cross-sections are in the ratio 12 : 2 2 = 1 : 4 . As the cylinders have the same perpendicular height, the volumes they enclose will also be in the ratio 1 : 4. Therefore the larger cylinder encloses a volume of 4× 100 =400 . © 2020 UK Mathematics Trust www.ukmt.org.uk 2 and

Senior Kangaroo 2020 Solutions5. Let �� = 1 8 . What is the value of �− 3� ? Solution 512We are given that � � = 1 8 . Taking the reciprocal we get � − � = 8; and then cubing gives �− 3� = 83 = 512 . 6. For what value of �does the expression �2 − 600 �+ 369 take its minimum value? Solution 300 Complete the square on �2 − 600 �+ 369 to obtain (� − 300 )2 − 300 2 + 369 . The minimum value of this expression will occur when (� − 300 )2 = 0. This is when � = 300 . 7. Margot writes the numbers 1,2,3,4,5,6,7 and 8 in the top row of a table, as shown. In the second row she plans to write the same set of numbers, in any order. Each number in the third row is obtained by finding the sum of the two numbers above it. In how many different ways can Margot complete row 2 so that every entry in row 3 is even? Solution 576 For an entry in row 3 to be even we need the corresponding entry in row 2 to have the same parity as the entry in row 1 (that is: both are odd or both are even). Columns 1, 3, 5 and 7 must therefore have odd entries in row 2. There are four odd numbers to arrange in these cells, with 4× 3× 2× 1= 24 ways to arrange these. Similarly, columns 2, 4, 6 and 8 must have even entries in row 2. There are four even numbers to arrange in these cells, with 4× 3× 2× 1= 24 ways to arrange these. Therefore there are 24×24 =576 ways in which Margot can complete the table in this way. 8. The number (2222 )5 × ( 5555 )2 is � digits long. What is the largest prime factor of �? Solution 101 The number (2 222 )5 × ( 5555 )2 = 21110 ×51110 =10 1110 . The number 101110 is1111 digits long, so �=1111 =11 ×101 . Both 11and 101 are prime. Hence the largest prime factor of 1111is101 . © 2020 UK Mathematics Trust www.ukmt.org.uk 3 1 2 3 4 5 6 7 8

Senior Kangaroo 2020 Solutions9. The radii of two concentric circles are in the ratio 1 : 3. ��is a diameter of the larger circle. �� is a chord of the larger circle and is tangent to the smaller circle. � �has length 140. What is the radius of the larger circle? Solution 210 Let � be the centre of the circles, let � be the point where the chord �� meets the smaller circle and let �be the radius of the smaller circle. ∠ �� � =90 °by the tangent-radius theorem. ∠ � � � =90 °by the angle in a semicircle theorem. Therefore, △�� � is similar to △� � � since both are right-angled and they share ∠��� . Now, ��:� � =�� :� � =1 : 2 . However, � �=140 , so �= 70 and the radius of the larger circle is 3× 70 =210 . 10. What is the smallest 3-digit positive integer �such that 2� +1is a multiple of 5? Solution 102 The powers of 2 (namely 2,4,8,16 ,32 ,64 , . . . ) have units digits which follow a sequence 2,4,8,6, . . . which repeat every four terms. We may calculate that 2100 ,2101 and 2102 have units digits of 6,2and 4respectively. Therefore the first 3-digit power of 2which is one less than a multiple of 5is 2102 . 11. A circle is drawn inside a regular hexagon so that it touches all six sides of the hexagon. The area of the circle is �× 64 √ 3 . What is the area of the hexagon? Solution 384 Let �be the centre of the hexagon. Let � �be an edge of the hexagon with midpoint �. For the circle we have ��2 = �× 64 √ 3 . Therefore �2 = 64 √ 3 . In △� � � ,� � =� � �� � 30°= 2 � √ 3 . The area of the equilateral △� � � is 1 2 × � � ×� � =1 2 × � � ×� � , since � �=� � . Therefore, area △� � � =1 2 × 2 � √ 3 × �= � 2 √ 3 The area of the hexagon is 6× � 2 √ 3 = 6× 64√ 3 √ 3 = 6× 64 = 384 . © 2020 UK Mathematics Trust www.ukmt.org.uk 4

