UK INTERMEDIATE MATHEMATICAL CHALLENGE February …

1 1UK INTERMEDIATE MATHEMATICAL CHALLENGEF ebruary 2nd 2012 EXTENDED SOLUTIONST hese solutions augment the printed solutions that we send to schools. For convenience, the solutionssent to schools are confined to two sides of A4 paper and therefore in many cases are rather short. Thesolutions given here have been extended. In some cases we give alternative solutions, and we haveincluded someExtension Problemsfor further INTERMEDIATE MATHEMATICAL CHALLENGE (IMC) is a multiple choice contest, in which you arepresented with five alternative answers, of which just one is correct. It follows that often you can findthe correct answers by working backwards from the given alternatives, or by showing that four ofthem are not correct. This can be a sensible thing to do in the context of the IMC, and we often givefirst a solution using this , this does not provide a full MATHEMATICAL explanation that would be acceptable if you werejust given the question without any alternative answers.

2 So usually we have included a completesolution which does not use the fact that one of the given alternatives is correct. Thus we have aimedto give full solutions with all steps explained. We therefore hope that these solutions can be used as amodel for the type of written solution that is expected in the INTERMEDIATE MATHEMATICAL Olympiad andsimilar welcome comments on these solutions, and, especially, corrections or suggestions for improvingthem. Please send your comments,either by e-mail by post toIMC Solutions, UKMT Maths Challenges Office, School of Mathematics,University of Leeds, Leeds LS2 Marking Guide12345678910111213141516171819202122 232425 BDECDACABDCBDAEBECADAECDBS upported by UKMT, 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 andstudents of the school or college, print out and distribute copies within the school or college, and usethem within the classroom.

3 If you wish to use them in any other way, please consult us at the addressgiven many of the following four numbers are prime?3333333333A 0B 1C 2D 3E 4 Solution:BThe number 3 is prime, but the other numbers listed are not prime as11333 ,1113333 and111133333 .Extension problemsIn general, a positive integer whose digits are all 3s is divisible by 3, .Hence, except for the number 3 itself, such an integer is not prime. A similar remark applies if 3 isreplaced 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 whosedigits are 1s. This case is considered in the following Check which of the numbers 1, 11, 111 and 1111, if any, are Show that a positive integer all of whose digits are 1s, and which has an even number of digits, isnot Show that a positive integer all of whose digits are 1s, and which has a number of digits which isa multiple of 3, is not Show that a positive integer all of whose digits are 1s, and which has a number of digits which isnot a prime number, is itself not a prime It follows from that a number all of whose digits are 1s can be prime only if it has a primenumber of these digits.

4 However numbers of this form need not be prime. Thus 11 with 2 digitsis prime, but 111 with 3 digits is not. Determine whether 11111, with 5 digits, is a computer quite a lot of arithmetic is needed to answer question As numbers get largerit becomes more and more impractical to test by hand whether they are prime. Using a computer wecan see that 1111111, 11111111111, 1111111111111 and 11111111111111111 with 7, 11, 13 and 17digits, respectively, are not prime. In fact,46492391111111 ,5132392164911111111111 ,26537165379531111111111111 and5363222357207172311111111111111111 , where thegiven factors are all prime next largest example after 11 of a number all of whose digits are 1s and which is prime is1111111111111111111 with 19 digits. The next largest has 23 digits, and the next largest after thathas 317 digits. It is not known whether there are infinitely many prime numbers of this form.

5 We havetaken this information from the bookThe Penguin Dictionary of Curious and Interesting NumbersbyDavid Wells, Penguin Books, positive integers are all different. Their sum is 7. What is their product?A 12B 10C 9D 8E 5 Solution:DIt can be seen that7421 and8421 , so assuming that there is just one solution, theanswer must be 8. In the context of the IMC, that is enough, but if you are asked to give a fullsolution, you need to give an argument to show there are no other possibilities. This is not supposea,bandcare three different positive integers with sum 7, and thatcba . If2 a,then3 band4 c, and so9 cba. So we must have that1 a. It follows that6 cb. If3 bthen4 cand hence743 cb. So2 b. Since1 aand2 b, it follows that4 equilateral triangle, a square and a pentagon all have the same side triangle is drawn on and above the top edge of the square and the pentagonis drawn on and below the bottom edge of the square.

