Organised by the

United Kingdom Mathematics Trust

MATHEMATICAL OLYMPIAD

UK I NTERMEDIATE

Cayley Question Papers and Solutions

2008 to 2010

i

UKMT

UKM

TUKM

T

UK Intermediate Mathematical Olympiad2008 to 2010

Cayley Question Papers and Solutions

Organised by the United Kingdom Mathematics Trust

Contents

Background ii

Rules and Guidelines 1

2008 paper 3

2009 paper 4

2010 paper 5

2008 solutions 6

2009 solutions 12

2010 solutions 18

©UKMT 2011

ii

Background

The IntermediateMathematicalOlympiad and Kangaroo(IMOK)arethefollow-up competitionsfor pupilswho do extremelywell inthe UKMT IntermediateMathematicalChallenge(about1 in 200are invited to take part). The IMOK was established in 2003.

Therearethreewritten papers(Cayley,Hamilton,Maclaurin)andtwo multiple-choicepapers(the Pink and Grey Kangaroo). Thewrittenpaperseachtaketwo hoursandcontainsix questions.BothKangaroo papers are one hour long and contain 25 questions.

The Cayley paper is for pupils in: Y9 or below (England andWales); S2 or below (Scotland); School Year 10 or below(Northern Ireland).

1

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo(IMOK)

Olympiad Cayley Paper

All candidates must be in School Year 9 or below (England and Wales), S2 or below(Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING

1. Time allowed: 2 hours.

2. The use of calculators, protractors and squared paper is forbidden.Rulers and compasses may be used.

3. Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the topleft corner with the Cover Sheet on top.

4. Start each question on a fresh A4 sheet.You may wish to work in rough first, then set out your final solution with clear explanationsand proofs.

Do not hand in rough work.

5. Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols such as ,fractions, or square roots, if appropriate, but NOT decimal approximations.

π

6. Give full written solutions, including mathematical reasons as to why your method is correct.Just stating an answer, even a correct one, will earn you very few marks; also, incomplete orpoorly presented solutions will not receive full marks.

7. These problems are meant to be challenging! The earlier questions tend to be easier; the lasttwo questions are the most demanding.Do not hurry, but spend time working carefully on one question before attempting another.Try to finish whole questions even if you cannot do many: you will have done well if you handin full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO!

The United Kingdom Mathematics Trust is a Registered Charity.

Enquiries should be sent to: Maths Challenges Office,School of Maths Satellite, University of Leeds, Leeds, LS2 9JT.

(Tel. 0113 343 2339)

http://www.ukmt.org.uk

2

Advice to candidates

• Do not hurry, but spend time working carefully on one question before attemptinganother.

• Try to finish whole questions even if you cannot do many.

• You will have done well if you hand in full solutions to two or more questions.

• Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols suchas , fractions, or square roots, if appropriate, but NOT decimal approximations.π

• Give full written solutions, including mathematical reasons as to why your method iscorrect.

• Just stating an answer, even a correct one, will earn you very few marks.

• Incomplete or poorly presented solutions will not receive full marks.

• Do not hand in rough work.

3

2008

1. How many four-digit multiples of 9 consist of four different odd digits?

2. A hexagon is made by cutting a small equilateral triangle fromeach corner of a larger equilateral triangle. The sides of thesmaller triangles have lengths 1, 2 and 3 units. The lengths of theperimeters of the hexagon and the original triangle are in theratio 5 : 7. What fraction of the area of the original triangle remains?

3. In the rectangle the midpoint of is and . The point is suchthat triangle is equilateral, with and lying on opposite sides of the line .

ABCD AB M AB : AD = 2 : 1 XMDX X A MD

Find the value of .∠XCD

4. The number is the product of the first 99 positive integers. The number is theproduct of the first 99 positive integers after each has been reversed. That is, forexample, the reverse of 8 is 8; of 17 is 71; and of 20 is 02.

N M

Find the exact value of .N ÷ M

5. A kite has sides and of length 25 cmand sides and of length 39 cm. Theperpendicular distance from to is 24cm.

AB ADCB CD

B AD

The perpendicular distance from to is cm.