Senior Kangaroo 2020 Solutions12. What is the value of √ 20212020 ×20202021 −20212021 ×20202020 ? Solution 100 Let �= 20202020 . We may express the quantity under the square-root as ( � + 10000 ) × (�+ 1) − ( �+ 10001 ) ×�= �2 + 10001 �+ 10000 −�2 − 10001 �= 10000 . The square root of 10000is100 . 13. How many ordered triples of positive integers (�, �, � )satisfy (� � ) � = 1024 ? Solution 009 Note that 1024=210 , so we require � �to be a factor of 10.Hence the solutions are (1024 ,1,1),(32 ,2,1),(32 ,1,2),(4,5,1),(4,1,5),(2,10 ,1),(2,1,10 ), (2 ,5 ,2 ), (2 ,2 ,5 ). 14. Let �,�,�and � be distinct positive integers such that � + � ,�+ � and �+ � are all odd and are all square. Let � be the least possible value of �+ �+ �+ � . What is the value of 10�? Solution 670 The numbers �,� ,� and �are distinct, so �+ �,� + �and �+ �must also be distinct. The smallest three odd squares which may be formed in this way are 9,25 and 49. Therefore, (� + �) + ( �+ �) + ( �+ �) = 9+ 25 +49 =83 . We may write �=�+ �+ �+ � = 83 −2� . So we may minimise �by maximising �. Since �+ � = 9, the largest possible value for �is 8. This means that �=83 −2× 8= 67 . We must check that a solution exists. The values �= 8,� = 1,�= 17 and � = 41 satisfy all the conditions provided. If one started with a different set of odd squares, then at least one would be 81 or larger. However, �+ �+ �+ � is greater than each of the odd squares involved. Hence 67 is indeed the least possible value. Therefore �is indeed 67and 10�=670 . © 2020 UK Mathematics Trust www.ukmt.org.uk 5

Senior Kangaroo 2020 Solutions15.On an island, kangaroos are always either grey or red. One day, the number of grey kangaroos increased by 28% while the number of red kangaroos decreased by 28% . The ratios of the two types of kangaroos were exactly reversed. By what percentage did the total number of kangaroos change? Solution 004 Let the initial populations of grey and red kangaroos be �and �respectively. After the change, the new populations are 1.28 � and 0.72 �respectively. As the ratios are reversed, we have 1 .28 � 0 .72 � = � � . Therefore �2 = 128 72 × �2 = 16 �2 9 and hence �= 4 � 3 . The ratio of new population:old population is ( 1 .28 � +0.72 �) :(� +�) = (1 .28 +0.72 � � ) :(1 + � � ) = (1 .28 +0.72 ×4 3 : (1 + 4 3 ) = (3 .84 +2.88 ):(4 + 3) = 6.72 : 7 =0.96 : 1 . Therefore there is a reduction of 4% in the population. 16. A square fits snugly between the horizontal line and two touching circles of radius 1000, as shown. The line is tangent to the circles. What is the side-length of the square? Solution 400 Let � be the centre of the left circle, � be the top-left vertex of the square, � be the point at which the left circle meets the tangent and � be the foot of the perpendicular from �to � � . Let the square have side 2�. For simplicity, we let �denote the radius 1000of the circles. The line �� has length �− 1 2 × 2� = �− � , since the common tangent to the circles (shown) is a line of symmetry of the square. Considering △� �� , we have �2 = (� − �)2 + ( �− 2�)2 . This leads to the quadratic 0= ( 5�− �) ( �− �) , which has solutions � = �and �= 0.2 �. Since � < �, we have �= 0.2 � = 0.2 × 1000 =200 . Therefore the square has side-length 2× 200 =400 . © 2020 UK Mathematics Trust www.ukmt.org.uk 6

