Slide 1

Howard Groves

Chairman UKMT JMC/IMC Problems Group Email: chgroves@tiscali.co.uk

A Problem

12 + 34 + 5 + 6 + 78 + 9 = 144In how many other ways is it possible to make a sum of 144 using only addition signs and all of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 in that order?

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

So 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, 8 4, 7 1, 3, 7 1, 4, 6

So, apart from 12 + 34 + 5 + 6 + 78 + 9 = 144, there are 3 other ways:

1 + 2 + 34 + 5 + 6 + 7 + 89 = 144

1 + 2 + 3 + 45 + 6 + 78 + 9 = 144

1 2 + 3 + 45 + 67 + 8 + 9 = 144

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The 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).

TEAMQu1234567891011121314Pts33334457555689TotalA6336881014121618104B33334925C336618D3610928The first team to answer a question correctly first time receives double points

Head to HeadHead to Head2 7 12 9 5 42 150 999 5 15 251985 30111 1999f(n) = 2n + 1

Head to Head2 5 10 4 7 30 0.5 200 7 28 103199033.2552 40 003f(n) = n2 + 3

Head to Head4 6 14 8 15 11 98 128 2 3 7 2 11 7 5 2f(n) = largest prime factor of n

Head to Head8 12 16 24 40 6 99 5000 2 3 4 4 2 9 6 70f(n) = integer part of square root of n

Head to Head6 12 16 1 3 8 11 20 3 6 35 65 6Six Twelve Sixteen One Three Eight Eleven Twenty f(n) = number of letters in the word n

7More ideas:

http//maths.anu.edu.au/files/swiss2012.pdf

Same total problems

122001 JMO QB1

The 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?

Let 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:12xyz1235764123476512635471237763

985*Seb has been challenged to place the numbers 1 to 9 inclusive in the nine regions formed by the Olympic rings s that there is exactly 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 6B 4C 3D 2E 1.

IMC 2012 Q8

Referring 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 not

possible. So in the only possible solution, * is replaced by 6.985cabdef2006 JMO QB6

The 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?

acbfedLet 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 = 1 + 2 + 3 = 6 and its maximum value = 4 + 5 + 7 = 16.

So 46 . Since 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 18.2004 JMO QB6

The 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

2004 JMO QB6

The total of the numbers in the four rows = 78 + 5 + 6 + 12 + x= 101 + x

So 101 + 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 28.56312

x2004 JMO QB6

If x = 7 then each row totals 27, so y = 12.(Impossible as 12 already allocated.)

56312

7y2004 JMO QB6

If x = 11 then each row totals 28, so y = 9.

56312

11y2004 JMO QB6

56312

119Numbers remaining: 1, 2, 4, 7, 8, 10Total 11Total 11

Total 10

822004 JMO QB6

56312

119Numbers remaining: 1, 4, 7, 10Total 11Total 11

Total 10

28Number of different ways = 16Digit / Divisibility Problems

1999 JMC Q17The 8-digit number 1234*678 is a multiple of 11. Which digit is represented by * ?A 1B 3C 5D 7E 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 18B 23C 25D 27E 30

2002 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 21B 33C 45D 57E 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 0B 1C 2D 3 E 4

2009 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 are multiples 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.

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?

Find all four digit numbers of the form 'aabb' which are perfect squares.

Let N be a positive integer less than 102013. 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 1500 B 1503 C 1512 D 1521 E 1539

The diagram represents the addition of three a b c3-digit numbers, which between them use alle f g of the digits from 1 to 9. Which of the following h i jcannot be obtained as the answer to the addition? * * * *A 1500 B 1503 C 1512 D 1521 E 1539

The 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.

The 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

2006 JMC Q17McBrides Law

For every difficult problem that you cant do, there is a similar, easier problem which you can do.

3258Put On/Off in front of all the one-bulb configurationsPut Off in front of all the two-bulb configurationsPut On/Off in front of all the two-bulb configurationsPut Off in front of all the three-bulb configurations232 + 33 + 5Compositions of a positive integerA composition of an integer n is a representation of n as a sum of positive integers, forexample the eight compositions of 4 are as follows:4; 3 + 1; 1 + 3; 2 + 2; 2 + 1 + 1; 1 + 2 + 1; 1 + 1 + 2; 1 + 1 + 1 + 1.

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 2 + 1 + 1, 1 + 2 + 1 and 1 + 1 + 2 are threedistinct compositions, but are all considered to be the same partition of 4.

The number 3 can be expressed as the sum of one or more positive integers in four different ways:3 1 + 22 + 11 + 1 + 1In how many ways can the number 5 be so expressed?

SMC 2012 Q12

5 (1); 2 + 3 (2); 1 + 4 (2); 1 + 2 + 2 (3); 1 + 1 + 3 (3); 1 + 1 + 1 + 2 (4); 1 + 1 + 1 +1 + 1 (1).

Number of different ways = 16Positive Integer Number of Compositions

3 4

4 8

516

n ?

11111++++corresponds to 2 + 2 + 1+/++/++/++/+So number of compositions of n = 2n-1

3 1 2 1 3 2

4 1 3 1 2 4 3 2

Proving its impossible

Langford Sequences(Named after the Scottish mathematician C. Dudley Langford)IMC 2002 Q9

A 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 1214233 B 41312432 C 14132342 D 32432141 E 31213244 Langford's Problem

Arrange 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 231213, and the unique (again modulo reversal) solution is 23421314. 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 231213, 23421314, 14156742352637, 14167345236275, 15146735423627, ... The number of solutions for , 4, 5, ... (modulo reversal of the digits) are 1, 1, 0, 0, 26, 150, 0, 0, 17792, 108144, ... No formula is known for the number of solutions of a given order .

http://mathworld.wolfram.com/LangfordsProblem.html

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Langford Sequences112233_ _ _ _ _ _ _ _ _ _

Langford Sequences112233441 _ _ or _ _2 _ and _ 3 _ _ or _ _4 _ and _ 5 _ _ or _ _65Can the seven tetrominoes be used to fill the 7 x 4 rectangle?

66

67

68

69ABFind a route from room A to room B which goes through every other room exactly once.

UKMT website: www.ukmt.org.uk .

Extended challenge solutions, including some extension material.

Resources website: www.ukmt-resources.org.uk

All Challenge papers since 2004; many other resources.chgroves@tiscali.org.uk