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NATIONAL MATHEMATICAL CENTRE

NIGERIAN MATHEMATICS OLYMPIAD

SAMPLE QUESTIONS 1


SAMPLE QUESTIONS FOR


PREPARIAD BYPREPARIAD BYPREPARIAD BYPREPARIAD BY RAMAZAN INCERAMAZAN INCERAMAZAN INCERAMAZAN INCE

(MATHEMATICS TEACHER)(MATHEMATICS TEACHER)(MATHEMATICS TEACHER)(MATHEMATICS TEACHER)

BOOK 10

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SAMPLE QUESTIONS 2

FORWARDFORWARDFORWARDFORWARD

The Nigerian International Mathematics and Sciences Olympiads are taking a new positive dimension in Nigeria. From the first Nigerian Mathematics Olympiad Competitions last year, a lot of improvements have been introduced. Since the International Olympiads do not have syllabus like the secondary schools, it is therefore an added advantage to make available problems that candidates can practice before the competitions. It is advisable that students work the problems. This is not to say that the competitive examinations will take the form of multiple choices. The standard International Mathematics Olympiad are in theory form. You may also find useful the compilation of past IMO problems produced by the Olympiad unit last year. National Mathematical Centre is appreciative of the collaboration between her and Nigerian Turkish International College. This work is a welcomed contribution.

Prof. S. O Ale, mni, OFR Director General NMC




SAMPLE QUESTIONS 3

INTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTION The multiple choice problems cover the various areas tested in International Mathematics Olympiad like:

Combinatorics Number Theory Geometry Analysis Algebra etc

Work out the problems and compare with the answers given. You may be able to have the full solutions later. Once more, the Nigerian Turkish International College is recognized for their contributions, especially Mr. Ramazan Ince concerning this present work.

S. D. OLUWANIYI Head Olympiad Unit NMC, Abuja.




SAMPLE QUESTIONS 4

CONTENT FORWARD………………………………………………………………...….……….2 INTRODUCTION……………………………………………………....…….………3 CONTENT………………………………………………….…………….....…………4 LIST OF THE NIGERIAN TEAM FOR IMO AND SENEGAL 2006…..…5 SECTION 1 ( Introduction)………………...................……………….6-13 SECTION 2 (INEQUALITIES)…………..................................….14-17 SECTION 3 (VIETA THEOREM)………………………………...……….18-19 SECTION 4 (Elementary Factorization and Number Theory)….20-23

SECTION 5 (Factorization for n na b± )………………………........…24-27

SECTION 6 (Equations of the form xy ax by c+ + = )…….......…28-29

SECTION 7 (Conjugate Irrationalities)………………………….…….30-33 SECTION 8 (MIXTURA- Mixed Algebraic Problems)…………..….34-38 SECTION 9 (Rate of Growth, Inequalities in Number Theory).39-43 SECTION 10 (Inequalities)……………………….......……….….….….44-48 SECTION 11 (Fillings and Colourings)………………………..….……49-57




SAMPLE QUESTIONS 5

NIGERIAN TEAM FOR PAMO 2005, ALGERIA

1. SEGUN ARIYIBI, NIGERIAN TURKISH INTERNATIONAL COLLEGES, LAGOS HONORABLE MENTION 2. SAMUEL CHUKUMWA, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA 3. JACINTE EBELE LOYALA JESUIT COLLEGE, ABUJA 4.AHIGBE METTHEUS LOYALA JESUIT COLLEGE, ABUJA

NIGERIAN TEAM FOR PAMO 2006, SENEGAL

1. UCHENDU NDUBISI, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA SILVER MEDAL 2. MUAZZAM IDRIS, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA 3. SEGUN ARIYIBI, NIGERIAN TURKISH INTERNATIONAL COLLEGES, LAGOS 4. EKWUE WINNER, CHRIST THE KING COLLEGE, ANAMBRA

NIGERIAN TEAM FOR IMO 2006, SLOVENIA

1. UCHENDU NDUBISI, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA 2. MUAZZAM IDRIS, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA 3. SEGUN ARIYIBI, NIGERIAN TURKISH INTERNATIONAL COLLEGES, LAGOS 4. EKWUE WINNER, CHRIST THE KING COLLEGE, ANAMBRA 5. OLUWAKAYODE JOHN, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA 6. OLURKOYO F. ADETUNJI, REGIANA P.G.S.S., ABUJA




SAMPLE QUESTIONS 6

SECTION 1 ( Introduction)

1. Prove that 53.83.109 + 40.66.96 is a composite number. Solution: 53.83.109 + 40.66.96 = 732931. We cannot find a prime divisor easily by using arithmetic.

( ) ( ) ( ). . 149 . 149 . 149x y z z y x= + − − − xyz= xyz− ( )149 ......+ . Therefore our number is not a prime.

2. Prove the following numbers are prime or not: a) 19982 – 1 = (1998-1).(1998+1) b) 2003+2013 = (200+201).(2002 – 200.201+2012) c) 89999 = 90000 – 1 = 3002 – 1 = ..... d) 216001 = 216000+1 = 603+1 3. (National Olympiad) Prove if 57599 is prime or composite. Solution: 57599 = 57600 – 1 = 2402 – 1 = 239.241

4. Prove that 10 6 6 122 2 .5 5+ + is composite.

Solution:

( )210 6 6 12 5 62 2 .5 5 2 5+ + = +

5. Prove that 10 122 5+ is composite.

Solution:

( ) ( )2 210 12 10 12 6 6 6 6 5 6 3 32 5 2 5 2 .5 2 .5 2 5 2 .5+ = + + − = + −

6. Calculate the sum 2 2 2 2 2 2 2 2

3 5 7 19...

1 .2 2 .3 3 .4 9 .10+ + + +

Solution:

( )( )

( ) ( )

2 2

2 2 222 2

12 1 1 1

1 1 1

k kk

kk k k k k

+ −+= = −

+ + +

2 2 2 2 2 2 2 2 2 2

3 5 7 19 1 1...

1 .2 2 .3 3 .4 9 .10 1 2⇒ + + + + = −

2

1

2+

2

1

3−

2

1

3+

2

1

4−

2

1...

9+ +

2

1

10−

1 991 0.99

100 100= − = =

7. Calculate the sum 2 2 2 2 2 2 2 2 21000 998 996 ... 4 2 999 997 ... 3 1+ + + + + − − − − −




SAMPLE QUESTIONS 7

Solution:

( ) ( ) ( ) ( )2 2 2 2 2 2 2 21000 999 998 997 ... 4 3 2 1− + − + + − + − 1999 1995 1991 ... 7 3= + + + + +

8. Calculate the sum 1 1 1 1

...1 2 2 3 3 4 99 100

+ + + ++ + + +

Solution:

2 1 3 2 4 3 100 99...

2 1 3 2 4 3 100 99

− − − −= + + + +

− − − − 2= 1 3− + 2− 4+ 3− ... 100 99+ + −

100 1 10 1 9= − = − =

9. Prove that for every natural number n

2 2

1 1 1 11 ...

1 2 2 3 3 4 1n n

n n− < + + + + <

+ + + + +

Solution:

2

2 2

1 1

1 1 1 11 ...

1 2 2 3 3 4 1

A n

n nn n

= + −

− < + + + + <+ + + + +��

.

Then we must prove that 21 1 1n n n− < + − < .

Since 2 1n n< +

21 1 1n n⇒ − < + − ü. So L.H.S is clear.

2 2 2 21 1 1 1 1 2 1 0 2n n n n n n n n+ − < ⇒ + = + ⇒ + < + + ⇒ < ü. So R.H.S is also clear.

10. Prove that 1 1 1 1

... 241 3 3 5 5 7 9997 9999

+ + + + >+ + + +

.

Solution: Hint: Multiply each fraction by its conjugate.

11. (IMO) It is known that ( )7

2 1 57122 57121+ = + . Prove ( )7

2 1 57122 57121− = − .

Solution:

( )7

71 1 12 1 2 1

2 1 2 1 57122 57121

− = ⇒ − = = + + +

( )7

2 1 57122 57121⇒ − = −

12. Calculate the expression 11 2 18 11 2 18− + +




SAMPLE QUESTIONS 8

Solution 1:

2 22 11 2 18a ab b− + = −2 2 11

18

a b

ab

+ =⇒

=

( ) ( ) ( ) ( )2 2 2 2

2 2. 2. 9 9 2 2. 2. 9 9= − + + + + ( ) ( )2 2

2 9 2 9= − + +

2 9 2 9 9 2= − + + = − 2+ 9 6+ =

Solution 2:

211 2 18 11 2 18 11 2 18x x− + + = ⇒ = − 11 2 18+ + ( ) ( )2 11 2 18 . 11 2 18+ − +

2 22 2 121 72 22 2 49 36x⇒ = + − = + =

6 6x x⇒ = ± ⇒ =

13. Calculate the expression (IMO was if the following number is rational or irrational)

3 320 14 2 20 14 2+ + −

Solution:

I. ( )3

3 2 2 320 14 2 2 3 . 2 3 .2 2 2.a b a a b ab b+ = + = + + +3 2

2 3

6 20

3 2 14

a ab

a b b

+ =⇒

+ = but this not easy to solve.