B CDhFind the value of .h

A

B

C

D

39cm

39cm 25cm24cm

25cm

hcm

6. A regular tetrahedron has edges of length 2 units. The midpoint of the edge is and the midpoint of the edge is .

ABCD ABM CD NFind the exact length of the segment .MN

4

2009

1. An aquarium contains 280 tropical fish of various kinds. If 60 more clownfish wereadded to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium?

2. The boundary of the shaded figure consists of foursemicircular arcs whose radii are all different. The centreof each arc lies on the line , which is 10 cm long. ABWhat is the length of the perimeter of the figure? A B

3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle.The length of the perimeter of the large rectangle is of the total perimeter of theoriginal two rectangles.

23

Prove that the final rectangle is in fact a square.

4. In the rectangle , the side has length and the side has length 1. Let thecircle with centre and passing through meet at .

ABCD AB 2 ADB C AB X

Find (in degrees).∠ADX

5. Two candles are the same height. The first takes 10 hours to burn completely whilst thesecond takes 8 hours to burn completely.Both candles are lit at midday. At what time is the height of the first candle twice theheight of the second candle?

6. Teams A, B, C and D competed against each other once. The results table was asfollows:

Team Win Draw Loss Goals for Goals against

A 3 0 0 5 1

B 1 1 1 2 2

C 0 2 1 5 6

D 0 1 2 3 6

(a) Find (with proof) which team won in each of the six matches.

(b) Find (with proof) the scores in each of the six matches.

5

2010

1. The sum of three positive integers is 11 and the sum of the cubes of these numbersis 251.

Find all such triples of numbers.

2. The diagram shows a square and anequilateral triangle . The point lies on so that .

ABCDABE F BC

EC = EFCalculate the angle .BEF

A B

CD

E

F

3. Find all possible solutions to the ‘word sum’ on the right. Each letter stands for one of the digits 0−9 and has thesame meaning each time it occurs. Different letters standfor different digits. No number starts with a zero.

O D D+ O D DE V E N

4. Walking at constant speeds, Eoin and his sister Angharad take 40 minutes and60 minutes respectively to walk to the nearest town.Yesterday, Eoin left home 12 minutes after Angharad. How long was it before hecaught up with her?

5. A square sheet of paper is folded along, as shown, so that the corner is folded onto

the midpoint of .

ABCDFG B

M CDProve that the sides of triangle havelengths in the ratio 3 : 4 : 5.

GCM

A B

CD M

G

F

6. A ‘qprime’ number is a positive integer which is the product of exactly twodifferent primes, that is, one of the form , where and are prime and .q × p q p q ≠ pWhat is the length of the longest possible sequence of consecutive integers all ofwhich are qprime numbers?

6

2008 Solutions

1. How many four-digit multiples of 9 consist of four different odd digits?

First solutionThere are five odd digits: 1, 3, 5, 7 and 9. The sum of the four smallest odd digits is 16 and the sum of the four largest is 24. Hencethe digit sum of any four-digit number with different odd digits lies between 16 and 24,inclusive.However, the sum of the digits of a multiple of 9 is also a multiple of 9, and the onlymultiple of 9 between 16 and 24 is 18. Hence the sum of the four digits is 18.Now , so that the four digits can be 1, 3, 5 and 9. If 7 is one of thefour digits then the sum of the other three is 11, which is impossible. So 7 cannot be oneof the digits and therefore the four digits can only be 1, 3, 5 and 9.

1 + 3 + 5 + 9 = 18

The number of arrangements of these four digits is . Hence thereare 24 four-digit multiples of 9 that consist of four different odd digits.

4 × 3 × 2 × 1 = 24

Second solutionThe sum of all five odd digits is . Subtracting 1, 3, 5, 7 and 9 in turn we get 24, 22, 20, 18 and 16, only one of which is amultiple of 9, namely . Since the sum of the digits of a multiple of 9 is alsoa multiple of 9, it follows that the four digits can only be 1, 3, 5 and 9.

1 + 3 + 5 + 7 + 9 = 25

18 = 25 − 7

The number of arrangements of these four digits is . Hence thereare 24 four-digit multiples of 9 that consist of four different odd digits.

