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1 UK INTERMEDIATE MATHEMATICAL CHALLENGE February 2nd 2012 EXTENDED SOLUTIONS These solutions augment the printed solutions that we send to schools. For convenience, the solutions sent to schools are confined to two sides of A4 paper and therefore in many cases are rather short. The solutions given here have been extended. In some cases we give alternative solutions, and we have included some Extension Problems for further investigations. The Intermediate Mathematical Challenge (IMC) is a multiple choice contest, in which you are presented with five alternative answers, of which just one is correct. It follows that often you can find the correct answers by working backwards from the given alternatives, or by showing that four of them are not correct. This can be a sensible thing to do in the context of the IMC, and we often give first a solution using this approach. However, this does not provide a full mathematical explanation that would be acceptable if you were just given the question without any alternative answers. So usually we have included a complete solution which does not use the fact that one of the given alternatives is correct. Thus we have aimed to give full solutions with all steps explained. We therefore hope that these solutions can be used as a model for the type of written solution that is expected in the Intermediate Mathematical Olympiad and similar competitions. We welcome comments on these solutions, and, especially, corrections or suggestions for improving them. Please send your comments, either by e-mail to enquiry@ukmt.co.uk or by post to IMC Solutions, UKMT Maths Challenges Office, School of Mathematics, University of Leeds, Leeds LS2 9JT. Quick Marking Guide 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 B D E C D A C A B D C B D A E B E C A D A E C D B Supported by UKMT, 2012. These solutions may be used freely within your school or college. You may, without further permission, post these solutions on a website which is accessible only to staff and students of the school or college, print out and distribute copies within the school or college, and use them within the classroom. If you wish to use them in any other way, please consult us at the address given above.
2 1. How many of the following four numbers are prime? 3 33 333 3333 A 0 B 1 C 2 D 3 E 4 Solution: B The number 3 is prime, but the other numbers listed are not prime as 11333 , 1113333 and 111133333 . Extension problems In general, a positive integer whose digits are all 3s is divisible by 3, since 111...1113333...333 . Hence, except for the number 3 itself, such an integer is not prime. A similar remark applies if 3 is replaced by any of the digits 2, 4, 5, 6, 7, 8 and 9 (except that in the cases of the digits 4, 6, 8 and 9, the number consisting of a single digit is also not prime). This leaves the case of numbers all of whose digits are 1s. This case is considered in the following problems. 1.1 Check which of the numbers 1, 11, 111 and 1111, if any, are prime. 1.2 Show that a positive integer all of whose digits are 1s, and which has an even number of digits, is not prime. 1.3 Show that a positive integer all of whose digits are 1s, and which has a number of digits which is a multiple of 3, is not prime. 1.4 Show that a positive integer all of whose digits are 1s, and which has a number of digits which is not a prime number, is itself not a prime number. 1.5 It follows from 1.4 that a number all of whose digits are 1s can be prime only if it has a prime number of these digits. However numbers of this form need not be prime. Thus 11 with 2 digits is prime, but 111 with 3 digits is not. Determine whether 11111, with 5 digits, is prime. Without a computer quite a lot of arithmetic is needed to answer question 1.5. As numbers get larger it becomes more and more impractical to test by hand whether they are prime. Using a computer we can see that 1111111, 11111111111, 1111111111111 and 11111111111111111 with 7, 11, 13 and 17 digits, respectively, are not prime. In fact, 46492391111111 , 5132392164911111111111 , 26537165379531111111111111 and 5363222357207172311111111111111111 , where the given factors are all prime numbers. The next largest example after 11 of a number all of whose digits are 1s and which is prime is 1111111111111111111 with 19 digits. The next largest has 23 digits, and the next largest after that has 317 digits. It is not known whether there are infinitely many prime numbers of this form. We have taken this information from the book The Penguin Dictionary of Curious and Interesting Numbers by David Wells, Penguin Books, 1986.