II. 3 320 14 2 20 14 2 x+ + − =

( )3 3 2 2 3 3 33 3 3x a b x a a b ab b a b ab a b= + ⇒ = + + + = + + +

3 20 14 2x⇒ = + 20 14 2+ − ( ) ( )2

233 20 14 2 . x+ − 3 340 3 400 392.x x⇒ = + − 3 40 3.2x x⇒ = +

3 6 40 0x x⇒ − + = 4x⇒ = is a root. 3

3 2 2

2

2

46 40

4 4 10

4 6 40

4 16

10 40

xx x

x x x x

x x

x x

x

−− −

− − + +

− −

− −

−

( ) ( )3 2

0

6 40 4 4 10x x x x x

∆<

⇒ − − = − + +��

4x⇒ =

14. (Olympiad Problem) Prove if the number 3 310 10

2 23 3 3 3

+ + − is rational or irrational.

Solution:

3 102

3 3x = +

102

3 3+ − ( )3

10 103 2 . 2 .

3 3 3 3x

+ + −

33

1004 3 4 .

27x x⇒ = + − 3 2

4 3.3

x x⇒ = +

3 2 4 0x x⇒ − − = ( 1 2x = is a root)

3 22 4

.... ....

xx x −− −

−




SAMPLE QUESTIONS 9

15. Let 1 1a = and 1

1n n

n

a aa

+ = + . Prove that 10014 18a< <

Solution: L.H.S

2 2 2 2

1 12

12 2n n n n

n

a a a aa

+ += + + ⇒ > + . Now we need to prove that 100 14a > .

2 2 2 2

100 99 98 1

99

2 2 2 ... 2 2 ... 2a a a a> + > + + > > + + + +��

2

100

100

199

0

a

a

>

> 100 199 196 14a⇒ > > =

R.H.S

We need to prove 100 18a < . ( )na is an increasing sequence so 1n na a +< and 2

11

na≤ .

2

11 1 1n

n

n aa

= ⇒ > ⇒ < 2 2 2

1 2

12 3n n n

n

a a aa

+⇒ = + + ≤ + . ( “=” holds only when n = 1)

2 2 2 2 2

100 99 98 2 1

99

3 3 3 ... 3 ... 3 3 ... 3a a a a a< + < + + < < + + + + ≤ + + + +��

2 2

100 1001 3.99 298a a≤ + ⇒ ≤ 100 298 324 18a⇒ ≤ < =

16. Determine all the necessary values of n such that 4 2 1n n+ + is prime.

Solution:

( ) ( ) ( )24 2 4 2 2 2 2 2 21 2 1 1 1 . 1n n n n n n n n n n n+ + = + + − = + − = + + − + .

Let’s suppose that the number is prime ( )

( )( )

( )

( )( )

2 2

11

1

1

1 . 1

pp

p

p

n n n n

− −− −

⇒ + + − +��

. We have four cases:

1) 2

01 1

1

nn n

n

=− + =

=

û

ü

2) 2

01 1

1

nn n

n

=+ + =

= −

ü

û

3) 2 21 1 2 0 0n n n n− + = − ⇒ − + = → ∆ <

4) 2 21 1 2 0 0n n n n+ + = − ⇒ + + = → ∆ <

Then n = {1,– 1}

A USEFUL FACTORIZATION

( ) ( )( ) ( ) ( )

( ) ( )

2 24 4 2 2 2 2 2 2

2 2 22 2 2 2 2 2

2 2 2 2

4 2 2 . 2 .

2 4 2 2

2 2 . 2 2

x y x y x y x y

x x x y x x xy

x y xy x y xy

+ = + + −= + − = + −= + + + −




SAMPLE QUESTIONS 10

17. Find integral nubers ,x y∈� such that 4 44x y+ ∈Ρ .

Solution:

( )

( )( )

( )

( )( )

4 4 2 2 2 2

11

1

1

4 2 2 . 2 2

pp

p

p

x y x xy y x xy y

− −− −

+ = − + + +��

( )( ) ( )( )2 24 4 2 2

0 0

4 .x y x y y x y y

> >

+ = − + + +��

Without losing geberality we can choose , 0x y ≥ because of 4th power. Then ( ) ( )2 22 2

1 p

x y y x y y− + ≤ + +��

So ( )2 2

0

111

1, 0

0

x y

yx y yx y

x yx yinvalid

y

− = =− + =

= = − = ± ≥

=

. Therefore the solutions are {(1;1),(-1;1),(-1,-1),(1;-1)}

18. Calculate

4 4 4 4 4 4

4 4 4 4 4 4

1 1 1 1 1 11 3 5 7 9 11

4 4 4 4 4 4

1 1 1 1 1 12 4 6 8 10 12

4 4 4 4 4 4

+ + + + + + + + + + + +

Solution:

( )( )( )( )( )( )( )( ) ( )( )( )( )

4 4 4 4 4 4

4 4 4 4 4 4

4.1 1 4.3 1 4.5 1 4.7 1 4.9 1 4.11 1

4.2 1 4.4 1 4.6 1 4.8 1 4.10 1 4.12 1

+ + + + + +=

+ + + + + +

( ) ( )2 21 2.1 2.1 1 2.1 2.1− + + +=

( )21 2.3 2.3− + ( )21 2.3 2.3+ + ( )2... 1 2.11 2.11− + ( )21 2.11 2.11+ +

( )21 2.2 2.2− + ( )21 2.2 2.2+ + ( )21 2.4 2.4− + ( )21 2.4 2.4+ + ( )2... 1 2.12 2.12− + ( )21 2.12 2.12+ +

( ) ( )2 22 1 2 1 1 2 2 1a a a a+ − + + = + +

2

2

2.1 2.1 1 1

2.12 2.12 1 313

− += =

+ +

19. Let n be arbitrary natural number. Prove that ( ) ( )( )1 2 3 1n n n n+ + + + is a perfect square.

Solution:

( )( )( ) ( )( )2 21 2 3 1 3 3 2 1n n n n n n n n+ + + + = + + + +

Let 2 3n n k+ = . Then ( ) ( ) ( )222 22 1 2 1 1 3 1k k k k k n n+ + = + + = + = + +

OR

Let ( )( )2 2 23 1 1 1 1 1 1t n n t t t t= + + ⇒ − + + = − + =




SAMPLE QUESTIONS 11

20. What is the maximum area of the rectangles whose perimeter is a constant real number P? Solution:

2

Pa b+ = = constant

22

4 4 16

P P PS x x x

⇒ = − + = −

. In order to get the maximum area x must be 0.

21. Let , ,a b a b+ ∈� . Is it true that ,a b∈� too?

Solution:

,a b a b∈ ⇒ − ∈� � and ( )( )a b a b a b− = − +

I. 0a b+ =

II. 0a b

a b a ba b

−+ > ⇒ − = ∈

+�

22

22

x ya x y a

a b x

x ya b y b x y b

+ = + ⇒ = ∈ − = ∈ ⇒

−+ = ∈ = − ⇒ = ∈

��

� �

YES

22. Find all integers x and y such that 2 2 86x y− =

Solution:

( ) ( )( )( )

( ) ( )

1 . 86

2.43

1.86

2 . 43

x y x y

− −

− + =

− −

( ) ( ) ( )2x y x y x even+ + − = . So both of them are of the same parity. However there is no case in this form.

Consequently there is no solution.

Theorem: 2 2x y n− = has integral solutions if and only if

4

n is odd

n

By using this theorem, since 86 = 4.21+2 there is no solution.

23. Prove that 2 2 3x y a− = has integral solutions for any a∈� .

Solution 1:

( )2

3 3 31

1 2 ...2

n nn

+ + + + =

,

( )11 2 ...

2

n nn

++ + + = ( )23 3 31 2 ... 1 2 ...n n⇒ + + + = + + +

if a = 0 ü

if a > 0

1 2 3 ...

1 2 3 ... 1

x a

y a

= + + + +

= + + + + − ( ) ( )( )222 2 1 2 3 ... 1 2 3 ... 1x y a a− = + + + + − + + + + −

( )( )33 3 3 3 3 3 31 2 3 ... 1 2 3 ... 1a a= + + + + − + + + + − 2 2 3x y a⇒ − =




SAMPLE QUESTIONS 12

if a < 0 take 1a a= −

1

1

1 2 3 ... 1

1 2 3 ...

x a

y a

= + + + + −

= + + + + ( )( ) ( )

2 22 2

1 11 2 3 ... 1 1 2 3 ...x y a a− = + + + + − − + + + +

( ) ( )33 3 3 3 3 3 3

1 11 2 3 ... 1 1 2 3 ...a a= + + + + − − + + + + 2 2 3

1x y a⇒ − = −

Solution 2:

( )( ) 3x y x y a− + =

I. a is odd. 3

3

3 3 3

11 2 1 2

2 1 1

2

ax

x y x a

x y a y a ay

+= ∈− = = +

+ = = − − = ∈

�

�

ü

ü

II. a is even 2

2

2 2 2

2 2

2

2

a ax

x y a x a a

x y a y a a a ay

+= ∈− = = +

+ = = − − = ∈

�

�

ü

ü

24. Solve the equation 2 3 4x y z+ = in prime numbers. ( ), ,x y z∈Ρ

Solution 1:

(i)

2

2 3

1z x

z x y

− =

+ = ( ) ( )

2 3

3 2

2 1

2 1 1 1

z y

x y y y y

= +

= − = − + +

( ) ( )

( )

2

2

11

2

1 .

. 1 1 0

yx y y

x

x y y

−= + +

→∅ + + >

( )11 3 13

2

yy x

−= ⇒ = ⇒ = but

22 10z = →∅

(ii)

2

2 2

z x y

z x y

− =

+ = ( )

2 2

2

2

2 1

z y y

x y y y y

= +

= − = −

( ) ( )1

. . 12 2

. 1 1 .

y yx y y

x x

∅

−= = −

��

. So there is no solution.