4 × 3 × 2 × 1 = 24

2. A hexagon is made by cutting a small equilateral trianglefrom each corner of a larger equilateral triangle. Thesides of the smaller triangles have lengths 1, 2 and 3units. The lengths of the perimeters of the hexagon andthe original triangle are in the ratio 5 : 7. What fraction of the area of the original triangleremains?

First solutionLet the side length of the large equilateral triangle be units; this triangle therefore has aperimeter of length units.

x3x

Now consider the hexagon, which has sides of lengths and units. Hence the hexagon has perimeter length units.

1, x − 3, 2, x − 5, 3 x − 43x − 6

Since the ratio of the perimeter lengths of the hexagon and the large triangle is 5 : 7, wehave

3x − 63x

=57

.

Rearranging and solving for we obtainx

x = 7. (∗)

7

Now, in order to find the area of the large equilateral triangle, we determine the height units using Pythagoras' theorem:

h

h2 = 72 − (72)2

= 49(1 − 14) = 49 × 3

4.

h = 49 ×34

= 7 ×3

2.Hence

Therefore the area of the large equilateral triangle is

12

× 7 ×7 32

=49 3

4.

We may find the areas of the three small equilateral triangles in a similar way. Theseareas are

34

,4 3

4and

9 34

.

The area of the hexagon is the area of the large equilateral triangle minus the areas of thethree small equilateral triangles, that is,

49 34

− ( 34

+4 3

4+

9 34 ) =

35 34

.

Finally, the fraction of the original equilateral triangle remaining is

35 34

÷49 3

4=

57

.

Second solutionHaving established that the large triangle has sides of length 7 (equation (*) in thesolution above), we may proceed as follows:The four equilateral triangles in the problem are similar. Now the ratio of the areas ofsimilar figures is equal to the ratio of the squares of their sides. Hence the four triangleshave areas in the ratio .12 : 22 : 32 : 72 = 1 : 4 : 9 : 49Hence the ratio of the areas of the hexagon and the large triangle is

49 − (1 + 4 + 9) : 49 = 35 : 49 = 5 : 7.This may be illustrated by dividing the large triangle into 49 small triangles, as shown.

Note:The observant reader will have noticed that the answer to this problem is surprising: theratio of the areas is the same as the ratio of the perimeters. There is no reason to expectthis to happen.

8

3. In the rectangle the midpoint of is and . The point is suchthat triangle is equilateral, with and lying on opposite sides of the line .

ABCD AB M AB : AD = 2 : 1 XMDX X A MD

Find the value of .∠XCD

SolutionThe key to this solution is to draw and consider triangle .MC MCXWe are given that is a rectangle, so that and .ABCD BC = AD ∠DAM = 90° = ∠MBCWe are also given that and that is the midpoint of . Therefore

.AB = 2AD M AB

DA = AM = MB = BC

A B

CD

M

X

It follows that triangles and are congruent (SAS) and we deduce that.

DAM MBCDM = MCBut triangle is equilateral, so and hence . In other words,triangle is isosceles.

MDX MX = DM MX = MCMCX

Now consider the angles at .M1. Triangle is right-angled with . It is also isosceles, so

, since the angle sum is .DAM ∠DAM = 90°

∠AMD = ∠ADM = 45° 180°2. Similarly, from triangle , .MBC ∠CMB = 45°3. Finally, because triangle is equilateral, .MDX ∠DMX = 60°

∠CMX = 180° − 45° − 45° − 60°Hence

= 30°since angles on a straight line add up to .180°

Lastly, we consider the angles at .CWe know that triangle is isosceles and that . Hence each base angleis ; in particular, .

XCM ∠CMX = 30°12 (180° − 30°) = 75° ∠XCM = 75°

Also, is a rectangle, so , and triangle is right-angled andisosceles, so . Therefore

ABCD ∠DCB = 90° MBC∠MCB = 45°

∠DCM = ∠DCB − ∠MCB

= 90° − 45°= 45°.

We can now calculate the value of .∠XCD

∠XCD = ∠XCM − ∠DCMWe have

= 75° − 45°= 30°.

9

4. The number is the product of the first 99 positive integers. The number is theproduct of the first 99 positive integers after each has been reversed. That is, forexample, the reverse of 8 is 8; of 17 is 71; and of 20 is 02.