3 2. Three positive integers are all different. Their sum is 7. What is their product? A 12 B 10 C 9 D 8 E 5 Solution: D It can be seen that 7421 and 8421 , so assuming that there is just one solution, the answer must be 8. In the context of the IMC, that is enough, but if you are asked to give a full solution, you need to give an argument to show there are no other possibilities. This is not difficult. For suppose a,band care three different positive integers with sum 7, and that cba . If 2a , then 3b and 4c , and so 9 cba . So we must have that 1a . It follows that 6 cb . If 3b then 4c and hence 743 cb . So 2b . Since 1a and 2b , it follows that 4c . 3. An equilateral triangle, a square and a pentagon all have the same side length. The triangle is drawn on and above the top edge of the square and the pentagon is drawn on and below the bottom edge of the square. What is the sum of the interior angles of the resulting polygon? A 0 18010 B 0 1809 C 0 1808 D 0 1807 E 0 1806 Solution: E The sum of the interior angles of the polygon is the sum of the angles in the triangle, the square and the pentagon. The sum of the interior angles of the triangle is 0 180 , and the sum of the angles of the square is 00 1802360 , and the sum of the angles of the pentagon is 00 1803540 . So the sum of the angles is 00 1806180)321( . Note: There is more than one way to see that the sum of the angles of a pentagon is 0 540 . Here is one method. Join the vertices of the pentagon to some point, say P, inside the pentagon. This creates 5 triangles whose angles sum to 0 1805 .The sum of the angles in these triangles is the sum of the angles in a pentagon plus the sum of the angles at P, which is 00 1802360 . So the sum of the angles in the pentagon is 000 180318021805 . Extension Problems 3.1 What is the sum of the angles in a septagon? 3.2 What is the sum of the angles in a polygon with nvertices? 3.3 Does your method in 3.2 apply to a polygon shaped as the one shown where you cannot join all the vertices by straight lines to a point inside the polygon? If not, how could you modify your method to cover this case? P
4 4. All four digits of two 2-digit numbers are different. What is the largest possible sum of two such numbers? A 169 B 174 C 183 D 190 E 197 Solution: C To get the largest possible sum we need to take 9 and 8 as the tens digits, and 7 and 6 as the units digits. For example, 381 68 79 Extension Problem 4.1 All nine digits of three 3-digit numbers are different. What is the largest possible sum of three such numbers? 5. How many minutes will elapse between 20:12 today and 21:02 tomorrow? A 50 B 770 C 1250 D 1490 E 2450 Solution: D From 20:12 today until 20.12 tomorrow is 24 hours, that is 14406024 minutes. There are 50 minutes from 20:12 tomorrow to 21:02 tomorrow. This gives a total of 1490501440 minutes. 6. Triangle QRS is isosceles and right-angled. Beatrice reflects the P-shape in the side QR to get an image. She reflects the first image in the side QS to get a second image. Finally, she reflects the second image in the side RS to get a third image. What does the third image look like? A B C D E Solution: A The effect of the successive reflections is shown in the diagram. S Q R Q R S 1st reflection – in QR 2nd reflection – in SQ 3rd reflection –in RS
5 7. The prime numbers pand qare the smallest primes that differ by 6. What is the sum of pand q? A 12 B 14 C 16 D 20 E 28 Solution: C Suppose q p . Then .6 pq The prime numbers are 2, 3, 5, 7, …. . With 2p , 8q , which is not prime. Similarly if 3p , 9q , which is also not prime. However, when 5p , 11q , which is prime. So, 5p , 11q gives the smallest primes that differ by 6. Then q p 16115 . 8. Seb has been challenged to place the numbers 1 to 9 inclusive in the nine regions formed by the Olympic rings so that there is exactly one number in each region and the sum of the numbers in each ring is 11. The diagram shows part of his solution. What number goes in the region marked * ? A 6 B 4 C 3 D 2 E 1 Solution: A We let u,v,w,x,yand zbe the numbers in the regions shown. Since the sum of the numbers in each ring is 11, we have, from the leftmost ring, that 119 u and so 2u . Then, from the next ring, 1152 v and so 4v . From the rightmost ring, 118 z and so 3z . We have now used the digits 2, 3, 4, 5, 8 and 9, leaving 1, 6 and 7 . From the middle ring we have that 114 xw , and so 7 xw . From the second ring from the right 113 yx , and so 8 yx . So we need to solve the equations 7 xw and 8 yx , using 1, 6 and 7. It is easy to see that the only solution is 1x , 7y and 6w . So 6 goes in the region marked *. 9. Auntie Fi’s dog Itchy has a million fleas. His anti-flea shampoo claims to leave no more than 1% of the original number of fleas after use. What is the least number of fleas that will be eradicated by the treatment? A 900 000 B 990 000 C 999 000 D 999 990 E 999 999 Solution: B Since no more than 1% of the fleas will remain, at least 99% of them will be eradicated. Now 99% of a million is 00099000010990000001 100 99 . 9 5 * 8 9 5 w * 8 v u v x y z
6 10. An ‘abundant’ number is a positive integer N , such that the sum of the factors of N (excluding N itself) is greater than N. . What is the smallest abundant number? A 5 B 6 C 10 D 12 E 15 Solution: D In the IMC, it is only necessary to check the factors of the numbers given as the options. However, to be sure that the smallest of these which is abundant, is the overall smallest abundant number, we would need to check the factors of all the positive integers in turn, until we find an abundant number. The following table gives the sum of the factors of N (excluding N itself), for 121 N . N 1 2 3 4 5 6 7 8 9 10 11 12 factors of N, excluding N - 1 1 1,2 1 1,2,3 1 1,2,4 1,3 1,2,5 1 1,2,3,4,6 sum of these factors 0 1 1 3 1 6 1 7 4 8 1 16 From this table we see that 12 is the smallest abundant number. Extension Problems 10.1. Which is the next smallest abundant number after 12? 10.2. Show that if nis a power of 2, and 2n (that is, n 4, 8, 16, .. etc) then 3 nis an abundant number. 10.3 Prove that if nis an abundant number, then so too is each multiple of n. 10.4 A number, N,is said to be deficient if the sum of the divisors of N, excluding N itself, is less than N. Prove that if Nis a power of 2, then Nis a deficient number. 10.5 A number, N, is said to be perfect if the sum of the divisors of N, excluding N itself, is equal to N. We see from the above table that 6 is the smallest perfect number. Find the next smallest perfect number. Note: It follows from Problems 10.2 and 10.4 that there are infinitely many abundant numbers and infinitely many deficient numbers. It remains an open question as to whether there are infinitely many perfect numbers. In Euclid’s Elements (Book IX, Proposition 36) it is proved that even integers of the form )12(2 1 pp , where 1 2 p is a prime number are perfect (for example, the perfect number 6 corresponds to the case where 2p ). Euclid lived around 2300 years ago. It took almost 2000 years before the great Swiss mathematician Leonard Euler showed that, conversely, all even perfect numbers are of the form )12(2 1 pp , where 1 2 p is prime. Euler lived from 1707 to 1783, but his theorem about perfect numbers was not published until 1849. It is still not known whether there are infinitely many even perfect numbers, as we don’t know whether there are infinitely many primes of the form 1 2 p . It is also not known whether there are any odd perfect numbers.