( )( )

( )( )

�

}

0

3 4 2 2 2

3

3

3

3

2 2

2

1

1

1

1

even

y z x z x z x

i y

ii y

y

y

y y z x y

y y

>

= − = + −

− − ∅

− −

→ + = ∅

∅

��

��




SAMPLE QUESTIONS 13

Solution 2:2 3 4x y z+ = All of x, y, z cannot be odd. Any two of x, y, z cannot be even and there is only one

even prime, 2. So we have three 2’s or one 2.

(i) 2 3 42 2 2+ ≠

(ii) if z = 2. Then the minimum value of 2 3x y+ is

2 3 43 3 36 2 16+ = > =

if x = 2 2

31x⇒ ≡

3 4 2y z x⇒ = −

if 33 81 4 77z y y= ⇒ = − = ⇒ ∉Ρ

if 4

33 1z z≠ ⇒ ≡ . Then

4 2

3 30 0 3z x y y− ≡ ⇒ ≡ ⇒ =

4 427 4 31z z z⇒ = − ⇒ = ⇒ ∉Ρ

if y = 2 3 3 4 2

32y y z x⇒ ≡ ⇒ = − . This equation has solution if z = 3 and

3x ≠ .2 28 81 73x x x= − ⇒ = ⇒ ∉Ρ . There is no prime solution.

25. Given that 3 3 , ,a b a b ++ ∈� . Prove that 3 3,a b +∈� .

Solution:

( )2

3 32 23 3 32a b a ab b ++ = + + ∈� ........(1)

3 32 23

3 33 3

a b a ba ab b

a ba b

++

+

+ ∈ += − + ∈

++ ∈

��

�........(2)

By (1) & (2)

( ) ( ) ( )3 3 3 32 2 2 21 2 3 a b a b+ ++ = + ∈ ⇒ + ∈� � ........ (3)

( ) ( ) 31 2 3 ab +− = ∈� .......(4)




SAMPLE QUESTIONS 14

SECTION 2 (INEQUALITIES)

26. Prove that for every ,x y∈� 2 2 2x y xy+ ≥

Solution:

( )2 2 2 2 20 2 0 2x y x y xy x y xy− ≥ ⇒ + − ≥ ⇒ + ≥

27. Prove that 2

a bab

+≥ for , 0a b ≥ .

Solution:

We know that

2 2

2

x yxy

+≥ . Let

2

x a a bab

y b

= +⇒ ≥

=

28. Prove that 2 2 2x y z xy yz zx+ + ≥ + + for every , ,x y z∈� .

Solution:

( ) ( )

2 2

2 2

2 2

2 2 2

2

2

2

2 2

x y xy

y z yz

z x zx

x y z xy yz zx

+ ≥

+ ≥ + ≥ + + ≥ + +

2 2 2x y z xy yz zx⇒ + + ≥ + +

29. (Olympiad Problem) Prove that 3 3 3 3 3 34 10 25 6 9 15− + > − + .

Solution:

?3 3 3 3 3 34 9 25 6 10 15+ + > + +

?3 3 3 3 3 34 9 25 2.3 2.5 3.5⇒ + + > + +

3

2 2 23

3

2

3

5

x

y x y z xy yz zx

z

=

= ⇒ + + ≥ + +

=

30. Solve the system

2

2

2

1

2

4

x yz

y zx

z xy

− =

− = − = −

Solution:

Add sides 2 2 2

0

1x y z yz zx xy

≥

+ + − − − = −�� . So there is no solution.




SAMPLE QUESTIONS 15

31. Prove that ( ) ( )3 3 3 2 2 23x y z xyz x y z x y z xy yz zx+ + − = + + + + − − −

32. Prove that 3 3 3 3x y z xyz+ + ≥ for , , 0x y z ≥ .

Solution: We know that

( ) ( )3 3 3 2 2 2

0 0

3x y z xyz x y z x y z xy yz zx

≥ ≥

+ + − = + + + + − − −��

3 3 3 3 0x y z xyz⇒ + + − ≥ 3 3 3 3x y z xyz⇒ + + ≥

“=” holds if and only if x = y = z

33. Prove that 3 . .3

a b ca b c

+ +≥ for , , 0a b c ≥ .

Solution:

Let 3x a= , 3y b= , 3z c= . We know that 3 3 3 3x y z xyz+ + ≥

3 33 . . . .3

a b ca b c a b c a b c

+ +⇒ + + ≥ ⇒ ≥ .

34. Solve the equation 3 3 9 27x y xy+ = − in

a) � b) � c) { }, 0x x+ = ∈ ≥� �

Solution:

a) 3 3 33 3. . .3x y x y+ + = . We know that

3 3 33 3. . .3x y x y+ + ≥ Equality holds when x = y = 3

b) 3 3 33 3. . .3 0x y x y+ + − = ( )( )2 2 2

*

3 3 3 3 0x y x y xy y x⇒ + + + + − − − =��

I. (*) 0 3x y= ⇒ = =

II. 3 0x y+ + = . x-arbitrary and 3y x= −

35. Solve 3 3 1 3x y xy+ + = in +� .

36. (Olympiad Problem) Solve 3 3 3 3 1x y z xyz+ + = + in � .

Solution:

3 3 3 3 1x y z xyz+ + − = ( ) ( )2 2 2 1x y z x y z xy yz zx⇒ + + + + − − − =

(1,0,0), (0,1,0), (0,0,1)

37. Solve 3 3 3 3 1x y z xyz+ + = + in

+� .

Solution:




SAMPLE QUESTIONS 16

( ) ( )2 2 2

3

1x y z x y z xy yz zx

≥

⇒ + + + + − − − =��

. So there is no solution.

38. Solve 3 3 3 3 1x y z xyz+ + = + in � .

Solution:

( ) ( )2 2 2

0

1x y z x y z xy yz zx

≥

⇒ + + + + − − − =��

. So 2 2 2

1

1

x y z

x y z xy yz zx

+ + =

+ + − − − =

2 2 2 1x y z xy yz zx⇒ + + − − − = ( ) ( ) ( )2 2 22x y y z z x⇒ − + − + − = . Therefore two of them is 1 or – 1 and

the other one is 1. Let y = z. Then ( )22 2 1x y x y− = ⇒ − = −

2 1 3 2x y y+ = ⇒ =

(1,0,0), (0,1,0), (0,0,1)

39. Solve ( ) ( ) ( )3 3 330x y y z z x− + − + − = in � .

Solution:

( )( )3 3 3 2 2 23a b c abc a b c a b c ab bc ca+ + − = + + + + − − − . Let a x y= − , b y z= − , c z x= − .

3 3 30 3a b c a b c abc⇒ + + = ⇒ + + = ( ) ( ) ( ) ( ) ( )( )3 3 3

30

3x y y z z x x y y z z x⇒ − + − + − = − − −��

( ) ( ) ( ) 10x y y z z x⇒ − − − =

. . 10

0

a b c

a b c

=

+ + =

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

10 1 . 1 . 10 .......... 1

10 1 . 2 . 5 ........... 2

= ± ± ±

= ± ± ±

10 1 1

5 2 1

± > ± + ±

± > ± + ± there is no solution.

40. Solve ( ) ( ) ( )3 3 318x y y z z x− + − + − = in � .

Solution:

( )( ) ( )3 18x y y z z x− − − = ( ) ( )( ) 6x y y z z x⇒ − − − =

. . 6

0

a b c

a b c

=

+ + = ( ) ( ) ( ) ( ) ( ) ( )6 1 . 1 . 6 1 . 2 . 3= ± ± ± = ± ± ± but 6 1 1± > ± + ± . Therefore we must choose

( ) ( ) ( )6 1 . 2 . 3= ± ± ±

i)

( )1 1

2 1 .......... 1

3 3 3

x y y x x

y z y x

z x z x z x

− = − ⇒ = + ∈

− = − = + − = ⇒ = + = +

�

and by symmetry ( )1 .......... 2

3

y

x y

z y

∈

= + = +

�

and

( )1 .......... 3

3

z

x z

y z

∈

= + = +

�

.




SAMPLE QUESTIONS 17

ii)

( )1 1

2 1 .......... 4

3 3 3

x y y x x

y z y x

z x z x z x

− = ⇒ = − ∈

− = = − − = − ⇒ = − = −

�

and by symmetry ( )1 .......... 5

3

y

x y

z y

∈

= − = −

�

and

( )1 .......... 6

3

z

x z

y z

∈

= − = −

�

.

Another Method:

( ) ( )( ) ( )3 3 3 3 3x y z x y z x y y z z x+ + − − − = − − −

( ) ( ) ( )3 3 330

a b c

x y y z z x− + − + − =��

so 0a b c+ + =

( ) ( ) ( )( )3 3 3 3

0

3a b c a b c a b b c c a+ + − − − = − − −��

( ) ( )( )3 3 3 3a b c a b b c c a⇒ + + = − − − −

Then the solution follows like the previous one.




SAMPLE QUESTIONS 18

SECTION 3 (VIETA THEOREM)

2 0x px q+ + = , 1 2

1,2

1 2.

x x px

x x q

+ = −

=. ( )( )2

1 2x px q x x x x+ + = − −

2 0, 0ax bx c a+ + = ≠1 2

1,2

1 2.

bx x

ax

cx x

a

− + = =

41. Let x1 and x2 be the roots of 2 6 1 0x x− + = . Prove that 1 2

n nx x+ ∈� for any n∈� .

Solution:

1,2

6 36 4 323 3 2 2

2 2x

± −= = ± = ± ( ) ( )

?