Senior Kangaroo 2020 Solutions17. How many solutions does equation | | |�− 1| − 1| − 1|= 1have?The modulus function |� | evaluates the absolute value of a number; for example |6 |= | − 6|= 6. Solution 004 Since | | |�− 1| − 1| − 1|= 1we have | |� − 1| − 1| − 1= ±1. Therefore | |� − 1| − 1|= ±1 + 1= 0or 2. Continuing in this way, |� − 1| − 1 = ± 0 or ±2, so |� − 1|= 3,1 or −1, (of which −1 is not possible). Therefore �− 1= ±3 or ±1, so �= −2,0 ,2 or 4. 18. The operation ⋄is defined on two positive whole numbers as the number of distinct prime factors of the product of the two numbers. For example 8⋄ 15 =3. What is the cube of the value of (720 ⋄1001 )? Solution 216 We have the prime factorisations 720=24 × 32 × 5and 1001 =7× 11 ×13 . Therefore 720×1001 will have 6distinct prime factors (namely 2,3 ,5,7 ,11 and 13). The cube of (720 ⋄1001 )is 63 = 216 . 19. A random number generator gives outputs of 1,2 ,3 ,4 and 5with equal probability. The values of �,� and �are each chosen by running the generator once. The probability that � × �+ � is even can be written as a fraction in its lowest terms as � � . What is the value of 10�+�? Solution 715 The expression �× �+ �is even if � �and �are either both odd, or both even. The product � �is odd only if �and �are both odd. This occurs with probability 3 5 × 3 5 = 9 25 . Therefore the probability that both � �and �are odd is 9 25 ×3 5 = 27 125 . The product � �is even with probability 1− 9 25 =16 25 . Therefore the probability that both � �and �are even is 16 25 ×2 5 = 32 125 . The total probability that � �+�is even is 27 125 +32 125 =59 125 . Therefore �=59 ,� =125 and10�+� =10 ×59 +125 =715 . © 2020 UK Mathematics Trust www.ukmt.org.uk 7

Senior Kangaroo 2020 Solutions20.Each square in this cross-number can be filled with a non-zero digit such that all of the conditions in the clues are fulfilled. The digits used are not necessarily distinct. What is the answer to 3 ACROSS? ACROSS DOWN 1. A multiple of 7 1. A multiple of a square of an odd prime; neither a square nor a cube 3. The answer to this Question 2. The internal angle of a regular polygon; the exterior angle is between10� and 20� 5. More than 10 4. A proper factor of 5 ACROSS but not a proper factor of 1 DOWN Solution 961 2 DOWN has possible answers of 162,165 and168. 1 ACROSS must end with a ’1’ so can only be 21or91. 1 DOWN must start with a ’2’ or a ’9’. The squares of odd primes are 9, 25, 49. The only multiples of these which are neither square nor cube and which start ’2’ or ’9’ are 90 (which cannot be correct since 3 ACROSS cannot start with a 0), 98 and 99 . Hence 1 ACROSS is 91. 5 ACROSS could start with a ’2’, ’5’ or ’8’. However, the first digit cannot be ’2’ since none of 21 . . . 29 (for 4 DOWN) has a proper factor which is two digits long and shares the same units digit. Similarly, the first digit cannot be ’5’ since none of 51 . . . 59 (for 4 DOWN) has a proper factor which is two digits long and shares the same units digit. Hence 2 DOWN is 168. Of 81 . . . 89 the only number to have a proper factor which is two digits long and shares the same units digit is 84with the factor 14. So 5 ACROSS is 84 and 4 DOWN is 14. For 1 DOWN we now know that 14 cannot be a factor, so this is 99 . Therefore the answer to 3 ACROSS is 961. © 2020 UK Mathematics Trust www.ukmt.org.uk 8 1 2 3 4 5