7 11. In the diagram, PQRS is a parallelogram; 0 50 QRS ; 0 62 SPT and PTPQ . What is the size of TQR ? A 0 84 B 0 90 C 0 96 D 0 112 E 0 124 Solution: C Because PQRS is a parallelogram, 0 50 QRSSPQ . Therefore 00 112)5062( TPQ . Therefore, as the angles in a triangle add up to 0 180 , 000 68112180 PTQPQT . Because PQ=PT , the triangle QPT is isosceles, and so PTQPQT . Therefore 0 34 PTQPQT . Because PQRS is a parallelogram, 0 180 QRSPQR , and therefore 000 13050180 PQR . Therefore, .9634130000 PQTPQRTQR 12. Which of the following has a different value from the others? A 18% of £30 B 12% of £50 C 6% of £90 D 4% of £135 E 2% of £270 Solution: B We have that 40.5£)30£(30£of%18 10018 . Similarly, 12% of £50 is £6.00, and 6% of £90 is £5.40. We already see that option B must be the odd one out. It is easy to check that 4% of £135 and 2% of £270 are also both £5.40. 13. Alex Erlich and Paneth Farcas shared an opening rally of 2 hours and 12 minutes during their table tennis match at the 1936 World Games. Each player hit around 45 shots per minute. Which of the following is closest to the total number of shots played in the rally? A 200 B 2000 C 8000 D 12 000 E 20 000 Solution: D Since they each hit about 45 shots in one minute, between them they hit about 90 shots per minute. Now 2 hours and 12 minutes is 132 minutes. So the total number of shots in the match is 13290 , and 13290 is approximately 00012120100 . Extension Problem 13.1 Note that 90 is 90% of 100 and 132 is 110% of 120. What is the percentage error in approximating 13290 by 120100 ? 620 50 0 P Q R S T
8 14. What value of xmakes the mean of the first three numbers in this list equal to the mean of the last four? 15 5 x 7 9 17 A 19 B 21 C 24 D 25 E 27 Solution: A The mean of the first three numbers in the list is )515( 31 x and the mean of the last four is )1797( 41 x . Now, )1797(3)515(4)1797()515( 41 31 xxxx 993480 xx 19 x . An alternative method in the context of the IMC would be just to try the given options in turn. This runs the risk of involving a lot of arithmetic, but here, as the first option is the correct answer, the gamble would pay off. 15. Which of the following has a value that is closest to 0? A 4 1 3 1 2 1 B 4 1 3 1 2 1 C 4 1 3 1 2 1 D 4 1 3 1 2 1 E 4 1 3 1 2 1 Solution: E When working out the values of these expressions it is important to remember the convention (sometimes known as BODMAS or BIDMAS) that tells us that Divisions and Multiplications are carried out before Additions and Subtractions. Some work can be saved by noting that the expressions A and B have values greater than 21, whereas the value of expression E lies between 0 and 21. So it must be C, D or E that has the value closest to 0. Now, noting that 3 4 1 4 3 1 4 1 3 1 , we obtain that the value of C is 3 2 3 4 2 1 4 1 3 1 2 1 ; that of D is 6 5 3 4 2 1 4 1 3 1 2 1 ; and that of E is 12 5 12 1 2 1 4 1 3 1 2 1 . From these calculations we see that E gives the value closest to 0. D 0 E 21A C B 6 5 12 5 12 7 3 2 6 11 [The value of A is 12 7 12 1 2 1 4 1 3 1 2 1 ; and that of B is B 6 11 3 4 2 1 4 1 3 1 2 1 .]