1 2 3 2 2 3 2 2n n

n nx x⇒ + = + + − ∈�

Base: 1 2 6x x+ = ∈� and ( )22 2

1 2 1 2 1 22 34x x x x x x+ = + − = ∈�

Step: Assume that 1 1

1 2

n nx x− −+ ∈� and 1 2

n nx x+ ∈� .

Then ( )( ) 1 1

1 2 1 2 1 2 1 2 2 1

n n n n n nx x x x x x x x x x+ ++ + = + + + ∈� .

( )( ) � ( )1 1 1 1

1 2 1 2 1 2 1 2 1 2

16

n n n n n nx x x x x x x x x x+ + − −

∈ ∈

+ = + + − +

� �

��

1 2

n nx x⇒ + ∈�

42. Let x1 and x2 be the roots of 2 1 0x px+ − = . Prove that if p is odd, tehen 1 2

n nx x+ and 1 1

1 2

n nx x+ ++

are relatively prime(coprime). Solution:

1 2

n n

nS x x= +

( )( ) ( )1 1 1 1

1 2 1 2 1 2 1 2 1 2

n n n n n nx x x x x x x x x x+ + − −+ = + + − + ( )1 1 1 2 1. ...... *n n nS S S x x S+ −⇒ = −

1 2

1 2 1

x x p

x x

+ = −

= −1S p⇒ = − ∈� and ( )2 2

2 1 2 1 22 2S x x x x p= + − = +

If 1nS − and ( ) 1*

n nS S +∈ ⇒ ∈� � .

We want to prove that ( )1, 1n nS S + = .

( ) ( )2

1 2, , 2S S p p= − + . Let’s suppose that p d and 2 2p d+ .

( )2

2 2

22 2 2 1

2

p dp p d d d or d

p d

⇒ + − ⇒ ⇒ = = +

but 2 |d = p because p is odd so d = 1.

Step: Let’s suppose ( )1, 1n nS S− = and suppose that 1

n

n

S d

S d+

then we need d = 1.

� �1 1 1 1 1. .n n n n n n n

d d

S p S S S S p S S d+ − − + −= − + ⇒ = + ⇒

.

However ( )1, 1n nS S− = so d = 1.




SAMPLE QUESTIONS 19

43. Let x1 and x2 be the roots of 2 1x ax b+ + = ( )1 2, ,a b and x x +∈ ∈� � . Prove that

2 2a b+ is

composite. Solution:

1 22

1 2

1 0. 1

x x ax ax b

x x b

+ = −+ + − =

= −

( )1 2

1 21 .

a x x

b x x

= − +⇒

= −( ) ( )2 22 2

1 2 1 21a b x x x x⇒ + = + + −

( )22 2

1 2 1 2 1 2 1 22 1 2x x x x x x x x= + + + − + ( )22 2

1 2 1 21x x x x= + + + ( ) ( )2 2

1 2

2 2

1 1x x

≥ ≥

= + +��

. So 2 2a b+ is composite.

44. (Olympiad Problem) Let x1 and x2 be real roots of 2 0x cx c− − = . Prove that

3 3 3 3

1 2 1 2 0x x x x+ + ≥ .

Solution: ( ) ( ) ( )33 3 3 3 2 2

1 2 1 2 1 2 1 1 2 2 1 2x x x x x x x x x x x x+ + = + − + + ( ) ( )( ) ( )2 3

1 2 1 2 1 2 1 23x x x x x x x x= + + − +

( )( ) ( )32 23 3 0c c c c c= − − + − = ≥

45. Let x1 and x2 be real roots of 2 0, , , ,x px q p q p q n+ + = ∈� . Is it true that

3 3 2

1 2x x n+ ?

Solution:

( ) ( )( )( )2

23 3 2

1 2 1 2 1 2 1 2

3

. 3

p np q n

x x x x x x x x n

−−

+ = + + −

��




SAMPLE QUESTIONS 20

SECTION 4 (Elementary Factorization and Number Theory)

46. (Olympiad Problem) Prove that 1003 1012 1 3+ .

Solution:

3 12 1 3n n++ .

n = 0 03 1 12 1 2 1 3 3+ = + =

n = 1 13 22 1 8 1 9 3+ = + =

n = 2 23 9 32 1 2 1 513 3+ = + =

Let’s suppose that 3 12 1 3 ,n n++ 23n+

.

( ) ( )1

1

33 3 .3 3 3

3

2 1 2 1 2 1 2 1n n n n

n

+

+

+ = + = + = +

( ) ( ) ( )2

2 23 3 3 3 3

3

2 2 1 2 1 2 1 3.2n n n n n

n+

− + = + + − ��

(By induction)

47. (Olympiad Problem) Prove that 10032 1+ 1023 .

48. (Olympiad Problem) Prove that 2002 9992 22 1 2 1− − .

49. Let 22 1n

na = − . Then if n mm n a a< ⇒ . So, prove that 2 22 1 2 1n m

− − .

Solution:

( ) ( )( ) ( )( )( )1 1 1 1 2 2 122 2 .2 2 2 2 2 2 22 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 ...n n n n n n n n

na− − − − − − −

= − = − = − = − + = − + + = After n – m steps

we have ( ) ( ) ( )( ) ( )1 2 12 2 2 2 22 1 2 1 2 1 2 1 ... 2 1m m m m n+ + −

= − + + + + . So 2 22 1 2 1n m

− − , n > m.

50. Find all integer n such that 3 3 3n n+ + .

Solution:

( )( )3 3 23 3 3 9n n n n+ = + − + so 3 27 3n n+ + for every n.

3

3

27 3

3 3

24 3

n n

n n

n

+ +

− + + +

3 24, 12, 8, 6, 4, 3, 2, 1 , 1, 2, 3,..., 24

21, 9, 5, 3, 1, 0, 1, 2, 4, 5, 6,..., 27

n

n

+ = − − − −

= − − − − − −

51. Find all possible values of n such that 3 27 2n n− − .




SAMPLE QUESTIONS 21

52. Find all possible values of n such that 3 23 1n n n+ − + .

53. Solve in positive inetegers 24 65x y+ =

Soution:

( ) ( )2 2 24 65 2 65 2 2 65x x x xy y y y− = ⇒ − = ⇒ − + =

2 66 332 1.

2.2 64 52 65

x

xx

y yyI

xy

= ⇒ = − =

= ⇒ =+ = and

2 18 92 5.

2.2 8 22 13

x

xx

y yyII

xy

= ⇒ = − =

= ⇒ =+ =

54. Solve in positive inetegers 22 65x y+ =

Solution: (We know the solution of 24 65x y+ = )

2

10 10

2

10 10

2

10 10 22 2 2 10

10 10 10

2

10 10

2

10 10

0 0

1,9 1

2,8 4...0,1, 4,5,6,9

.. 3,7 9

4,6 6

5 5

n n

n n

n nn an

n a n n

n n

n n

≡ ⇒ ≡≡ ⇒ ≡≡ ⇒ ≡= ⇒ ≡ ≡ ≡ ⇒ ≡

≡ ⇒ ≡

≡ ⇒ ≡

1 2 3 4 5 6

10

10

2 2 2 2 2 2

2 4 8 6 2 4

65 7 9 3 1 7 9

N

N

N

=

≡

+ ≡

So we have 2 4x k= + or 4 4x k= + . Hence x is even.

Let ( )2. 2 2 22. 2 65 2 65 2 65 4 65X

x X Xx X y y= ⇒ + = + ⇒ + = ⇒ + =

Therefore 2 4

9 9

X x

y y

= =

= = and

5 10

33 33

X x

y y

= =

= =

(4;9), (10;33)

55. (St. Petersburg Olympiad Problem) Solve in positive inetegers 23 55x y+ =

Solution:

According to given equaion y must be even. So the last digit of 2y is 0, 4 or 6. 1 2 3 4

10 10 10 103 3, 3 9, 3 7, 3 1≡ ≡ ≡ ≡ . So x must be even. (55 + 3 = 58, 55 + 7 = 62. They are not the cases)

( ) ( ) ( )( )1 1 1 1 12 2

2 2 23 55 3 55 3 55 3 3 55x x x x xy y y y+ = + = ⇒ − = ⇒ − + =

1

11

283 1.

3 63 55

x

x

yyI

x xy

=− =

= ⇒ =+ = and

1

11

83 5.

1 23 11

x

x

yyII

x xy

=− =

= ⇒ =+ =

(2;8), (6;28)




SAMPLE QUESTIONS 22

56. Solve in positive inetegers 22 1x n+ =

Solution:

( )( )2 1 1x n n= − + .

2 3 12 1, 2, 2 , 2 ,..., 2x x− . ( ) ( )1

2

1 2 21 1 2 3

1 2 4

k

k

are of the same parity

nn n n

n↓ ↓

− = =+ − − = ⇒ ⇒ =

+ = =��

. So 3 22 1 3+ = .

57. Let’s suppose that A = sum of two perfect squares and B = sum of two perfect squares. Prove that A.B = sum of two perfect squares too.

Example:

2 2

2 2

2 2

1 2. 10 3 1

1 1

AA B

B

= + = = +

= + or

2 2

2 2 2 2

2 2

2 3. 325 18 1 10 15

3 4

AA B

B

= + = = + = +

= +

58. Show that ( )( ) ( ) ( )2 22 2 2 2x y z t xz yt xt yz+ + = + + −

Solution:

?2 2 2 2 2 2 2 2 2 2 2 2 2x z x t y z y t x z x t xyzt+ + + = + + 2 2 2 2 2y z y t xyzt+ + −

59. Show that ( )( ) ( ) ( )2 22 2 2 2x y z t xz yt xt yz+ + = − + +

(Solution 12 continued)

2 2 2 2,A x y B z t= + = + ( ) ( )2 22 2 2 2 2 2 2 2.AB x z x t y z y t xz yt xt yz⇒ = + + + = + + −

60. 2 2

1 1 1

2 2 2 2

2 2 2 1 2

2 2

3 3 3

.