N M

Find the exact value of .N ÷ M

First solutionFrom the given definition we have

N = (1 × 2 ×… × 9) × 10 × (11 × 12 ×… × 19) × 20 ×… × 90 × (91 ×… × 99),which rearranges to

N = (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (10 × 20 ×… × 90).Also

M = (1 × 2 ×… × 9) × 01 × (11 × 21 ×… × 91) × 02 ×… × 09 × (19 ×… × 99)which rearranges to

M = (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (01 × 02 ×… × 09)= (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (1 × 2 ×… × 9).

Comparing these arrangements for and , we see that M has the same terms as except that the product is replaced by the product .

M N N10 × 20 ×… × 90 1 × 2 ×… × 9

Thus when we divide by all the common terms cancel and we are left withN M

NM

=10 × 20 ×… × 90

1 × 2 ×… × 9

= 109.Second solutionWe may place the numbers from 1 to 99 into three categories, determined by how theyare transformed when they are reversed:1. single digit numbers are unchanged;‘a’2. a two-digit number , where neither nor is zero, is transformed to the two-

digit number ; and‘ab’ a b

‘ba’3. a multiple of 10 such as is transformed to , a single-digit number.‘a0’ ‘ 0a’ = aThus there is a correspondence between the factors in and , as shown in the table:N M

N M

‘a’ ‘a’

‘aa’ ‘aa’‘ ab’ and ‘ba’ ‘ba’ and ‘ab’

‘a0’ ‘a’

Single-digit numbers are unchanged; two-digit numbers with a repeated digit areunchanged; pairs of two-digit numbers, with different digits and neither digit zero, areunchanged as a pair; the multiples of 10 in are replaced by single-digit numbers in .N MThus when we divide by all the identical factors cancel and we are left withN M

N

M=

10 × 20 ×… × 90

1 × 2 ×… × 9

= 109.

10

5. A kite has sides and of length 25 cm andsides and of length 39 cm. Theperpendicular distance from to is 24 cm.

AB ADCB CD

B ADThe perpendicular distance from to is cm.

B CD h

Find the value of .h

A

B

C

D

39cm

39cm 25cm24cm

25cm

hcm

First solutionAs shown in the figure below, let the perpendicular from to the line meet the line

at the point ; let the perpendicular from to the line meet the line at thepoint and let the distance be cm.

B ADAD X B CD CD

Y DY y

A

B

C

D

X

Y39 − y

y

25

24

25

39

h

Considering triangle and using Pythagoras' Theorem we obtainABX

AX = 252 − 242 cm

= 7 cm.Similarly, from triangle we haveBDX

BD = 242 + (25 − 7)2 cm

= 30 cm. (∗)Now from triangles and , again by Pythagoras' theorem, we deduce thatBDY BCY

= h2 + (39 − y)2 392

and = . (1)h2 + y2 302

Subtract to get = ,(39 − y)2 − y2 392 − 302

which simplifies to = 900,78y

so that = .y15013

Finally, by substituting in equation (1), we find

h =36013

.

Second solutionAnother solution uses the length of obtained in (*) above to find the area of isoscelestriangle . Once the area is known the value of the height may be found from

BDBCD h

area .= 12 × 39 × h

Can you see how to find the area of triangle and so complete the solution?BCD

Note: Triangle is a ‘5, 12, 13’ triangle.DYB

11

First solutionWe make use of the following result.Theorem (Median of isosceles triangle): The line joining the apexto the midpoint of the base of an isosceles triangle is perpendicularto the base. That is, in the following figure, .∠PSR = 90°

SP

R

Q

Applying the theorem to triangle , we find that . Similarly, in triangle, .

ABC ∠AMC = 90°ABD ∠AMD = 90°Now applying Pythagoras' theorem to the triangles and we getAMC AMD

CM2 = AC2 − AM2 = 22 − 12

and DM2 = AD2 − AM2 = 22 − 12.Hence and , so triangle is isosceles.CM = 3 DM = 3 CMDNow apply the theorem to triangle to obtain .CMD ∠CNM = 90°Then by Pythagoras' theorem in triangle CNM

MN2 = CM2 − CN2

= 3 − 1.Therefore .MN = 2

A

B

C

D

M

N

6. A regular tetrahedron has edges of length 2 units. ABCDThe midpoint of the edge isAB

Find the exact length of the segment .MN and the midpoint of the edge is .M CD N

Second solutionA tetrahedron may be formed by joining face diagonals of a cube, as shown below.Since the faces of the cube are congruent squares the face diagonals have equal lengthand so the tetrahedron is regular.Now and are midpoints of opposite edges of the tetrahedron. Therefore they aremidpoints of opposite face diagonals of the cube, that is, centres of opposite faces of thecube. Hence .