6 What is the sum of theinterior angles of the resulting polygon?A018010 B01809 C01808 D01807 E01806 Solution:EThe sum of the interior angles of the polygon is the sum of the angles in the triangle, the square andthe pentagon. The sum of the interior angles of the triangle is0180, and the sum of the angles of thesquare is001802360 , and the sum of the angles of the pentagon is001803540 .So the sum of the angles is001806180)321( .Note:There is more than one way to see that the sum of the angles of a pentagon is one method. Join the vertices of the pentagon to some point, sayP, inside the pentagon. Thiscreates 5 triangles whose angles sum to01805 .The sum of the angles in these triangles is the sum ofthe angles in a pentagon plus the sum of the angles atP, which is001802360 . So the sum of theangles in the pentagon is000180318021805 .Extension is the sum of the angles in a septagon? is the sum of the angles in a polygon withnvertices?

7 Your method in apply to a polygon shaped as the one shownwhere you cannot join all the vertices by straight lines to a point insidethe polygon? If not, how could you modify your method to cover thiscase? four digits of two 2-digit numbers are different. What is the largest possible sum of twosuch numbers?A 169B 174C 183D 190E 197 Solution:CTo get the largest possible sum we need to take 9 and 8 as the tens digits, and 7 and 6 as the unitsdigits. For example,3816879 Extension nine digits of three 3-digit numbers are different. What is the largest possible sum of threesuch numbers? many minutes will elapse between 20:12 today and 21:02 tomorrow?A 50B 770C 1250D 1490E 2450 Solution:DFrom 20:12 today until tomorrow is 24 hours, that is14406024 minutes. There are 50minutes from 20:12 tomorrow to 21:02 tomorrow. This gives a total of1490501440 isosceles and reflects the P-shape in the sideQRto get an reflects the first image in the sideQSto get a second , she reflects the second image in the sideRSto get athird does the third image look like?

8 ABCDES olution:AThe effect of the successive reflections is shown inthe reflection inQR2nd reflection inSQ3rd reflection prime numberspandqare the smallest primes that differ by 6. What is the sum ofpandq?A 12B 14C 16D 20E 28 Solution:CSupposeqp . pqThe prime numbers are 2, 3, 5, 7, .. With2 p,8 q, which isnot prime. Similarly if3 p,9 q, which is also not prime. However, when5 p,11 q, which isprime. So,5 p,11 qgives the smallest primes that differ by 6. Then qp16115 . has been challenged to place the numbers 1 to 9inclusive in the nine regions formed by the Olympicrings so that there is exactly one number in each regionand the sum of the numbers in each ring is 11. Thediagram shows part of his number goes in the region marked * ?A 6B 4C 3D 2E 1 Solution:AWe letu,v,w,x,yandzbe the numbers in the regions shown. Sincethe sum of the numbers in each ring is 11, we have, from theleftmost ring, that119 uand so2 u.

9 Then, from the next ring,1152 vand so4 v. From the rightmost ring,118 zand so3 have now used the digits 2, 3, 4, 5, 8 and 9, leaving 1, 6 and 7 .From the middle ring we have that114 xw, and so7 xw. From the second ring from theright113 yx, and so8 yx. So we need to solve the equations7 xwand8 yx,using 1, 6 and 7. It is easy to see that the only solution is1 x,7 yand6 w. So 6 goes in theregion marked *. Fi s dog Itchy has a million fleas. His anti-flea shampoo claims to leave no more than1% of the original number of fleas after use. What is the least number of fleas that will beeradicated by the treatment?A 900 000B 990 000C 999 000D 999 990E 999 999 Solution:BSince no more than 1% of the fleas will remain, at least 99% of them will be eradicated. Now 99% ofa million is0009900001099000000110099 .95*895w* abundant number is a positive integerN, such that the sum of the factors ofN(excludingNitself) is greater What is the smallest abundant number?

10 A 5B 6C 10D 12E 15 Solution:DIn the IMC, it is only necessary to check the factors of the numbers given as the options. However, tobe sure that the smallest of these which is abundant, is the overall smallest abundant number, wewould need to check the factors of all the positive integers in turn, until we find an abundant following table gives the sum of the factors ofN(excludingNitself), for121 ofN,excludingN-111,211,2,311,2,41,31,2,5 11,2,3,4,6sum of thesefactors0113161748116 From this table we see that 12 is the smallest abundant Which is the next smallest abundant number after 12? Show that ifnis a power of 2, and2 n(that is, n4, 8, 16, .. etc) then 3nis an Prove that ifnis an abundant number, then so too is each multiple A number,N,is said to bedeficientif the sum of the divisors ofN, excludingNitself, is that ifNis a power of 2, thenNis a deficient A number,N, is said to beperfectif the sum of the divisors ofN, excludingNitself, is equal see from the above table that 6 is the smallest perfect number.