Transcript - UKMT Gateway

Howard Groves
Chairman UKMT JMC/IMC Problems Group
Email:
[email protected]
A Problem
12 + 34 + 5 + 6 + 78 + 9 = 144
In 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
UKMT Activities
Maths Challenges: Junior, Intermediate and Senior
Follow 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 600 000 pupils from more than 4000 schools take
the Challenges each year.
UKMT 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).
UKMT Activities
Mentoring schemes for able pupils. Fourth talk today – James Cranch
Summer Schools: two week-long residential schools each year for able pupils in
the UK ; two extra summer schools in 2013
New for 2013: Girls’ Summer School, Oxford, August 2013
Publications - 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.
Testbase Special Offer
Normal Price: £60
Offer price: £40
Email Rachel Greenhalgh directly and tell her you were
at York Teachers’Meeting
[email protected]
UKMT 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 and
Gloucestershire. Two-day long non residential events packed with
mathematics beyond the curriculum.
Several more in 2013-14 (with DFE funding)
Senior Kangaroo : follow up to SMC but not as difficult as BMO papers
Resources on the UKMT Website
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.
Primary Team Challenge Resources
Primary Team Maths Resources
UKMT Resources Gateway
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).
Qu
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Pts
3
3
3
3
4
4
5
7
5
5
5
6
8
9
Total
A
6
3
3
6
8
8
10
14
12
16
18
104
B
3
3
3
3
4
9
25
C
3
3
6
D
3
6
TEAM
6
10
The first team to answer a question correctly first time receives double points
18
9
28
Head to Head
Head to Head
2
7
12
9
5
42
150 999
5
15
25
19
11
85
301 1999
f(n) = 2n + 1
Head to Head
2
5
10
4
7
28
103
19
7
30
0.5
200
52 903 3.25 40 003
f(n) = n2 + 3
Head to Head
4
6
14
8
15
11
98
128
2
3
7
2
5
11
7
2
f(n) = largest prime factor of n
Head to Head
8
12
16
24
40
6
99
5000
2
3
4
4
6
2
9
70
f(n) = integer part of square root of n
Head to Head
6
12
16
1
3
8
11
20
3
6
7
3
5
5
6
6
Six
Twelve
Sixteen One
Three
Eight Eleven Twenty
f(n) = number of letters in the word n
More ideas:
http//maths.anu.edu.au/files/swiss2012.pdf
‘Same total’ problems
2001 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?
1
2
1
x
z
y
2
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:
1
5
1
4
3
6 2
4
1
7
3
7
1
3
7
6 2
5
6
5
4 2
7
3
7
6 2
3
IMC 2012 Q8
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 numbers
in 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
9
*
5
8
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.
9
b
a
5
c
d
e
f
8
2006 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 region
is empty and the total of the
numbers in each circle is T.
What values of T are possible?
a
d
e
b
f c
Let 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  3T  56 .
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?
3
6
5
12
2004 JMO QB6
The total of the numbers in the four rows
= 78 + 5 + 6 + 12 + x
= 101 + x
x
3
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.
6
5
12
2004 JMO QB6
If x = 7 then each row totals 27, so y = 12.
(Impossible as 12 already allocated.)
7
3
y
6
5
12
2004 JMO QB6
If x = 11 then each row totals 28, so y = 9.
11
3
y
6
5
12
2004 JMO QB6
Total 11
11
3
Numbers remaining: 1, 2, 4, 7, 8, 10
9
6
5
2
8
Total 11
12
Total 10
2004 JMO QB6
Total 11
11
3
Numbers remaining: 1, 4, 7, 10
9
6
5
Number of different ways = 16
8
Total 11
2
12
Total 10
Digit / Divisibility Problems
1999 JMC Q17
The 8-digit number 1234*678 is a multiple of 11.
Which digit is represented by * ?
A 1
B 3
C 5
D 7
E 9
2000 JMC Q17
The 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
2002 JMC Q22
When 26 is divided by a positive integer N, the
remainder is 2. What is the sum of all the possible
values of N?
A 21
B 33
C 45
D 57
E 70
2005 JMC Q15
There 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 2
D 3
E 4
2009 McLaurin Q2
Find the possible values of the digits p and q given that
the five-digit number p543q is a multiple of 36.
Similar Problems
Find 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
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
a
e
h
* *
b c
f g
i j
* *
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 Problem
In 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 Q17
McBride’s Law
For every difficult problem that you can’t do,
there is a similar, easier problem which you can
do.
2
3
5
8
2
3
Put On/Off in front of all the one-bulb configurations
Put Off in front of all the two-bulb configurations
2+3
Put On/Off in front of all the two-bulb configurations
Put Off in front of all the three-bulb configurations
3+5
Compositions of a positive integer
A 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; 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 three
distinct compositions, but are all considered to be the same
partition of 4.
SMC 2012 Q12
The number 3 can be expressed as the sum of one or more positive
integers in four different ways:
3
1+2
2+1
1+1+1
In how many ways can the number 5 be so expressed?
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 = 16
Positive Integer
Number of Compositions
3
4
4
8
5
16
n
?
+/
1
+
+
+/
1
+
+ 1
+/
+
+
+/
1
+
+ 1
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 it’s 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
http://mathworld.wolfram.com/LangfordsProblem.html
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 .
_1 _ _1 _ _ _ _ _ _ _
Langford Sequences
_ _1 _ _1 _ _ _ _ _ _
Langford Sequences
_ 1_ _ 1_ 2
_ _ _ _2 _ _
Langford Sequences
_ _1 _ 1_ 2
_ 3_ _ 2_ _ _3
Langford Sequences
_ 1_ _3 1_ 2
_ _ 3_ _2 _ _
Langford Sequences
Langford Sequences
4_ _1 _3 _1 2
_ 4_ 3_ 2_ _ _
1 _ _ or _ _ 2 _ and _
3 _ _ or _ _ 4 _ and _
5 _ _ or _ _
Can the seven tetrominoes be used to fill the 7 x 4 rectangle?
B
A
Find 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.
[email protected]