M N

MN = ARLetting the sides of the cube have length , from Pythagoras' theorem in triangle weget

a ARC

so that

Hence and therefore .a = 2 MN = 2

A

B

C

D

P

Q

R

S

M

N

AC2 = AR2 + RC2

22 = a2 + a2

= 2a2.

12

2009 Solutions

1. An aquarium contains 280 tropical fish of various kinds. If 60 more clownfish wereadded to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium?

SolutionLet there be clownfish in the aquarium. xIf 60 clownfish are added there are clownfish and 340 tropical fish in total.x + 60Since the proportion of clownfish is then doubled, we have

2 ×x

280=

x + 60

340.

Multiplying both sides by 20 we get

x7

=x + 60

17and hence

17x = 7 (x + 60) .It follows that and thus there are 42 clownfish in the aquarium.x = 42

2. The boundary of the shaded figure consists of foursemicircular arcs whose radii are all different. Thecentre of each arc lies on the line , which is 10 cmlong.

AB

What is the length of the perimeter of the figure? A B

SolutionThe centre of the large semicircular arc lies on , so we know that is a diameter ofthe large semicircle. But is 10 cm long, so the radius of the large semicircle is 5 cm.

AB ABAB

Let the radii of the other three semicircles be cm, cm and cm. The centres ofthese arcs also lie on , so the sum of their diameters is equal to the length of . Itfollows that and hence .

r1 r 2 r 3AB AB

2r 1 + 2r 2 + 2r3 = 10 r 1 + r2 + r3 = 5Now the lengths, in cm, of the semicircular arcs are , , and . Therefore theperimeter of the figure has length, in cm,

5π πr1 πr2 πr3

5π + πr1 + πr 2 + πr3 = π (5 + r1 + r2 + r3)= π (5 + 5)= 10π.

Hence the perimeter of the figure has length cm.10π

13

3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle.The length of the perimeter of the large rectangle is of the total perimeter of theoriginal two rectangles.

23

Prove that the final rectangle is in fact a square.

First solutionSince the smaller rectangles are placed together edge-to-edge, they have a side length incommon. Let this side have length and let the other sides have lengths and asshown.

y x1 x2

y

x1 x2

The perimeters of the smaller rectangles are and , so the totalperimeter of the two smaller rectangles is .

2x1 + 2y 2x2 + 2y2x1 + 2x2 + 4y

The perimeter of the large rectangle is .2 (x1 + x2) + 2y = 2x1 + 2x2 + 2yWe are given that the length of the perimeter of the large rectangle is of the totalperimeter of the two original rectangles. Hence we may form the equation

23

2x1 + 2x2 + 2y = 23 (2x1 + 2x2 + 4y) .

We may simplify this equation by multiplying both sides by 3 and expanding thebrackets, to obtain

6x1 + 6x2 + 6y = 4x1 + 4x2 + 8y,which simplifies to

x1 + x2 = y.This means that the length and width of the large rectangle are the same. In other words,the rectangle is actually a square.

Second solutionThe total perimeter length of the original two rectangles is equal to the perimeter lengthof the large rectangle added to the lengths of the two edges which are joined together.

P

But the perimeter length of the large rectangle is and hence the two edges which arejoined together have total length .

23P

13P

However, the two edges which are joined together are parallel to two sides of the largerectangle and have the same length as them. Hence these two sides of the large rectanglehave total length .1

3PSince the perimeter length of the large rectangle is , the other two sides of the largerectangle also have total length . It follows that all the sides of the rectangle are equalin length, in other words, the rectangle is a square.

23P

13P

14

4. In the rectangle , the side has length and the side has length 1. Let thecircle with centre and passing through meet at .