9 16. The diagram shows a large equilateral triangle divided by three straight lines into seven regions. The three grey regions are equilateral triangles with sides of length 5 cm and the central black region is an equilateral triangle with sides of length 2 cm. What is the side length of the original large triangle? A 18 cm B 19 cm C 20 cm D 21 cm E 22cm Solution : B Let P, Q, R, S, T, U and V be the points shown. All the angles in all the triangles are 0 60 . So PSUQRT and hence RT is parallel to SU. Similarly, as TUVRSV ,RS is parallel to TU. Therefore RSUT is a parallelogram. Therefore RS has the same length as TU , namely, 752 cm. Similarly PQ has length 7 cm. So the length of PS which is the sum of the lengths of PQ, QS and RS is 19757 cm. 17. The first term in a sequence of positive integers is 6. The other terms in the sequence follow these rules: if a term is even then divide it by 2 to obtain the next term; if a term is odd then multiply it by 5 and subtract 1 to obtain the next term. For which values of nis the nth term equal to n? A 10 only B 13 only C 16 only D 10 and 13 only E 13 and 16 only Solution: E Since the options refer only to the 10th, 13th and 16th terms of the sequence, as far as this IMC question is concerned it is only necessary to check the first 16 terms in the sequence. These are as shown in the table below: From this we see that the 13th term is 13, and the 16th term is 16, and that these are the only cases where the nth term is equal to n. However, a complete answer requires a proof that for all 16n , the nth term is not equal to n. It can be seen that after the 16th term the sequence continues 8, 4, 2, 1, 4, 2, 1… with the cycle 4, 2, 1 now repeating for ever. It follows that, for 17n , the only values taken by the nth term are 8, 4, 2 and 1. We deduce that for 16n , the nth term is not equal to n. P Q R S T U V 1616 1313 232422848 32264158415177 64151314172346 226341575 262521272144 52210411141533 1041521103262 21242961 th termth term nnnn
10 18. Peri the winkle starts at the origin and slithers anticlockwise around a semicircle with centre )0,4(. Peri then slides anticlockwise around a second semicircle with centre )0,6(, and finally clockwise around a third semicircle with centre )0,3(. Where does Peri end this expedition? A )0,0( B )0,1( C )0,2( D )0,4( E )0,6( Solution: C As may be seen from the diagram, Peri first moves along the semicircle with centre )0,4(from the point )0,0(to the point )0,8(, then along the semicircle with centre )0,6(to the point )0,4(, and finally along the semicircle with centre )0,3(to end up at the point )0,2(. 19. The shaded region shown in the diagram is bounded by four arcs, each of the same radius as that of the surrounding circle. What fraction of the surrounding circle is shaded? A 1 4 B 4 1 C 2 1 D 3 1 E it depends on the radius of the circle Solution: A Suppose that the surrounding circle has radius r. In the diagram we have drawn the square with side length 2 rwhich touches the circle at the points where it meets the arcs. The square has area 22 4)2(rr . The unshaded area inside the square is made up of four quarter circles with radius r, and thus has area 2r . Hence the shaded area is 222)4(4rrr . The circle has area 2r .So the fraction of the circle that is shaded is 1 44)4( 2 2 r r . 20. A rectangle with area 125 cm 2has sides in the ratio 4:5. What is the perimeter of the rectangle? A 18 cm B 22.5 cm C 36 cm D 45 cm E 54 cm Solution: D Since the side lengths of the rectangle are in the ratio 4:5, they are 4 acm and 5 acm, for some positive number a. This means that the rectangle has area 2 2054 aaa cm 2. Hence 12520 2 a . So 2a 4 25 20 125 , and hence 2 5a . Hence the rectangle has perimeter 2 5 1818)54(2 aaa 45 cm. 3
11 21. The parallelogram PQRS is formed by joining together four equilateral triangles of side 1 unit, as shown. What is the length of the diagonal SQ ? A 7 B 8 C 3 D 6 E 5 Solution: A Let Tbe the foot of the perpendicular from Q to the line SR extended. Now RQT is half of an equilateral triangle with side length 1. Hence the length of RT is 21and hence ST has length 25 21 11 . By Pythagoras’ Theorem applied to the right angled triangle RQT , 22 21 2 )(1 QT . Therefore 43 41 221 22 1)()1( QT . Hence, by Pythagoras’ Theorem applied to the right angled triangle SQT , 7)( 43 425 43 225 222 QTSTSQ . Therefore, 7 SQ . 22. What is the maximum possible value of the median number of cups of coffee bought per customer on a day when Sundollars Coffee Shop sells 477 cups of coffee to 190 customers, and every customer buys at least one cup of coffee? A 1.5 B 2 C 2.5 D 3 E 3.5 Solution: E Put the set of numbers of cups of coffee drunk by the individual customers into numerical order with the smallest first. This gives an increasing sequence of positive integers with sum 477. Because 190 is even, the median of these numbers is the mean of the 95th and 96th numbers in this list. Suppose these are aand b, respectively. Then the median is )( 21 ba . We note that ba 1 . Also, each of the first 94 numbers in the list is between 1 and a, and each of the last 94 numbers is at least b. So if we replace the first 94 numbers by 1, and the last 94 numbers by b, we obtain the sequence of numbers 9594 .....,,,,,,1...,1,1,1, bbbba (1) whose sum does not exceed 477, the sum of the original sequence. Therefore 4779594 ba (2) As a1 , it follows that 47795949595 bab , hence 3829547795 b and therefore P Q R S P Q S R T
