3In how many other ways is it possible to make a sum of 144 = 144In how many other ways is it possible to make a sum of 144using only addition signs and all of the digits 1, 2, 3, 4, 5, 6,7, 8 and 9 in that order?= 45So we need to find an extra 99.The two-digit number ‘ab’ = 10a + b so changing a + b to ‘ab’ adds 9a to the total .An extra eleven 9s are required.The possibilities are 3, , , 3, , 4, 6
4So, apart from 12 + 34 + 5 + 6 + 78 + 9 = 144, there are 3 other ways: = 144= 144= 144
5UKMT ActivitiesMaths Challenges: Junior, Intermediate and SeniorFollow on rounds:(Junior Mathematical Olympiad - JMO;Intermediate Mathematical Olympiad and Kangaroo (suite of 3 Olympiad and 2 multiple choice European Kangaroo papers) – IMOK;British Mathematical Olympiad Round 1 – BMO1 and the Senior Kangaroo ( held on the same day as BMO1 - not multiple choice but single answers marked by a machine).Over pupils from more than 4000 schools take the Challenges each year.
6UKMT Activities Team Challenges: Team Challenge (years 8 and 9) and Senior Team Challenge (years 12 and 13), a series of regional events run throughout the UK (Feb, March and April for TMC, Oct and Nov for STMC) with a National Final held in London for both events. Teams of four students, a range of tasks for the day, very popular and great fun.Training and selection of the team to represent the UK at the International Mathematical Olympiad (BMO1 leads to BMO round 2, and training events in Oxford, Hungary, Trinity College Cambridge, and Oundle School). Also UK teams at the Balkans Mathematical Olympiad, Romanian Masters in Mathematics and the new European Girls’ Mathematical Olympiad (inaugural event run by UKMT and held at Murray Edwards College Cambridge in April 2012).
7UKMT ActivitiesMentoring schemes for able pupils. Fourth talk today – James CranchSummer Schools: two week-long residential schools each year for able pupils in the UK ; two extra summer schools in 2013New for 2013: Girls’ Summer School, Oxford, August 2013Publications - past papers, and UKMT publications. UKMT are European agents for the ‘Art of Problem Solving’ books.Testbase. A CD ROM containing 825 questions from the Junior, Intermediate and Senior challenges from 1997 to 2008 and can perform searches on Challenge, topic, difficulty of the questions and allows people to create their own question sheets from them.UKMT Yearbook – sent to all schools which participate in the Challenges.
email@example.com Testbase Special Offer Normal Price: £ Offer price: £40Rachel Greenhalgh directly and tell her you were at York Teachers’Meeting
9UKMT Activities Teacher Meetings Best in School events: run in limited locations in partnership with the RI.Recent Developments Primary Team Maths Resources Mathematical Circles: two trial events took place in 2012 in Glasgow andGloucestershire. Two-day long non residential events packed with mathematics beyond the curriculum.Several more in (with DFE funding)Senior Kangaroo : follow up to SMC but not as difficult as BMO papers
10Resources on the UKMT Website UKMT website: .Extended challenge solutions, including some extension material.Resources website:All Challenge papers since 2004; many other resources.Primary Team Challenge Resources
13The Millfield Team Competition (with thanks to Ceri Fides)I collect about 15 questions on a particular topic using the testbase software (or you could even just use a whole IMC or SMC paper) and then print out sets of questions on coloured paper. Each team gets one set of questions. Each question has three boxes at the bottom for answers (for this reason I normally remove the multiple choice option and just look for numerical answers. The questions are each worth different amounts of points and the team to answer a question first gets double points. The points for each question and current leader board are displayed throughout the lesson. The teacher just sits at the desk and students bring questions up. If they are correct hold onto the slips and if they are incorrect cross out one of the boxes and hand the slip back. When the teacher gets a chance they can record the scores on the score board (this is where the colours are invaluable as you can see which team handed in what and I keep all the slips in a pile so I can see which came in first).