ABCD AB 2 ADB C AB X

Find (in degrees).∠ADX

SolutionWe begin with a diagram showing the information given in the question. We have usedthe fact that is a rectangle, so that and , andangles , and are right angles.

ABCD BC = AD = 1 DC = AB = 2ABC BCD CDA

A B

C

X

D

1

2

Since and are both radii of the circle, also has length 1. This means thattriangle is isosceles and so .

BX BC BXXBC ∠BXC = ∠BCX

Furthermore, since is a right angle, and are both equal to .∠ABC ∠BXC ∠BCX 45°From the fact that is a right angle, it follows that .∠BCD ∠XCD = 90° − 45° = 45°We may use Pythagoras' theorem in triangle to obtainXBC

XC2 = BX2 + BC2

= 12 + 12

= 2

and so .XC = 2We are given that also has length and so triangle is isosceles. This meansthat and are equal, and so each is equal to .

DC 2 XCD∠CXD ∠CDX (180° − 45°) ÷ 2 = 671

2°Lastly, we use the fact that is a right angle to conclude that

.∠CDA ∠ADX = 90° − 671

2°= 221

2°

15

5. Two candles are the same height. The first takes 10 hours to burn completely whilst thesecond takes 8 hours to burn completely.Both candles are lit at midday. At what time is the height of the first candle twice theheight of the second candle?

SolutionLet the initial height of each candle be cm. In one hour the first candle will burn cmand the second candle will burn cm. Thus in hours, the candles will burn

h h10

h8 t

ht10

cm ht8

cm,and

respectively.If both candles are lit at midday, then hours after midday the heights of the first andsecond candles will be

t

(h −ht10) cm (h −

ht8 ) cm,and

respectively.We are asked to find the time at which the height of the first candle is twice the height ofthe second candle. We therefore need to find the value of such thatt

h −ht10

= 2 (h −ht8 ) .

We may divide every term by , since we know that is not zero, and expand thebrackets to obtain the equation

h h

1 −t

10= 2 −

t4

.

Multiplying both sides by 20, we get

20 − 2t = 40 − 5t,and so

t =203

= 623.

Hence the height of the first candle is twice that of the second after 6 hours and 40minutes, in other words, this happens at 18:40.

16

6. Teams A, B, C and D competed against each other once. The results table was asfollows:

Team Win Draw Loss Goals for Goals against

A 3 0 0 5 1

B 1 1 1 2 2

C 0 2 1 5 6

D 0 1 2 3 6

(a) Find (with proof) which team won in each of the six matches.

(b) Find (with proof) the scores in each of the six matches.

Solution(a) Team A won all three games and so beat teams B, C and D. Of the three games that team C played, the one that was lost can only have been againstteam A. Therefore team C drew against teams B and D.If we consider the three games that team B played, the game against team A was lost, thegame against team C was a draw and so the remaining game, that team B won, wasagainst team D.In summary:

A beat B, A beat C, A beat D;B drew with C, B beat D; andC drew with D.

(b) Consider the following table in which the rows give the number of goals scored foreach team and the columns give the number of goals against each team.

Goals against

A B C D All

A − z + 1 5

Goals B − x t 2

for C z x − y 5

D y − 3

All 1 2 6 6 15

We have let the number of goals scored by team C against team B be , so that thenumber of goals scored by team B against team C is also , since their match was a draw.Similarly, we have let the number of goals scored by team C against team D be , so thatthis is also the number scored by team D against team C.

xx

y

Furthermore, we have let the number of goals scored by team C against team A be , sothat the number of goals scored by team A against team C is since the differencebetween the number of goals scored and conceded by team C is 1.

zz + 1

Finally, we have let the number of goals scored by team B against team D be . Then isat least 1 since team B beat team D.

t t

17

We observe that the row for C now means that (which agrees with thecolumn for C).

x + y + z = 5

From the column for A we see that is at most 1, since the total in that column is 1.Similarly, from the row for D, we see that is at most 3, and from the row for B we seethat is at most 1 since is at least 1.

zy

x tBut we have , so that the only possibilities are , and .It follows that .

x + y + z = 5 x = 1 y = 3 z = 1t = 1

Therefore the table is:

Goals against

A B C D All

A − 2 5

Goals B − 1 1 2

for C 1 1 − 3 5

D 3 − 3

All 1 2 6 6 15

We may now complete the table by, for example, first noting that all other entries in thecolumn for A are 0, and then filling in the rows from the bottom.