12 95 382b . Therefore, since bis an integer, 4b . When 4b , it follows from (2) that 47738094 a , giving 3a . This shows that the maximum possible values for aand bare 3 and 4, respectively. We can see that these values are possible, as, if we substitute these values in (1), we obtain a sequence of numbers with sum 4774953194 . So 3.5 is the maximum possible value of the median. Extension Problem 22.1 What is the maximum possible value of the median number of cups of coffee bought per customer on a day when the Sundollars Coffee Shop sells 201 cups of coffee to 100 customers, and every customer buys at least one cup of coffee? 23. In the triangle PQR , 2 PS ; 1 SR ; 0 45 PRQ ; Tis the foot of the perpendicular from Pto QS and 0 60 PST . What is the size of QPR ? A 45 0 B 60 0 C 75 0 D 90 0 E 105 0 Solution: C In the triangle PST , 0 90 PTS and 0 60 PST . Therefore 0 30 TPS and the triangle PST is half of an equilateral triangle. It follows that 1 21 PSST .Therefore triangle RST is isosceles, and hence SRTSTR .By the Exterior Angle Theorem, PST SRTSTR .Therefore 0 30 SRTSTR .Hence 000 153045 SRTPRQQRT . Using the Exterior Angle Theorem again, it follows that QRTTQRSTR , and hence 00 3045 QRTSTRTQR 0 15 . Therefore the base angles of triangle TQR are equal. Hence TQR is an isosceles triangle, and so RTQT . We also have that the base angles in triangle TPR are both equal to 0 30 , and so RT PT . Therefore PTRTQT . So PTQ is an isosceles right-angled triangle. Therefore 0 45 QPT . Finally, we deduce that 000 753045 TPSQPTQPR . P Q R T S 600 450 2 1 P S Q R T 600 150 150300 300 300 2 1 450 450
13 24. All the positive integers are written in the cells of a square grid. Starting from 1, the numbers spiral anticlockwise. The first part of the spiral is shown in the diagram. Which number is immediately below 2012? A 1837 B 2011 C 2013 D 2195 E 2210 Solution: D The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4. The squares of the even numbers have the form 2)2(n , that is, 2 4n .We see that the number 14 2 n occurs to the left of the cell containing 2 4n . Below 14 2 n there occur the numbers 124,..,34,24 222 nnnn , and then in the cells to the right of the cell containing 124 2 nn , there occur the numbers 144,...,324,224 222 nnnnnn 2)12( n . Now 193644 2 and 202545 2 . Thus 2011 is in the same row as 2025 and to the left of it, in the sequence 1981,….,2012,…,2025, and below these occur the numbers 2163,…., 2 472209 , with 2208 below 2025 as shown below. It follows that 2195 is the number below 2012. In the diagram below the square numbers are shown in bold. … 32 31 17 16 15 14 13 30 18 5 4 3 12 29 19 6 1 2 11 28 20 7 8 9 10 27 21 22 23 24 25 26 1937 1936 … … … … … … … … … … … … … … 32 31 17 16 15 14 13 30 18 5 4 3 12 29 19 6 1 2 11 28 20 7 8 9 10 27 21 22 23 24 25 26 1981 1982 … … … … … … … … 2012 … 2024 2025 … 2164 2165 … … … … … … … … 2195 … 2207 2208 2209
14 25. The diagram shows a ceramic design by the Catalan architect Antoni Gaudi. It is formed by drawing eight lines connecting points which divide the edges of the outer regular octagon into three equal parts, as shown. What fraction of the octagon is shaded? A 5 1 B 9 2 C 4 1 D 10 3 E 16 5 Solution: B We consider the triangular segment of the octagon formed by joining two adjacent vertices, Pand Q to the centre, O. For convenience, we show this segment, drawn on a larger scale, on the left, where we have added the lines RW ,ST ,TW and UV. These lines are parallel to the edges of the triangle POQ ,as shown and together with the lines RU and SV they divide the triangle OPQ into 9 congruent triangles, of which 2 are shaded. Thus 9 2 of the segment is shaded. The same holds for all the other congruent segments of the octagon. So 9 2 of the whole octagon is shaded. Extension Problem 25.1 In the solution we have said that the triangle OPQ is divided into 9 congruent triangles, but we have not justified the claim that the triangles are congruent. Complete the argument by giving a proof that these triangles are congruent. P Q O P Q O R S T U V W
22. E The median number of cups of coffee is the median of a sequence of 190 positive integers . Let the sum of these terms be . (t1,t2, ,t190) S The median of the 190 numbers is . The alternatives imply that the median cannot be greater than 3.5. The next lowest possible value for the median would be 4. For this to be possible, . 12(t95+ t96) t95+ t96= 8 If then the minimum value for would occur if all other values of were as small as possible, that is the first 94 values would all equal 1 and the last 94 values would all equal 4. In this case, , whereas we are told that 477 cups of coffee were sold. Any other values of and such that would produce a larger minimum value of . For example, if and then the minimum value of would be , that is 572. So the median of the 190 terms cannot be 4, but it is possible for it to be 3.5. If the first 94 terms all equal 1, and and the last 94 terms all equal 4 then as required and the median is . t95=t96=4 S t S=94×1+2×4+94×4=478 t95 t96 t95+ t96= 8 S t95= 3 t96= 5 S 94×1+3+5+94×5 t95= 3 t96= 4 S = 477 12(3+4)=3.5 So the maximum possible value of the median number of cups of coffee bought per customer is 3.5. 