Junior Mathematical Challenge - UKMT

1 UKMTUKMTUKMTJ uniorMathematicalChallengeThursday 1 May 2014 Organised by the United Kingdom Mathematics Trustsupported bySolutions and investigationsThese solutions augment the printed solutions that we send to schools. For convenience, thesolutions sent to schools are confined to two sides of A4 paper and therefore in many cases arerather short. The solutions given here have been extended. In some cases we give alternativesolutions, and we have included some exercises for further Junior Mathematical Challenge (JMC) is a multiple-choice paper. For each question, youare presented with five options, of which just one is correct. It follows that often you can find thecorrect answers by working backwards from the given alternatives, or by showing that four ofthem are not correct.

2 This can be a sensible thing to do in the context of the , this does not provide a full Mathematical explanation that would be acceptable if youwere just given the question without any alternative answers. So for each question we haveincluded a complete solution which does not use the fact that one of the given alternatives we have aimed to give full solutions with all steps explained. We therefore hopethat these solutions can be used as a model for the type of written solution that is expectedwhen presenting a complete solution to a Mathematical problem (for example, in the JuniorMathematical olympiad and similar competitions).These solutions may be used freely within your school or college.

3 You may, without futher permission,post these solutions on a website that is accessible only to staffand students of the school or college,print out and distribute copies within the school or college, and use them in the classroom. If you wishto use them in any other way, please consult us. UKMT May 2014We welcome comments on these solutions, and, especially, corrections or suggestions for improving send your comments to:JMC Solutions, UKMT, School of Mathematics Satellite,University of Leeds, Leeds LS2 9 JTT0113 343 May 2014 JuniorMathematicalChallenge2014 Solutions and is(999 99+9) 9?A 91B 99C 100D 101E 109 SolutionDWe can calculate the value in more than one method is to first work out the value of the expression in the brackets and then dividethe result by 9.

4 This gives(999 99+9) 9=909 9= , we can first divide each number in the bracket by 9 and then evaluate theresulting expression. This gives(999 99+9) 9=(111 11+1)= course, both methods give the same many minutes are there in112of a day?A 240B 120C 60D 30E 15 SolutionBThere are 24 hours in a day. So in112of a day the number of hours equals112 24=2. In 1hour there are 60 minutes. Hence the number of minutes in 2 hours is 2 60=120. Sothere are 120 minutes in112of a many seconds are there in1120of a day? fraction of a day is 3 hours? fraction of a day is 250 seconds? my row in the theatre the seats are numbered consecutively from T1 to T50.

5 I am sittingin seat T17 and you are sitting in seat T39. How many seats are there between us?A 23B 22C 21D 20E 19 SolutionCThe seats between us are numbered from T18 up to T38. So the seats between us arethe 38 seats from T1 up to T38, other than the 17 seats from T1 up to T17. So there are38 17=21 seats between April 2014 UKMT May 201421 May 2014 JuniorMathematicalChallenge2014 Solutions and investigationsFor many integers are there between 100 and 200 (not including either 100 or 200)? thatkis an integer greater than 27, and that there are 50 integers between 27andk, not including 27 andk. What is the value ofk? a formula for the number of integers in the sequencem,m+1.

6 ,n 1,n,wheremandnare integers, withm< number 987 654 321 is multiplied by many times does the digit 8 occur in the result?A 1B 2C 3D 4E 9 SolutionEThere does not seem to be any better way to answer this question than to do the multiplica-tion:1234567899 9888888888 From this we see that the digit 8 occurs 9 times in the the following multiplications which have surprising answers:(a) 127 9721,(b) 32 800 328 271,(c) 239 4649,(d) 21 649 513 some other examples of a similar is the difference between the smallest 4-digit number and the largest 3-digit number?A 1B 10C 100D 1000E 9899 SolutionAThe smallest 4-digit number is 1000.

7 The largest 3-digit number is 999. So their differenceis equal to 1000 999= April 2014 UKMT May 201431 May 2014 JuniorMathematicalChallenge2014 Solutions and diagram shows a square divided into strips of equal width. Threestrips are black and two are fraction of the perimeter of the square is grey?A15B14C425D13E25 SolutionATwo sides of the square are wholly black, and25of two sides are grey. So the length of theperimeter that is grey is equal to 2 25=45of the length of one side. The length of theperimeter is 4 times the length of one side. So the fraction of the perimeter that is grey is454= square is divided into seven strips of equal width.

8 Four of the strips are black and threeare grey. What fraction of the perimeter of the square is grey? square is divided into an odd number of strips of equal width. The strips are alternatelyblack and grey, with one more black strip than grey strip. The fraction of the perimeterthat is grey is625. How many strips are there? is 2014 4102?A 2012B 2088C 2092D 2098E 2112 SolutionBIt is easier to subtract the smaller number, 2014, from the larger number, 4102. Now4102 2014=2088and so2014 4102= April 2014 UKMT May 201441 May 2014 JuniorMathematicalChallenge2014 Solutions and many prime numbers are there in the list1,12,123,1234,12 345,123 456?A 0B 1C 2D 3E 4 SolutionAIt is important to remember that we donotregard 1 as a prime see that 12, 1234 and 123 456 are not prime numbers becauses they are divisible by2.

9 Also, 123 is not a prime number because it is divisible by 3, and 12 345 is not a primenumber because it is divisible by there are no prime numbers in the course, there are other ways to see that some of the numbers in the list are not primenumbers. For example, there is a useful test for divisibility by 3:An integer is divisible by 3 if, and only if, the sum of its digits is divisible by example, 1+2+3+4+5+6=21, and 21 is divisible by 3. It follows, using thistest, that 123 456 is divisible by , if the sum of the digits of a positive integer, other than 3 itself, is a number that isdivisible by 3, we can deduce that the integer is not a prime number.

10 (a) Is the number 123 456 789 divisible by 3?(b)Why does the test for divisibility by 3 in terms of the sum of the digits of an integerwork? 1 234 567 a prime number? [Hint: Look at (a).] the number 12 345 678 a prime number? a test for a number to be divisible by 9 in terms of the sum of its a test for a number to be divisible by 11 in terms of its digits. [Note:this test is alittle more complicated than those for divisibility by 3 and by 9.]NoteIt is only aconventionthat 1 does not count as a prime number. However, it is a very usefulconvention and one used by all reason, among many, why this is a useful convention is that it enables us to stateTheFundamental Theorem of Arithmeticin a simple way.