Goals against

A B C D All

A − 1 2 2 5

Goals B 0 − 1 1 2

for C 1 1 − 3 5

D 0 0 3 − 3

All 1 2 6 6 15

In summary, the scores in each match were as follows:

A beat B 1 − 0A beat C 2 − 1A beat D 2 − 0

B drew with C 1 − 1B beat D 1 − 0

C drew with D 3 − 3

18

2010 Solutions

1 The sum of three positive integers is 11 and the sum of the cubes of these numbers is 251.

Find all such triples of numbers.

Solution Let us calculate the first few cubes in order to see what the possibilities are:

13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216 and 73 = 343. (∗)The sum of the cubes of the positive integers is 251, which is less than 343, hence noneof the integers is greater than 6.

Now , therefore at least one of the integers is 5 or more. 2513 = 832

3 > 64 = 43

If one of the integers is 6, then the other two cubes add up to .From (*) above, is the only possibility. Also,

so that 6, 3 and 2 is a possible triple of numbers.

251 − 63 = 251 − 216 = 3533 + 23 = 27 + 8 = 35

6 + 3 + 2 = 11

If one of the integers is 5, then the other two cubes add up to. From (*) above is the only

possibility. Also, so that 5, 5 and 1 is a possible triple of numbers.251 − 53 = 251 − 125 = 126 53 + 13 = 125 + 1 = 126

5 + 5 + 1 = 11

Hence 2, 3, 6 and 1, 5, 5 are the triples of numbers satisfying the given conditions.

2 The diagram shows a square and an equilateraltriangle . The point lies on so that

.

ABCDABE F BC

EC = EFCalculate the angle .BEF

A B

CD

E

F

Solution The diagram seems to include several isosceles triangles,and we solve the problem by proving this is the case. Forexample, since the square and the equilateraltriangle share the side , all their sides are the samelength. That means the triangle is isosceles.

ABCDABE AB

BCE

Now, angle is (since it is thedifference between the interior angle of a square and theinterior angle of an equilateral triangle). Hence angles

and are each , becausethey are the base angles of an isosceles triangle.

EBC 90° − 60° = 30°

BCE CEB 12 (180° − 30°) = 75°

A B

CD

E

F

75°

30°

60°

We are also given that triangle is isosceles. Since we have worked out CEFthat angle , we deduce that angle . FCE = 75° CEF = 180° − (2 × 75°) = 30°Finally, we find that angle .BEF = ∠CEB − ∠CEF = 75° − 30° = 45°

19

3 Find all possible solutions to the ‘word sum’ on the right. Each letter stands for one of the digits 0−9 and has the samemeaning each time it occurs. Different letters stand for differentdigits. No number starts with a zero.

O D D+ O D DE V E N

Solution Firstly, it is clear that the three-digit number ‘ODD’ lies between 100 and 999.Therefore, since ‘EVEN’ = 2 × ‘ODD’, we have

200 < < 1998.‘EVEN’Hence the first digit E of ‘EVEN’ is 1 since it is a four-digit number.

We are left with the following problem: O D D+ O D D1 V 1 N

Now the same numbers are added in the tens and units columns, but N , otherwise Nand E would be equal. The only way for different totals to occur in these columns is forthere to be a ‘carry’ to the tens column, and the greatest possible carry is 1, so that N = 0.

≠ 1

There are two possible digits D that give N = 0, namely 0 and 5. But 0 is already takenas the value of N, so that D = 5. The problem is thus:

O 5 5+ O 5 51 V 1 0

Now, the digit O has to be big enough to produce a carry, but cannot be 5, which isalready taken as the value of D. So the possibilities are

6 5 5+ 6 5 5

1 3 1 0

7 5 5+ 7 5 5

1 5 1 0

8 5 5+ 8 5 5

1 7 1 0

9 5 5+ 9 5 5

1 9 1 0

but the second and fourth of these are not allowed since V repeats a digit used foranother letter. We are left with the two possibilities

6 5 5+ 6 5 5

1 3 1 0

8 5 5+ 8 5 5

1 7 1 0

and it is clear that both of these work.