23. C As in the solution for Q21, may be thought of as half an equilateral triangle, so S has length 1 unit. Therefore is isosceles and, as , . So . Using the exterior angle theorem in , . PTS T SRT TSR=120°SRT=STR=30° TRQ =45°-30°=15° TQR TQR=STR-TRQ =30°-15°=15° So is isosceles with . However, is also isosceles with since . Therefore , from which we deduce that is an isosceles right-angled triangle in which . So . TQR TQ =TR PRT PT=TR PRT=TPR=30° TQ =TP PQT PQT = QPT = 45° QPR = QPT + TPS = 45°+ 30°= 75° 24. D The nature of the spiral means that 4 is in the top left-hand corner of a square of cells, 9 is in the bottom right-hand corner of a square of cells, 16 is in the top left-hand corner of a square of cells and so on. To find the position of 2012 in the grid, we note that so 2025 is in the bottom right-hand corner of a square of cells and note also that . The table below shows the part of the grid in which 2012 lies. The top row shows the last 15 cells in the bottom row of a square of cells, whilst below it are the last 16 cells in the bottom row of a square of cells. 2×2 3×3 4×4 452=2025 45×45 472= 2209 45×45 47×47 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 2194 2195 2196 2197 2198 2199 2200 2201 2202 2203 2204 2205 2206 2207 2208 2209 So 2195 lies below 2012. 25. B The diagram shows part of the ceramic. and are vertices of the outer octagon, which has at its centre. The solid lines are part of the original figure, whilst the broken lines , , two broken lines which are parallel to and broken lines parallel to and respectively have been added. As can be seen, these lines divide into nine congruent triangles. The shaded A B O OA OB AB OA OB OAB O B A portion of triangle has area equal to that of two of the triangles. So of the area of has been shaded. Now the area of the outer octagon is eight times the area of and the area of shaded portion of the design is eight times the area of the shaded portion of so the fraction of the octagon which is shaded is also . 29 OAB OAB OAB 29 UKMT UKMT UKMT UK I NTERMEDIATE M ATHEMATICAL C HALLENGE THURSDAY 2nd FEBRUARY 2012 Organised by the United Kingdom Mathematics Trust from the School of Mathematics, University of Leeds http://www.ukmt.org.uk SOLUTIONS LEAFLET This solutions leaflet for the IMC 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. The UKMT is a registered charity 1. B 3 is the only one of the four numbers which is prime. The sums of the digits of the other three numbers are 6, 9, 12 respectively. These are all multiples of 3, so 33, 333, 3333 are all multiples of 3. 2. D The following triples of positive integers all sum to 7: (1, 1, 5), (1, 2, 4), (1, 3, 3); (2, 2, 3). In only one of these are the three integers all different, so the required integers are 1, 2, 4 and their product is 8. 3. E The diagram shows that the interior angles of the polygon may be divided up to form the interior angles of six triangles. So their sum is . 6× 180°
4. C The digits to be used must be 9, 8, 7, 6. If any of these were to be replaced by a smaller digit, then the sum of the two two-digit numbers would be reduced. For this sum to be as large as possible, 9 and 8 must appear in the tens column rather than the units column. So the largest possible sum is 97 + 86 or 96 + 87. In both cases the sum is 183. 5. D The difference between the two given times is 24 hours 50 minutes = ( 24 × 60 + 50) minutes = (1440 + 50) minutes = 1490 minutes. 6. A The diagram shows the result of the successive reflections. P2 P3 Q R S P P1 Q R S P1 P2 Q R S 7. C The primes in question are 5 and 11. The only primes smaller than 5 are 2 and 3. However neither 8 nor 9 is prime so and cannot be 2 and 8, nor 3 and 9. p q 8. A Referring to the diagram, ; ; . So the values of , and are 1, 6, 7 in some order. a=11-9=2 b=11-5-a=4 f=11-8=3 cd e If then , but then would need to be 2, not 7. c = 1 d = 6 e 9 8 5 c a b de f If , then and and this is a valid solution. Finally, if then would need to equal 0, which is not possible. So in the only possible solution, * is replaced by 6. c = 6 d = 1 e= 7 c = 7 d 9. B 1% of 1 000 000 = 1 000 000 ÷ 100 = 10 000. So the least number of fleas which will be eradicated is 1 000 000 10 000 = 990 000. 10. D The table shows the first 12 positive integers, , and the sum, , of the factors of excluding itself. As can be seen, 12 is the first value of for which this sum exceeds , so 12 is the smallest abundant number. N S N N N N N 1 2 3 4 5 6 7 8 9 10 11 12 S 0 1 1 3 1 6 1 7 4 8 1 16 (Note that for the sum also equals 6. For this reason, 6 is known as a perfect number. After 6, the next two perfect numbers are 28 and 496.) N = 6 11. C Opposite angles of a parallelogram are equal so . QPS = 50° Therefore and, as triangle is isosceles, . QPT=112° QPT PQT=(180°-112°)÷2=34° As is a parallelogram, . So . PQRS PQR=180°-50°=130° TQR=130°-34°=96° 12. B The values of the expressions are £5.40, £6.00, £5.40, £5.40, £5.40 respectively. 13. D In the rally, approximately 90 shots were hit per minute for a total of 132 minutes. As , D is the best alternative. 90× 130 = 11 700 14. A The mean of the first three numbers is ; the mean of the last four numbers is . Therefore , that is , so . 13(20+ x) 14(33+ x) 4(20+ x)= 3(33+ x) 80+ 4x = 99+ 3x x = 99- 80 = 19 15. E ; ; ; ; . Of the fractions , the closest to 0 is . 