14TEAMQu1234567891011121314PtsTotalA1618104B25CD28The first team to answer a question correctly first time receives double points
232001 JMO QB1The numbers from 1 to 7 inclusive are to be placed, one per square, in the figure on the right so that the totals of the three numbers in each of the three straight lines is the same.In how many different ways can this be done if the numbers 1 and 2 must be in the positions shown?12
2412xyzLet x, y and z be the numbers in the squares shown. Now the sum of the numbers from 1 to 7 inclusive is 28 and therefore the sum of the three equal totals will be 28 + x + 2 since x and 2 both appear in two of the lines of three numbers.Thus 30 + x must be a multiple of 3 and hence x must also be a multiple of 3,i.e. x = 3 or 6.If x =3, the total of each line = 33/ 3 = 11 and therefore y = 6 and z = 7.The two remaining squares are filled with 4 and 5 so this may be done in two different ways.If x = 6, the total of each line = 36 /3 = 12 and therefore y = 4 and z = 5.The two remaining squares are filled with 3 and 7 so this may also be done in two different ways.There are, therefore, four different ways of completing the grid:12357641234765126354712376
25IMC 2012 Q8Seb has been challenged to place the numbers 1 to 9 inclusivein the nine regions formed by the Olympic rings s that there isexactly one number in each region and the sum of the numbersin each ring is 11. The diagram shows part of his solution.What number should replace * ?A 6 B 4 C 3 D 2 E 1.985*
26Referring to the diagram, a = 11 – 9 = 2; b = 11 – 5 – a = 4; f = 11 – 8 = 3. So the values of c, d, e are 1, 6, 7 in some order.If c = 1 then d = 6, but then e would need to equal 2, not 7.If c = 6, then d =1, e = 7 and this is a valid solution.Finally, if c = 7 then d would need to equal 0, which is notpossible. So in the only possible solution, * is replaced by 6.985cabdef
272006 JMO QB6The numbers 1 to 7 are to be placed in the seven regions formed by three overlapping circles, with 6 in the central region, so that no regionis empty and the total of the numbers in each circle is T.What values of T are possible?
28acbfedLet the numbers inside the regions be a, b, c, d, e, f as shown.Then: a + d + e + 6 = T; b + d + f + 6 = T; c + e + f + 6 = T.Adding these equations gives a + b + c + 2d + 2e + 2f + 18 = 3T.Now a, b, c, d, e, f are 1, 2, 3, 4, 5, 7 in some order, so a + b + c + d + e + f = 22.Therefore 22 + d + e + f + 18 = 3T, that is 40 + d + e + f = 3T.Now the minimum value of d + e + f = = 6 and its maximum value = = 16.SoSince T is a positive integer, it cannot, then, take any values other than 16, 17 or 18.If T = 16, then d + e + f = 8 and the task may be completed with a = 7, b = 4, c = 3, d = 1, e = 2, f = 5.If T = 17, then d + e + f = 11 . However, a + d + e + 6 = T = 17, so a + d + e = 11.This requires a to equal f, which is impossible, so T cannot be 17.If T = 18, then d + e + f = 14 and the task may be completed with a = 5, b = 2, c = 1, d = 3,e = 4, f = 7.The only possible values of T, then, are 16 and 18.
29In how many different ways can the diagram be completed correctly? 2004 JMO QB6The diagram must be completed so that each square contains a different whole number from 1 to 12 inclusive and also so that the four numbers in the set of squares along each edge have the same total.In how many different ways can the diagram be completed correctly?56312
302004 JMO QB6The total of the numbers in the four rows = x= xSo x is a multiple of 4.Possible values of x are 3, 7, 11 but 3 is allocated.If x = 7 then each row totals 27.If x = 11 then each row totals 28.x56312
312004 JMO QB6 If x = 7 then each row totals 27, so y = 12. (Impossible as 12 already allocated.)756312y
322004 JMO QB6 If x = 11 then each row totals 28, so y = 9. 11 3 y 6 5 12y
361999 JMC Q17The 8-digit number 1234*678 is a multiple of 11. Which digit is represented by * ?A 1 B 3 C 5 D 7 E 92000 JMC Q17The first and third digits of the five-digit number d6d41 are the same. If the number is exactly divisible by 9, what is the sum of its five digits?A 18 B 23 C 25 D 27 E 30
372002 JMC Q22When 26 is divided by a positive integer N, the remainder is 2. What is the sum of all the possible values of N?A B C 45D 57 E 702005 JMC Q15There are six different three-digit numbers, each of which contains all the digits 1, 3 and 5.How many of these three-digit numbers are prime?A 0 B 1 C 2D E 4
382009 McLaurin Q2Find the possible values of the digits p and q given that the five-digit number p543q is a multiple of 36.Similar ProblemsFind all five-digit numbers of the form 7pppq which aremultiples of 45.Find all four-digit numbers of the form 5x3y which are multiples of 12.Are there any four-digit numbers of the form 5x2y which are multiples of 33? If so, find all such numbers.