4 Walking at constant speeds, Eoin and his sister Angharad take 40 minutes and60 minutes respectively to walk to the nearest town.Yesterday, Eoin left home 12 minutes after Angharad. How long was it before he caughtup with her?

Solution Let the distance from home to town be . Now in every minute Eoin travels one-

fortieth of the way to town: that is, a distance of . So after minutes, he has

travelled a distance

D kmD40

km t

tD40

km.

20Similarly, in every minute Angharad travels one-sixtieth of the way to town: that is, a

distance of . But she has had 12 minutes extra walking time. So after Eoin has

been walking for minutes, she has been walking for minutes and so hastravelled a distance

D60

km

t t + 12

(t + 12)D60

km .

We are asked how long Eoin has been walking when they meet. They meet when theyhave travelled equal distances, which is when

tD

40=

(t + 12)D

60.

We cancel the from each side and multiply both sides by 120 to obtain D

120 ×t

40= 120 ×

t + 1260

.

Simplifying, we get3t = 2 (t + 12) ,

which we solve to give .t = 24Thus Eoin catches up with Angharad after he has walked for 24 minutes.

5 A square sheet of paper is folded along , asshown, so that the corner is folded onto the midpoint

of .

ABCD FGB

M CDProve that the sides of triangle have lengths inthe ratio 3 : 4 : 5.

GCM

A B

CD M

G

F

Solution This problem does not give us units, and so we choose them so that the side length of thesquare is . Since is the midpoint of , we have . Then we define .Since , . But, as is the image of after folding, too.

2s M CD CM = s x = CGBC = 2s GB = 2s− x GM GB GM = 2s− x

A B

CD M

G

F

s s

x2s− x

2s− x

Now Pythagoras' theorem for triangle gives usMCG

s2 + x2 = (2s − x)2 .We multiply out to get

s2 + x2 = 4s2 − 4sx + x2.

21

Eliminating the terms and dividing by (which is not zero), we obtainx2 s

s = 4s − 4x,which has the solution .x = 3

4s

Thus the triangle has sides of length , and . Multiplying allthe sides by 4, we get , and , so the side lengths are in the ratio , asrequired.

GCM x = 34s s 2s − x = 5

4s3s 4s 5s 3 : 4 : 5

6 A ‘qprime’ number is a positive integer which is the product of exactly two differentprimes, that is, one of the form , where and are prime and .q × p q p q ≠ pWhat is the length of the longest possible sequence of consecutive integers all of whichare qprime numbers?

Solution To help to understand this problem, it is natural to test the first few numbers to see whichsmall numbers are qprime, and which are not:

1 is not a qprime since it has no prime factors. 2 and 3 are not qprimes since they are prime. 4 is not qprime since it is .2 × 25 is not qprime, since it is prime.

is the first qprime number.6 = 2 × 37 is not qprime. 8 is not qprime, since it is .2 × 2 × 29 is not, since it is .3 × 3

is another qprime.10 = 2 × 511 is not. 12 is not, since it is .2 × 2 × 3

Of course, we cannot prove a general result just by continuing the list, but it can guide usto a proof, such as the one that follows.

We note that no multiple of 4 is ever qprime, since a multiple of 4 is a multiple of .This means that a string of consecutive qprime numbers can be of length at most three,because any sequence of four or more consecutive integers includes a multiple of 4.

2 × 2

We are therefore led to ask whether any strings of three consecutive qprime numbersexist. We have looked as far as 12 and not found any, but we will continue searching,using the fact that none of the numbers is a multiple of 4:

For (13, 14, 15), the number 13 is prime and so not qprime. For (17, 18, 19), 17 is not qprime (nor are the others). For (21, 22, 23), 23 is not qprime. For (25, 26, 27), 25 is not qprime (nor is 27). For (29, 30, 31), 29 is not qprime (nor are the others). For (33, 34, 35), all three are qprime (being , and ).3 × 11 2 × 17 5 × 7

So we have found a sequence of three consecutive qprimes, and have also proved that nosequence of four (or more) consecutive qprimes exists.

Thus the longest possible sequence of consecutive integers all of which are qprimenumbers has length 3.