1 2+1 3×1 4=1 2+ 1 12= 7 12 1 2+1 3÷1 4=1 2+1 3×4 1=1 2+4 3=11 6 1 2×1 3÷1 4= 1 2×1 3×4 1=2 3 1 21 3÷1 4=1 21 3×4 1=1 2-4 3=-5 6 1 21 3×1 4=1 2- 1 12= 5 12 7 12,11 6,2 3,-5 6,5 12 5 12 16. B As triangle is equilateral, . From the symmetry of the figure, we may deduce that so triangle is equilateral. The length of the side of this equilateral triangle = length of = (5 + 2 + 5) cm =12 cm. So . By a similar argument, we deduce that = 7 cm, so the length of the side of triangle . ABC BAC = 60° AD = DE ADE DE AF = AD - AF = (12- 5)cm = 7cm BD ABC = (7+ 5+ 7)cm = 19cm A B C D E F 17. E The terms of the sequence are 6, 3, 14, 7, 34, 17, 84, 42, 21, 104, 52, 26, 13, 64, 32, 16, 8, 4, 2, 1, 4, 2, 1, . As can be seen, there will now be no other terms in the sequence other than 4, 2 and 1. It can also be seen that the only values of for which the th term = are 13 and 16. n n n 18. C After traversing the first semicircle, Peri will be at the point (8, 0); after the second semicircle Peri will be at (4, 0) and after the third semicircle, Peri will be at the point (2, 0). 19. A The diagram shows the original diagram enclosed within a square of side , where is the radius of the original circle. The unshaded area of the square consists of four quadrants (quarter circles) of radius . So the shaded area is . Therefore the required fraction is 2r r r 4r2- pr2 = r2(4- p) r2(4- p) pr2 = 4- p p = 4 p - 1. 20. D Let the sides of the rectangle, in cm, be and respectively. 4x 5x Then the area of the square is . So , that is . 4x×5xcm 2=20x2cm 2 20x2=125 x2=254 Therefore , but cannot be negative so the sides of the rectangle are 10 cm and 12.5 cm. Hence the rectangle has perimeter 45 cm. x = ±52 x 21. A In the diagram, is the foot of the perpendicular from to produced. Angles and are alternate angles between parallel lines so . Triangle has interior angles T Q SR PQR QRT QRT = 60° QRT P Q R T S of so it may be thought of as being half of an equilateral triangle of side 1 unit, since the length of is 1 unit. So the lengths of and are and units respectively. 90°, 60°, 30° QR RT QT 12 32 Applying Pythagoras' Theorem to , . So the length of is units. QST SQ2=ST2+QT2=(52)2+(32)2=254+34=7 SQ 7
4. C The digits to be used must be 9, 8, 7, 6. If any of these were to be replaced by a smaller digit, then the sum of the two two-digit numbers would be reduced. For this sum to be as large as possible, 9 and 8 must appear in the tens column rather than the units column. So the largest possible sum is 97 + 86 or 96 + 87. In both cases the sum is 183. 5. D The difference between the two given times is 24 hours 50 minutes = ( 24 × 60 + 50) minutes = (1440 + 50) minutes = 1490 minutes. 6. A The diagram shows the result of the successive reflections. P2 P3 Q R S P P1 Q R S P1 P2 Q R S 7. C The primes in question are 5 and 11. The only primes smaller than 5 are 2 and 3. However neither 8 nor 9 is prime so and cannot be 2 and 8, nor 3 and 9. p q 8. A Referring to the diagram, ; ; . So the values of , and are 1, 6, 7 in some order. a=11-9=2 b=11-5-a=4 f=11-8=3 cd e If then , but then would need to be 2, not 7. c = 1 d = 6 e 9 8 5 c a b de f If , then and and this is a valid solution. Finally, if then would need to equal 0, which is not possible. So in the only possible solution, * is replaced by 6. c = 6 d = 1 e= 7 c = 7 d 9. B 1% of 1 000 000 = 1 000 000 ÷ 100 = 10 000. So the least number of fleas which will be eradicated is 1 000 000 10 000 = 990 000. 10. D The table shows the first 12 positive integers, , and the sum, , of the factors of excluding itself. As can be seen, 12 is the first value of for which this sum exceeds , so 12 is the smallest abundant number. N S N N N N N 1 2 3 4 5 6 7 8 9 10 11 12 S 0 1 1 3 1 6 1 7 4 8 1 16 (Note that for the sum also equals 6. For this reason, 6 is known as a perfect number. After 6, the next two perfect numbers are 28 and 496.) N = 6 11. C Opposite angles of a parallelogram are equal so . QPS = 50° Therefore and, as triangle is isosceles, . QPT=112° QPT PQT=(180°-112°)÷2=34° As is a parallelogram, . So . PQRS PQR=180°-50°=130° TQR=130°-34°=96° 12. B The values of the expressions are £5.40, £6.00, £5.40, £5.40, £5.40 respectively. 13. D In the rally, approximately 90 shots were hit per minute for a total of 132 minutes. As , D is the best alternative. 90× 130 = 11 700 14. A The mean of the first three numbers is ; the mean of the last four numbers is . Therefore , that is , so . 13(20+ x) 14(33+ x) 4(20+ x)= 3(33+ x) 80+ 4x = 99+ 3x x = 99- 80 = 19 15. E ; ; ; ; . Of the fractions , the closest to 0 is . 1 2+1 3×1 4=1 2+ 1 12= 7 12 1 2+1 3÷1 4=1 2+1 3×4 1=1 2+4 3=11 6 1 2×1 3÷1 4= 1 2×1 3×4 1=2 3 1 21 3÷1 4=1 21 3×4 1=1 2-4 3=-5 6 1 21 3×1 4=1 2- 1 12= 5 12 7 12,11 6,2 3,-5 6,5 12 5 12 16. B As triangle is equilateral, . From the symmetry of the figure, we may deduce that so triangle is equilateral. The length of the side of this equilateral triangle = length of = (5 + 2 + 5) cm =12 cm. So . By a similar argument, we deduce that = 7 cm, so the length of the side of triangle . ABC BAC = 60° AD = DE ADE DE AF = AD - AF = (12- 5)cm = 7cm BD ABC = (7+ 5+ 7)cm = 19cm A B C D E F 17. E The terms of the sequence are 6, 3, 14, 7, 34, 17, 84, 42, 21, 104, 52, 26, 13, 64, 32, 16, 8, 4, 2, 1, 4, 2, 1, . As can be seen, there will now be no other terms in the sequence other than 4, 2 and 1. It can also be seen that the only values of for which the th term = are 13 and 16. n n n 18. C After traversing the first semicircle, Peri will be at the point (8, 0); after the second semicircle Peri will be at (4, 0) and after the third semicircle, Peri will be at the point (2, 0). 