A x y

A x y A A X Y

A x y

= +

= + = += +

Base: n = 2

61. Let 1 2,z z ∈� , 1 2 1 2Re , Re , Im , Imz z z z ∈� . (2

1 2,z z ∈� ). Then prove that 2

1 2.z z ∈� .

Solution:

2 2 2

1 1 1 1 1 1 1: .z z x i y z x y∈ = + ⇒ = +� and 2 2 2

2 2 2 2 2 2 2: .z z x i y z x y∈ = + ⇒ = +�

( ) ( )1 2 1 2 1 2 1 2 2 1. . . . .z z x x y y i x y x y⇒ = − + +

( )( ) ( ) ( )2 2 2 2 22 2 2 2

1 1 2 2 1 2 1 2 1 2 1 2 1 2 2 1. . . . . .x y x y z z z z x x y y x y x y+ + = = = − + +




SAMPLE QUESTIONS 23

62. 65 5R = . Prove that ( ) 2,x y ∈� such that ( ),x y T∈ .

63. ( ) ( )5 . 13 , ,m n

R m n += ∈� . Prove that there exist ineteger lattice points

such that the points lie on given circle. ( ) 2,x y∃ ∈� such that

2 2 2x y R+ =

Solution:

2 2 2

2 2

5 1 2

13 2 3

= +

= + and ( ) ( ) �

2

2 . 13 5 .13 5.....5.13.....13m n

m n

m n

R = 5 = = �� . Since each 5 can be written as the sum

of two squares and the product of two squares can be written as the sum of two squares the proof is O.K.




SAMPLE QUESTIONS 24

SECTION 5 (Factorization for n na b± )

64. ( )( )1 2 3 2 2 1...n n n n n n na b a b a a b a b ab b− − − − −− = − + + + + +

65. ( )( )1 2 3 2 2 1...n n n n n n n

n is odd

a b a b a a b a b ab b− − − − −+ = + − + − − +��

66. (Fermat’s Theorem) If 2 1n + ∈Ρ , then 2kn = .

Solution:

Let n has an odd divisor m. (m >1) Then 1.n m n= where 1n ∈� .

( ) ( )( )1 1 1 1. 12 1 2 1 2 1 2 1 2 ...m

m n n n nn −+ = + = + = + − . So 2 1n + is not prime. Therefore our assumption wrong

and n has no odd divisor. Therefore 2kn = .

67. (Mercenn’s Theorem) If 2 1n − ∈Ρ , then n∈Ρ . Solution:

Suppose that n is composite. Then 1 1 1. , , , 1, 1n m n m n m n= ∈ ≠ ≠� .

( ) ( )( )1 1 1. 12 1 2 2 1 2 1 ...m

m n n nn −− = = − = − . So 2 1n − is not prime. Therefore n is prime.

68. Prove that 100 1003 4+ is composite.

Solution:

( ) ( ) ( ) ( )5 5

100 100 20 20 20 203 4 3 4 3 4 . ...+ = + = +

69. Prove that

100 1003 5

2

+∉Ρ .

Solution:

( ) ( ) ( )( )5 5

20 20 20 20100 100 3 5 3 5 ...3 5

2 2 2

+ ++= = and

20 203 5

2

+∈� .

70. Determine if there exists n∈� such that ( )1 1987nnn n+ + .

Solution: Suppose that n is odd.

( )( ) ( ) ( ) ( )11 .... 2 1 .... 1987nn n n n−+ + − = + .

2 1 1987 993n n+ = ⇒ = which is odd.




SAMPLE QUESTIONS 25

71. Determine if there exists n∈� such that ( )1 1997nnn n+ +

Solution:

(1997 2 1 998n n= + ⇒ = but it is not odd)

1997.3 5991= and 2 1 5991 2995n n+ = ⇒ = .

( )( )2995 29952995 2996 2995 2996 ....+ = + . So if it is divisible by 5991, then it is also divisible by 1997.

72. Prove that 1999 19991997 1998+ is not prime.

Solution:

Since 1999 is odd ( ) ( )1999 19991997 1998 1997 1998 ...+ = +

73. Prove that the sum ( ) ( )1 2 3 4 5 6 7 1 2 3 4 5 6 7n n n n n n n+ + + + + + + + + + + + for any odd n.

Solution:

1 2 3 4 5 6 7 7.4+ + + + + + = 7

7

7

4

4

4

1 2 3 4 5 6 7n n n n n n n

+ + + + + +

��

��

��

( ) ( )( )( )( )( )

1 7 1 7 ...

2 6 2 6 ...

3 5 3 5 ...

4

n n

n n

n n

n

+ = +

+ = +

+ = +

All of them divisible by 4 and

( )( )( )( )( )( )

1 6 1 6 ...

2 5 2 5 ...

3 4 3 4 ...

7

n n

n n

n n

n

+ = +

+ = +

+ = +

All

of them divisible by 7. Since ( 7, 4 ) = 1, the number is divisible by 28. Remark: If we replace 7 by 4k + 3, then the solution is same.

74. Prove that we can factorize 582 1 . . , , , 1a b c a b c+ = > and , ,a b c∈� .

Solution:

( ) ( ) ( )( )2 258 29 29 29 29 15 29 152 1 2 1 2 1 2.2 2 2 1 2 2 1+ = + = + − = − + + + . Also ( ) ( )( )

292 22 1 2 1 ...+ = + . So it is

divisible by 5. Then 29 152 2 1 5− + or

29 152 2 1 5+ + . Therefore 582 1 5. .a b+ = .

75. (Euclid’s Theorem) If 2 1n − is prime, then ( )12 2 1n n− − is perfect.

Solution: The number has divisors.

( )12 2 1n n

p

− − →��

1 21 2 4 8 ... 2 2 4 ... 2 .n np p p p− −+ + + + + + + + + + ( )1 22 1 2 4 ... 2n np− −= + + + + +

( ) ( ) ( )( ) ( )( ) ( )1 1 1 1 12 2 1 2 1 2 1 2 1 2 1 2 1 1 2 1 .2n n n n n n n n np− − − − −= + − = − + − − = − − + = −




SAMPLE QUESTIONS 26

76. (Olympiad Problem) Prove that for any n∈� ( )2 1 3 2 3n

n n− − −

Solution 1:

( ) ( )( )� ( )2 1 2 2 3 2 1 2 ...... 2 3n

n n n n n

∈

− − + − = − − + −�

( )( ) ( ) ( )2 3 ... 2 3 2 3n n n= − + − −

Solution 2:

( ) � ( )2 3 2 3 3

n

nn

ba

a b

− + − = + − ��

1 1

1 1

2

. ... . 3n

n n n n

na A a b A ab b− −−= + + + + − ( ) � ( )....... 2 3 ....... 1n

a

a a= + − = +

Solution 3:

( )2 3 2 3

2 1 2 2 1 2n n

nn n n

− −− ≡ ⇒ − ≡ ( ) ( )

2 32 1 3 2 3 0 2 1 3 2 3

n

n nn n n n

−⇒ − − ≡ − ≡ ⇒ − − −

77. Let’s consider

1

10101...10101n zerosn units+

�� . Determine ( with a proof ) for which values of n the number is prime.

Solution:

210101...10101 1 100 10000 ... 1 100 100 ... 100n= + + + = + + + +

( ) ( ) ( ) ( )1 22 1 1 11 1 10 1 10 1 10 1 10 1100 1 100 1

100 1 99 99 99 99

nn n nn n

+ + + ++ + − − − +− −= = = = =

−

if

1

1

10 1 991

10 1 99

n

nn

+

+

− >> ⇒

+ > and

( )( )1 110 1 10 1

99

n n+ +− +∈� . After reduction we will have a product of two

numbers so our number is a composite number.

78. Let ( ), , , 0a b c d∈ >� and ab cd= . Prove that a b c d+ + + is a composite.

Solution:

( )( )2 c a c bab ac bc c aba b c d a b c

c c c

+ ++ + ++ + + = + + + = = . Let 1 2.c c c= and 1 2,c a c c b c+ + .

1 2

.a c b c

c c

+ += .

1a c

a c cc

++ > ⇒ > and

1

1a c a c

c c

+ +≥ > . By the same idea

2

1b c

c

+> then we have

��1 2

1 1

.a c b c

c c

> >

+ +. So

a+b+c+d is composite.

79. Let ( ), , , 0a b c d∈ >� and ab cd= . Prove that 1984 1984 1984 1984a b c d+ + + is a composite.

80. Let 2 1 , , 1k np p n+ = ∈Ρ > . Find all possible values of p, n and k.




SAMPLE QUESTIONS 27

Solution:

( )( )1 2 22 1 1 ... 1k n n n

n numbersn is even

p p p p p p− −= − = − + + + + +��

. Let n = 2.N. Then ( )( )2 1 1 1k n N Np p p= − = − + .

( ) ( )1 1 2N Np p− − + = So ( )( )

{1 2

3, 1 21 4

N

N

N

pp N n

p

− == = ⇒ =

+ =

So the solution is 3 22 1 3+ = .