39 Find a two-digit number 'ab' such that the difference between'ab' and its reverse 'ba' is a prime number.Gill has recently moved to a new house, which has a three-digit number. The sum of this house number and its three individual digits is 429. What is the house number?The digit 3 is written at the right of a certain two-digit number to make a three-digit number. The new number is 777 more than the original two-digit number. What was the original number?
40The diagram represents the addition of three * * * Find all four digit numbers of the form 'aabb' which are perfect squares. Let N be a positive integer less than When the digit 1 is placed after the last digit of N, the number formed is three times the number formed when the digit 1 is placed in front of the first digit of N. How many different values of N are there?The diagram represents the addition of three * * *3-digit numbers, which between them use all * * *of the digits from 1 to 9. Which of the following * * *cannot be obtained as the answer to the addition? * * * * A B C D E
41The diagram represents the addition of three a b c The diagram represents the addition of three a b c3-digit numbers, which between them use all e f gof the digits from 1 to 9. Which of the following h i jcannot be obtained as the answer to the addition? * * * * A B C D EThe sum = 100a + 10 b + c + 100e + 10f + g + 100h + 10i + j= 100(a + e + h) + 10 (b + f + i) + c + g + j=[ 99 (a + e + h) + a + e + h] + [9(b + f + i) + b + f + i] + c + g + j= 99 (a + e + h)] + 9(b + f + i) + a + e + h +b + f + i + c + g + j= 99 (a + e + h)] + 9(b + f + i) + 45 since a + b + c + d + e + f + g + h + i = 45= 9[ 11 (a + e + h)] + (b + f + i) + 5 ]So the sum is a multiple of 9 and the only alternative which is not a multiple of 9 is 1500,although a complete solution would check that the other 4 options are possible.
42The On / Off ProblemIn how many different ways can 20 lightbulbs be arranged in a straight line if it is not permitted for two adjacent bulbs to both be ‘On’?17 711
462 3 2 + 3 3 + 5 Put On/Off in front of all the one-bulb configurations Put Off in front of all the two-bulb configurations2 + 3Put On/Off in front of all the two-bulb configurationsPut Off in front of all the three-bulb configurations3 + 5
48A composition of an integer n is a representation of n as a sum of positive integers, for example the eight compositions of 4 are as follows:4; 3 + 1; 1 + 3; 2 + 2; ; ; ;A partition of n is a representation of n as a sum of positive integers where the order of the summands is considered irrelevant.Thus , and are threedistinct compositions, but are all considered to be the same partition of 4.
49In how many ways can the number 5 be so expressed? SMC 2012 Q12The number 3 can be expressed as the sum of one or more positiveintegers in four different ways:In how many ways can the number 5 be so expressed?
50Number of different ways = 16 5 (1); (2); (2); (3); (3);(4); (1).Number of different ways = 16
57Langford Sequences(Named after the Scottish mathematician C. Dudley Langford)
58IMC Q9A Langford number is one in which each digit occurs twice; the digits 1 are separated by one other digit, the digits 2 are separated by two others, and so on. Which of the following is a Langford number?A B CD E
59No formula is known for the number of solutions of a given order . Langford's ProblemArrange copies of the digits 1, ..., such that there is one digit between the 1s, two digits between the 2s, etc. For example, the unique (modulo reversal) solution is , and the unique (again modulo reversal) solution is Solutions to Langford's problem exist only if n = 0 or 3 (mod 4), so the next solutions occur for n = 7 .There are 26 of these, as exhibited by Lloyd (1971). In lexicographically smallest order (i.e., small digits come first), the first few Langford sequences are , , , , , ...The number of solutions for , 4, 5, ... (modulo reversal of the digits) are 1, 1, 0, 0, 26, 150, 0, 0, 17792, , ...No formula is known for the number of solutions of a given order .