19. A The diagram shows the original diagram enclosed within a square of side , where is the radius of the original circle. The unshaded area of the square consists of four quadrants (quarter circles) of radius . So the shaded area is . Therefore the required fraction is 2r r r 4r2- pr2 = r2(4- p) r2(4- p) pr2 = 4- p p = 4 p - 1. 20. D Let the sides of the rectangle, in cm, be and respectively. 4x 5x Then the area of the square is . So , that is . 4x×5xcm 2=20x2cm 2 20x2=125 x2=254 Therefore , but cannot be negative so the sides of the rectangle are 10 cm and 12.5 cm. Hence the rectangle has perimeter 45 cm. x = ±52 x 21. A In the diagram, is the foot of the perpendicular from to produced. Angles and are alternate angles between parallel lines so . Triangle has interior angles T Q SR PQR QRT QRT = 60° QRT P Q R T S of so it may be thought of as being half of an equilateral triangle of side 1 unit, since the length of is 1 unit. So the lengths of and are and units respectively. 90°, 60°, 30° QR RT QT 12 32 Applying Pythagoras' Theorem to , . So the length of is units. QST SQ2=ST2+QT2=(52)2+(32)2=254+34=7 SQ 7
22. E The median number of cups of coffee is the median of a sequence of 190 positive integers . Let the sum of these terms be . (t1,t2, ,t190) S The median of the 190 numbers is . The alternatives imply that the median cannot be greater than 3.5. The next lowest possible value for the median would be 4. For this to be possible, . 12(t95+ t96) t95+ t96= 8 If then the minimum value for would occur if all other values of were as small as possible, that is the first 94 values would all equal 1 and the last 94 values would all equal 4. In this case, , whereas we are told that 477 cups of coffee were sold. Any other values of and such that would produce a larger minimum value of . For example, if and then the minimum value of would be , that is 572. So the median of the 190 terms cannot be 4, but it is possible for it to be 3.5. If the first 94 terms all equal 1, and and the last 94 terms all equal 4 then as required and the median is . t95=t96=4 S t S=94×1+2×4+94×4=478 t95 t96 t95+ t96= 8 S t95= 3 t96= 5 S 94×1+3+5+94×5 t95= 3 t96= 4 S = 477 12(3+4)=3.5 So the maximum possible value of the median number of cups of coffee bought per customer is 3.5. 23. C As in the solution for Q21, may be thought of as half an equilateral triangle, so S has length 1 unit. Therefore is isosceles and, as , . So . Using the exterior angle theorem in , . PTS T SRT TSR=120°SRT=STR=30° TRQ =45°-30°=15° TQR TQR=STR-TRQ =30°-15°=15° So is isosceles with . However, is also isosceles with since . Therefore , from which we deduce that is an isosceles right-angled triangle in which . So . TQR TQ =TR PRT PT=TR PRT=TPR=30° TQ =TP PQT PQT = QPT = 45° QPR = QPT + TPS = 45°+ 30°= 75° 24. D The nature of the spiral means that 4 is in the top left-hand corner of a square of cells, 9 is in the bottom right-hand corner of a square of cells, 16 is in the top left-hand corner of a square of cells and so on. To find the position of 2012 in the grid, we note that so 2025 is in the bottom right-hand corner of a square of cells and note also that . The table below shows the part of the grid in which 2012 lies. The top row shows the last 15 cells in the bottom row of a square of cells, whilst below it are the last 16 cells in the bottom row of a square of cells. 2×2 3×3 4×4 452=2025 45×45 472= 2209 45×45 47×47 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 2194 2195 2196 2197 2198 2199 2200 2201 2202 2203 2204 2205 2206 2207 2208 2209 So 2195 lies below 2012. 25. B The diagram shows part of the ceramic. and are vertices of the outer octagon, which has at its centre. The solid lines are part of the original figure, whilst the broken lines , , two broken lines which are parallel to and broken lines parallel to and respectively have been added. As can be seen, these lines divide into nine congruent triangles. The shaded A B O OA OB AB OA OB OAB O B A portion of triangle has area equal to that of two of the triangles. So of the area of has been shaded. Now the area of the outer octagon is eight times the area of and the area of shaded portion of the design is eight times the area of the shaded portion of so the fraction of the octagon which is shaded is also . 29 OAB OAB OAB 29 UKMT UKMT UKMT UK I NTERMEDIATE M ATHEMATICAL C HALLENGE THURSDAY 2nd FEBRUARY 2012 Organised by the United Kingdom Mathematics Trust from the School of Mathematics, University of Leeds http://www.ukmt.org.uk SOLUTIONS LEAFLET This solutions leaflet for the IMC 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. The UKMT is a registered charity 1. B 3 is the only one of the four numbers which is prime. The sums of the digits of the other three numbers are 6, 9, 12 respectively. These are all multiples of 3, so 33, 333, 3333 are all multiples of 3. 2. D The following triples of positive integers all sum to 7: (1, 1, 5), (1, 2, 4), (1, 3, 3); (2, 2, 3). In only one of these are the three integers all different, so the required integers are 1, 2, 4 and their product is 8. 3. E The diagram shows that the interior angles of the polygon may be divided up to form the interior angles of six triangles. So their sum is . 6× 180°