SAMPLE QUESTIONS 28

SECTION 6 (Equations of the form xy ax by c+ + = )

81. Let ( ), 0a b∈ >� be the lengths of the sides of a rectangle. If the area and perimeter have the same

value, then find the possible values of a and b. Solution 1:

The squares at corner gives 4 more units to perimeter. So the area of the inner part must 4. Then we have two choices

Solution 2:

( )2 2 2 2 2 2ab a b ab a b a b b= + ⇒ − = ⇒ − =

. 2I b = �

( )2 2 42 2 4 4 4. 2

2 2 2 2

bb bII a

b b b b

− +− += = = = +

− − − −

2 4 6 3

2 2 4 4

2 1 3 6

2 1 1 2

b b a

b b a

b b a

b b a

− = ⇒ = ⇒ =

− = ⇒ = ⇒ =⇒

− = ⇒ = ⇒ =

− = − ⇒ = ⇒ = −

. So we have two

rectangles with side lengths (4;4) and (3;6) Solution 3:

( )1

2 2 0 2ab a b a

−

− − = ⇒ −

124

4−2−

( )4

2b

−

−��

421

1−2−

4= ⇒��

82. Show that , , :a b c xy ax by c∀ ∈ + + =� has integer solutions.

Solution:

( )( )x b y a c ab+ + = +

. 0x b x

I c ab ory y a

= − ∈ + =

∈ = −

�

� and

. 0II c ab+ ≠ . Let . , ,c ab α β α β+ = ∈� x b x b

ory a y a

α α

β β

+ = = − ⇒

+ = = −

83. Solve 1 1 1

, px y p+ = ∈Ρ in � .

a 3 4 6

b 6 4 3




SAMPLE QUESTIONS 29

Solution:

( )( ) 210

x yxy px py x p y p p

xy p

+= ⇒ − − = ⇒ − − =

2 2

2 2

1 1

1 1

x p p pp p

y p p p p p

− − − −

− − − −

2

2 2

11 2

2 1

px p p p p

y p p p p p p

−+ +⇒

+ + −

0

0

2p p−

1p −

84. Solve 1 1 1

, nx y n− = ∈� in � . Prove that !∃ solution in n⇔ ∈Ρ� .

Solution: (⇐ )

( )

2

1

1 1 1 10

n

n

y xny nx xy x n

x y n xy n

−− = ⇒ = ⇒ − − = ⇒ − + ( )

2

1

nn

n y+��

( )

2

1n y

n

+ >

=��

2 2

1 1

0

0

n x x n

n y n y n n

n x n xinvalid

n y n y

− = = −

+ = = − ⇒

− = = + = =

. So there

is only one solution when n is prime. (⇒ )

If ( )2 2

1 2 1 2 1 2. . 1n n n n n a a a a n∉Ρ⇒ = ⇒ = < < <

1 1

2 2

n x a x n a

n y a y a n

− = = − ⇒

+ = = − . So if n∉Ρ , then there is at least one solution diffrent than the first solution

because 1 1a > . Therefore if , !n∈Ρ ∃ solution.




SAMPLE QUESTIONS 30

SECTION 7 (Conjugate Irrationalities)

85. Let 2m

n< . Prove that

2

12 1

4

m

n n

< −

.

Solution:

2 2

2 1

4 2 2.n nε = = so

2

12 2

2 2.

m m

n n n< ⇒ < − .

( )2

2 2 2 2 22 2 2 2 1m

m n m n m nn< ⇒ < ⇒ < ⇒ ≤ − . We want to get

2

2 2 4

2 12 1

4 16

m

n n n

< − +

or

�

2 2

2

0

12 1

8m n

n>

< − + . Since 2 22 1m n≤ − ,

2 2

2

12 1

8m n

n< − +

86. Let 7m

n< . Prove that

17

m

n mn< − .

Solution: If only one of m and n is less than 0. It is obvious. Since , 0m n > and , 0m n < give the same equation let’s suppose that , 0m n > .

i) if m = n = 1, then 1 1 1

7, 71 1 1.1< < − ü

ii) if at least one of m and n is/are not equal to 1, then . 2m n ≤ .

a-) Let 2m

n< .

1 17 7 6.25 0.5 2.5 0.5 2

. 2mn⇒ − ≥ − > − > − =

12 7

m

n mn⇒ < < − .

b-) Let 2 7m

n≤ < .

22 2

27 7 7

m mm n

n n< ⇒ < ⇒ < 2 27m n a⇒ = − . This equation has no solution for

1a = and 2a = . 2 27m a n+ =

7

2

7

0 1 2 3 4 5 6

0 1 4 2 2 4 1

m

m

≡

≡.

2m a+ is not divisible by 7 for a = 1 and a = 2. Then 2 27 3m n≤ − .

2?2 2

2 2 2 2

1 2 7 1 2 7. 17 7 7

m m nm n

n mn n mn m n m m< − ⇒ < − + ⇒ < − − +

27n⇒?

23 7n− <?

2 2 2 2

2 7. 1 2 7. 13 0

n n

m m m m− + ⇒ − + >

27n⇒ ( )? ?

23 1 2 7 0 3 2 7 1 0m mn m m n⇒ + − > ⇒ − + > .

2 2m

m nn≥ ⇒ ≥ ( ) ( )

0

3 2 7 1 6 2 7 1 0m m n m n n

>

⇒ − + > − + >��

ü




SAMPLE QUESTIONS 31

87. Let 1 1, , ,a a b b ∈� and 1 12 2a b a b+ = + . Is it possible 1

1 1

1

2 2a a

a b a bb b

=+ = + ⇔

= ?

Solution:

( )1 1 1

1 1 11

1

0

20 2

b b b b and a a

a a b b a ab b CONTRADICTION

b b

− = ⇒ = =

− = − ⇒ − − ≠ ⇒ = ∈ −�

88. Let n is not a perfect square and 1

1 1

1

a aa b n a b n

b b

=+ = + ⇒

=

89. Let ( )1 2 2, ,n

a b a b+ = + ∈� . Prove that ( )1 2 2n

a b− = − .

Solution 1:

( ) ( ) ( ) ( )2 3

1 2 1 31 2 1 1 2 ... 2 1 1 2 ... 1 2 1 2 ...0 1 0 2 1 3

n nn n n n n n

n n n n n n n

n

− − − − + = + + + = + + + + + + +

2a b= +

( ) ( )� ( )�1 2 ... 2 ...n

a b

− = −

Solution 2:

( )1 2 2n

n na b+ = + and ( )1 2 2n

n na b− = − .

Base n = 1 ü

( ) ( ) ( ) ( ) ( )1

1 2 2 1 2 2 2n

n n n n n na b a b a b+

+ = + + = + + +

( ) ( ) ( ) ( ) ( )1

1 2 2 1 2 2 2n

n n n n n na b a b a b+

− = − − = + − +

90. Let ( ) �2001

1 2 1002 1 ........, ... ...integer

decimal partpart

a a a+ =��

be reprsented as decimal number. Find first hundred digits of the

decimal part. Solution:

We know that ( )( )

2001

2001

1 2 2

,

1 2 2

a b

a b

a b

+ = +

∈+ − = −

� ( ) ( )2001 2001

1 2 1 2 2 .........,00...0...a⇒ + + − = = and

( ) ( )2001 2001

1 2 2 1 2 .........,00...0...a+ − − = =

( )10

3

1 1 12 1 2 1

2 1024 10− < ⇒ − < < .




SAMPLE QUESTIONS 32

( ) ( ) �

599

2002001 2000

600

30.000...01

12 1 2 1 10

10

− ⇒ − < − < =

��

( ) � ( )2001 2001

0

1 2 2 2 1a∈

>

+ = + −� ��

. So 1 2 100... 0a a a= = = =

91. Let ( ) �2002

1 2 1002 1 ........, ... ...integer

decimal partpart

a a a+ =��

be reprsented as decimal number. Find first hundred digits of the

decimal part. Solution:

By the same way, we have ( ) ( )2002 2002

1 2 1 2 2 .....,0000....a+ + − = = and

( ) ( )2002 2002

1 2 2 1 2 .....,0000....a+ + − = =

( )( ) �

2002

1 2 100 600

2002

0

1 2 ......, ... ... ...

2 1 ......,00......0........0**

2 ,000......000

a a a a

a

≠

+ =

+ − =

=

In order to have 0’s 1 2 100... 9a a a= = = = by rounding.

92. Find the first hundred digits of the number ( )1985

3 2+ .

93. Prove that it is impossible to find , , ,a b c d ∈� such that ( ) ( )2 2

2 2 5 4 2k k

a b c d+ + + = +

holds. Solution: By conjugate construction (conjugate irrartionality)

( ) ( )2 2

00 0

2 2 5 4 2k k

a b c d<

> >

− + − = −�� CONTRADICTION!

94. Prove that it is impossible to find , , ,a b c d ∈� such that ( ) ( )4 4

3 3 1 3a b c d+ + + = + holds.

Solution:

( ) ( )4 4

3 3 1 3a b c d+ + + = + . Take conjugate ( ) ( ) �4 4

00 0

3 3 1 3a b c d<

> >

− + − = −��

. CONTRADICTION! So

there are no such a, b, c, d.




SAMPLE QUESTIONS 33

95. Prove that if ( )1 2 2n

a b+ = + , then 2 22 1a b− = ± .

Solution:

Take conjugate

( )( )

( ) 2 2

1 2 2

1 2 2

1 2

n

n

n

a b

a b

a b

+ = +

× − = −

− = −

. So depends on n, 2 22 1a b− = ± .

96. Prove that for any natural number n, k∃ ∈� such that ( )2 1 1n

k k− = + − .




SAMPLE QUESTIONS 34

SECTION 8 (MIXTURA- Mixed Algebraic Problems)

97. Let x0 be the real root of 3 2 1

3x x x+ + = − .

a) Is x0 rational or not? b) Find x0.

Solution:

Theorem: Let 0

px

q= and ( ), 1p q = be a root of [ ]1

1 1 0... 0,n n

n na x a x a x a x−−+ + + + = � ,

then 0a p and na q .

a) Now let’s use this theorem ,

3 2 3 2

11 13 3 3 1 0

33 1

px x x x x x

q

+ + = − ⇒ + + + = ⇒ ⇒ ±

and

1

3± are x0 but neither

1

1± nor

1

3± are roots of

the equation. So x0 is not rational.

b) ( )33 2 3 2 3 3 33 3 3 1 0 3 3 1 2 1 2 1 2x x x x x x x x x x x+ + + = ⇒ + + + = − ⇒ + = − ⇒ + = −3

1

1 2x

−⇒ =

+. So

solution is irrational.

98. Find the natural number solutions of y x x yx y x y+ = + .

Solution: i) if x y= ⇒ solution is all natural numbers.

ii) if ( ) ( )1 1y x x y x y y x y x y y x yx y x y x y x x y y x x y y− −> ⇒ + = + ⇒ − = − ⇒ − = − and

1 1

y y

x y x y

x y

x y− −

>

− > − after multiplying both of inequalities we get ( ) ( )1 1y x y y x yx x y y− −− > − because all of

them are positive. So no solution. iii) if y x> ⇒ same way as ii).

99. Let { }, , 0x y z ∈ +� . Then 1n n nx y z ++ = is valid for 0n n≥ . It means we have infinite system as

0 0 0

0 0 0

0 0 0

1

1 1 2

2 2 3

n n n

n n n

n n n

x y z

x y z

x y z

+

+ + +

+ + +

+ =

+ = + =

Solution:

i) if 0 0 0n nz x y x y= ⇒ + = ⇒ = =

ii) 0

n nx y

z zz z

≠ ⇒ + =

for

na

b

we have

1

0

n

n

n

n

n

aa b

b

aa b

b

aa b

b

→∞

→∞

= ⇒ =

< ⇒ →

> ⇒ →∞

. So if x > z , then left hand side

n→∞→∞ but z is constant and




SAMPLE QUESTIONS 35

if y > z, then L.H. S n→∞→∞ , but z is constant. Therefore these cases are impossible and

if x, y < z, then 0

n nx y

z z →∞

+ →

impossible because 0z ≠ . However

if x = z, then 1

ny

zz

+ = ⇒

y = z or y = 0. Therefore 1 1z x= ⇒ = OR

if y = z, then 1

nx

zz

+ = ⇒

x = z or x = 0. Therefore 1 1z y= ⇒ = .

Consequently the solutions are 1) x = y = z = 0 2) x = z = 1, y = 0 3) y = z = 1, x = 0 4) x = y = z = 2

100. Let n n nx y z+ = . Prove that the equation is not true if n z≥ for , ,x y z∈� .

Solution 1:

Theorem (Bernoulli Inequality): ( )1 1n

a na+ > + for a>0, { }0n∈ −�

Let’s suppose that it holds. We know that ,z x y> and ( ) ( ) ( )?

1 1 2 1n n nn n nx y z z z z+ ≤ − + − = − < .

( ) �1 1 1

2 1 2 1 1 21 1 1 1

n n nnn

Bernoulli

z z nn z z z

z z z z

− + ≥ ⇒ > − ⇒ > ⇒ = + > + > − − − −

. So

( )2 1nn n nz z x y> − ≥ + .

Solution 2:

( )1 1nnz z= − + and 1 ,z x y− ≥ because ,z x y> and w.l.o.g we can choose x y≥ ( because of

symmetry). So ( ) ( ) 11 1 1 ... ... 2n n n n n n n n nz x x nx x x x x y−− + ≥ + = + + ≥ + + > ≥ + . Therefore

n n nz x y> + .

101. (Quadratic function) Definition: 100 10abc a b c= + + and 0 , , 9a b c≤ ≤ .

Let abc∈Ρ . Prove that 2 4b ac− is not a perfect square.

Solution:

Let’s consider 2 0ax bx c+ + = and suppose that

2 24b ac n− = . It means that 2 4b ac− is a perfect square

and let’s try to get a contradiction. 2

1,2

4

2 2

b b ac b nx

a a

− ± − − ±= = and

2

2 2

b n b nax bx c a x x

a a

− − − + + + = − −

( because ( )( )2

1 2ax bx c a x x x x+ + = − − .) and now, 2.10 .10 10 10

2 2

b n b nabc a b c a

a a

+ − = + + = + +

from 2 24b ac n− = , b and n are of the same parity so

2

b n+ and

2

b n− are integers.




SAMPLE QUESTIONS 36

2 10 10 10 102 2 2 2

10 102 2

b n b n b n b na a a

b n b n a a a aa abc

a a a a

+ − + − + + + + + − ⇒ + + = = =

. So iti is

reducable and each factor is bigger than 10a but this is a contradicition because abc was prime.

102. If , ,a b c∈� and ( ) 0a a b c+ + < , then prove that 2 4b ac> .

Solution:

( ) ( )2 1f x ax bx c f a b c= + + ⇒ = + +

if

( ) ( ) ( )0 . 1 0 1 0a a a b c a f f> ⇒ + + = < ⇒ < . So there are two solutions. It means

2 20 4 0 4b ac b ac∆ > ⇒ − > ⇒ > OR

if ( ) ( ) ( )0 . 1 0 1 0a a a b c a f f< ⇒ + + = < ⇒ > ⇒ two solutions 0⇒ ∆ > .

103. If , ,a b c∈� and ( ) 0c a b c+ + < , then prove that 2 4b ac> .

Solution:

i) if ( ) 2f x cx bx a= + + , then the same solution as question 6.

ii) ( ) ( ) ( ) ( )0 0 . 1 0c f c a b c f f= ⇒ + + = < this means

( )0 0f < and ( )1 0f > .........(1) or

( )0 0f > and ( )1 0f < .........(2)

So the graph cuts x-axis and this means that it will cut x-axis again because the graph is a parabola.

....(1) ....(2)




SAMPLE QUESTIONS 37

104. Let

( )( )( )

2

1 1 1 1

2

2 2 2 2

2

3 3 3 3

f x a x b x c

f x a x b x c

f x a x b x c

= + +

= + +

= + +

be quadratic functions and 1 2 3, , 0a a a > and let’s suppose every

two of them has a common root. Prove that ( ) ( )( )2

1 2 3 1 2 3 1 2 34b b b a a a c c c+ + ≥ + + + + .

Solution: Algebraic way is very long.

i) if common root is same for each pair. It means x0 is common root of 1 2 3, ,f f f

1 0 2 0 3 0( ) ( ) ( ) 0f x f x f x⇒ + + = 2

1 2 3 1 2 3 1 2 3( ) ( ) ( ) 0a a a x b b b x c c c⇒ + + + + + + + + = has root. So its 0∆ > .

2

1 2 3 1 2 3 1 2 3( ) 4( )( ) 0b b b a a a c c c∆ = + + − + + + + > ü ii) if common root is different for each pair

1 2 3x x x< < . So ( ) ( ) ( ) ( )1 2 3F x f x f x f x= + + and for

( ) ( ) ( ) ( )2 1 2 2 2 3 2

01two of them areof them is negative

F x f x f x f x= + +��

( )2 0F x⇒ < and

( ) ( ) ( ) ( )2

1 2 3 1 2 3 1 2 3F x a a a x b b b x c c c= + + + + + + + + . ( Here 1 2 3 0a a a+ + >

is given.) ( ) ( ) ( )1 2 3 2 0 0F x

a a a F x⇒ + + < ⇒ ∆ > it means 0∆ > .

105. Solve ( )2 25 1 6 8x y x y+ + = + in real numbers.

Solution 1:

2 2

3 4 35 5 0

55 5x y x

− + − = ⇒ =

and

4

5y =

Solution 2:

Let’s consider the equation as a quadratic equation in x. ( )2 25 6 5 8 5 0x x y y⇒ − + − + = . Then

( ) ( )222

1,2

3 25 40 16 3 5 46 100 160 64

10 5 5

y y yy yx

± − − + ± − −± − + −= = = . In order to have solution

45 4 0

5y y− = ⇒ = and

3

5x = .

106. Solve 2 3

1

xz yt

xt zy

− =

+ = in � .

Solution:




SAMPLE QUESTIONS 38

2 2 2 2

2 2 2 2

4 4 9

2 1

x z y t xyzt

x t z y xyzt

+ − =

+ + = multiply 2nd equation by 2 and add eqn’s. We have

2 2 2 2 2 2 2 22 2 4 11x z x t y z y t+ + + = ( )( )2 2 2 22 2 11x y z t⇒ + + = . So ( )( )

2 2

2 2

2 1

2 11

x y

z t

+ =

+ =

or ( )( )

2 2

2 2

2 11

2 1

x y

z t

+ =

+ =

1, 0, 3, 1x y z t= ± = = ± = ± ⇒ check solution in original equation. Since y = 0 we have

1

3 0

1 3

1

x

xz y

xt z

t

== =

⇒ = =

=

or

1

0

3

1

x

y

z

t

= − =

= − = −

and

2 2

2 2

3

2 11 1 3

1 12 1

0

x

x y y xz

z yzz t

t

= ± + = = ± =

⇒ ⇒ = ± =+ =

=

3

1

1

0

x

y

z

t

= =

⇒ =

=

or

3

1

1

0

x

y

z

t

= − = −

= − =

107. { }, , 0k l m∈ −� are given. 3 3 3, ,k l l m m k ( cyclic symmetry ). Prove that

( )13k l m klm+ + .

Solution:

( )13

, ,

. . . , 13x y z

x y z

k l m A k l m x y z+ + = + + =∑

i) Let , , 1 x y zx y z k l m klm≥ ⇒

ii) Let ( )1 1

(?)

, 1, 0 .x y x y

m

x y z k l kl k l− −≥ = ⇒ =

��

if 11 3 yy l m−− ≥ ⇒

if 1 2 3 13 10y y x y x− ≤ ⇒ ≤ ⇒ + = ⇒ ≥

( )�

( )3 39 3 3 3

l

k k l k m= ⇒

.

By symmetry , 1, 0x z y≥ = ................

iii) Let 0, 13y z x= = = . ? ?

13 12k klm k lm⇒

� �12 9 3 12.

m l

k k k k lm= ⇒




SAMPLE QUESTIONS 39

SECTION 9 (Rate of Growth, Inequalities in Number Theory)

108. Let 1 2

1 2

2 ...

5 ...

n

k

n

l

a a a

b b b

=

= ai and bj are digits. Then find k + l.

Solution:

1 110 2 10 10 2 10k n k k n k− −≤ < ⇒ < < .....(1) (� �1

1 2

1

100...0 10 ... 99...92 10m n m

m

m m

x x x−

−

= < ≤ < )

Similarly 110 5 10l n l− < < ......(2)

Multiply (1) and (2) side by side 210 10 10 2 1 1k l n k l k l n k l n k l k l n+ − +< < ⇒ + − < < + ⇒ = + − ⇒ + = +

109. Let 2 ...

5 ...

n

n

a

a

=

= . Then find a.

Solution:

( )( )

( )22

.10 2 1 .10

.10 5 1 .10

.10 10 1 .10

m n m

l n l

m l n m l

a a

a a

a a+ +

< < +

< < +

< < +

( )22 10 1 3a a a⇒ < < + ⇒ = (5 52 32, 5 3125= = )

110. Solve 2 23m m k+ = in � .

Solution:

2 2 3m m m≤ +

I. 2 23 0, 0m m m m k+ = ⇒ = =

II. 2 2 20 2 1 3 4 4m m m m m m m> ⇒ + + ≤ + < + + ( ) ( )

2

2 221 3 2

k

m m m m⇒ + ≤ + < +��

2 22 1 3 1 2m m m m m k⇒ + + = + ⇒ = ⇒ =

111. (Olympiad Problem) Find all possible natural values of m and n such that 2m n+ and 2n m+

are perfect squares. Solution: I. m = n = 0

II. Let

2

2 22 2

0 00

0

by symmetry

m nn km

n k m kn n

= = = = ⇒

= =+ = ��

III. Let 2 2, 1,m n m n m≥ + >




SAMPLE QUESTIONS 40

If ( )

( )

22 2 2

22 2 2

1

1

m n x m n m

n m y n m n

+ = ⇒ + ≥ +

+ = ⇒ + ≥ +

( )

( )

22

22

2

1

1

m n m

n m n

m

+ ≥ +

+ + ≥ +

2n+ 2m n m+ + ≥ 2n+ ( )2 2m n+ + +

2 0m n⇒ + + ≤ but this is imposible for natural numbers.

112. Two boys play a game. They write a digit one after another. First boy wants not to get a perfect square but the second one wants. Is there any strategy for second boy to win? (They can write their digits at right or left.)

Solution:

0

I II

1

2 25

3 36

4

5 25

6

7

8 81

9

21 23

1,4,9,16,25,36,49,64,81,100,121,144��

(1) 7x (2) 8y (2) – (1) < 21 . So they cannot be a perfect square if the last digit is 7 or 8.

7 77 * , * 7

8 8

113. Solve 3 4 5n n n+ = in � .

Solution 1: i) n = 1 is not a solution.

ii) 2 2 23 4 5 2n+ = ⇒ = is a natural solution.

iii) 5 3k k> and 5 4k k> for 1k ≥ .

So ( )2 2 2 2 2 2 2 2 25 5 3 4 5 .3 5 .4 3 .3 4 .4 3 4k k k k k k k k+ + += + = + > + = + .

Therefore tehere is no solution except n = 2. Solution 2:

3 43 4 5 1

5 5

n n

n n n + = ⇒ + =

, 3 4

0 , 15 5

< < .




SAMPLE QUESTIONS 41

0 2 1

71 15

2 1 1

3 ... ...

Left Right

n

n

n

n

= >

= >

= =

≥ <

114. Solve 3 5

42

n nn+

= in � .

Solution 1:

3 5 2.4n n n+ =

i) 1 3 5 8n = ⇒ + = . So n = 1 is a solution.

ii) Assume that ( )3 5 2.4 1k k k k+ ≥ ≥ .

Then

( )1 13 5 3.3 5.5 4.3 4.5 3 5 4 3 5 5 3k k k k k k k k k k k k+ ++ = + = + − + = + + −1 1

0

4.2.4 5 3 2.4 5 3 2.4k k k k k k k+ +

>

≥ + − = + − >��

Then since for k = 1, 3 5 2.4k k k+ =

For k = 2, 3 5 2.4k k k+ > and since for 2, 3 5 2.4k k kk ≥ + ≥ there is only one solution n = 1.

Solution 2:

3 54

2 2

n nn+ = .

4 3n n> and ( ) ( )5 1 44 2.4 5 0.5 0.8

2 2 5

nnnn n n > ⇒ > ⇒ < ⇒ <

( )4 0.5 0,...4 .n A> ⇒ < � impossible. n = 1 ü

n = 2 �

n = 3 �

Solution 3:

3 5 3 5

2 25 4 5 4

n nn n

n n

+ = ⇒ + =

and 5 1

1 14 4 4

n nn = + > +

4 1 24

nn ≥ ⇒ + ≥

n = 1 ü

n = 2 �

n = 3 �

n = 4 �

Solution 4:

3 5 2.4 5 4 4 3n n n n n n n+ = ⇒ − = −




SAMPLE QUESTIONS 42

( ) ( ) ( ) ( )1 2 1 25 4 5 5 .4 ... 4 3 4 4 .3 ...n n n n− − − −⇒ − + + = − + +

1 2

1 2

5 5 .4 ...

4 4 .3 ...

n n

n n

AA B

B

− −

− −

= + + >

= + + 1 0n⇒ − ≥

n = 1

115. Solve ( ) ( )( )( )1 1 2 3m m n n n n+ = + + + in � .

Solution 1: ( ) ( )( )2 21 3 3 2m m n n n n+ = + + + .

Let 2 3n n k+ = . Then ( ) ( )1 2m m k k+ = + .

If 2 2 1k m k m m≥ ⇒ + ≥ + > + , then ( ) ( )2 1k k m m+ > + .

If ( ) ( )1 2 1 2 1k m k m k m k k m m< ⇒ ≤ − ⇒ + ≤ + ⇒ + < + .

Therefore the equation has no natural solution. Solution 2:

( )22 2 21 3 1m m n n k+ + = + + =

( )22 2 21 2 1 1m m m m m m< + + ≤ + + = + . “=” holds when m = 0.

( )22 2 1m k m< ≤ +

0 0m n= ⇒ =

Exercise: Solve the same problem in � .

116. Prove that 1986 19861981 30+ is not a perfect square.

Solution 1:

If modulo is used 2n ≡

3

2 and 2n ≡3

3 , ... it is O.K.

Solution 2:

( )29931986 2.993 1981= ⇒ and next consecutive perfect square is ( )2

993 1986 9931981 1 1981 2.1981 1+ = + + . If

we can show that ( ) ( )2 2993 1986 1986 9931981 1981 30 1981 1

x

< + < +�� , then it will be O.K.

( )9932 993 2 19861981 30 1981 30 30> ⇒ > = ( )2

993 1986 993 1986 19861981 1 1981 2.1981 1 1981 30⇒ + = + + > +

117. Solve 2xy yz zx xyz+ + = + in � .

Solution: W.l.o.g put number in order x y z≤ ≤

i) 0 2 1, 2x yz y z= ⇒ = ⇒ = = ( ) ( ) ( ) ( ) ( ) ( )0,1, 2 , 0, 2,1 , 1,0, 2 , 1,2,0 , 2,1,0 , 2,0,1⇒

ii) x = 1 y z yz⇒ + + yz= ( )2 2 1 1,1,1y z y z+ ⇒ + = ⇒ = = ⇒

iii) x = 2 2 2y z yz⇒ + + 2= 2 2 2 2yz yz y z+ ⇒ − − = − ( )( ) ( ) ( )2 11 2

2 2 2, 2 2y z y z

− −

⇒ − − = − ≤ −��




SAMPLE QUESTIONS 43

2 2 0y y− = − ⇒ =

⇒ 2 1 3

2 2 4

y y

z z

− = ⇒ = − = ⇒ =

( ) ( ) ( ) ( ) ( ) ( )2,3, 4 , 2,4,3 , 3, 2, 4 , 3, 4, 2 , 4,2,3 , 4,3, 2⇒

iv) 3 3 ,x x y z= ⇒ ≤ ≤

3 3 3 2y z yz yz+ + = +

3

3

3 3 3

y yz

z yz

yz yz

y z yz yz

≤ ≤ + ≤ + + ≤

3 3 3 2y z yz yz+ + < + . So our case is actually 3 ,x y z≤ ≤

118. Prove that 1978 1n − 1000 1n − .

Solution:

a b

b b

a b b

−

−

Let’s suppose that 1978 1 1000 1n n− − . Then

( ) ( )1978 1 1000 1 1000 1 1978 1000 1000 1n n n n n n− − − − ⇒ − − .

2 .989 2 .500 1000 1n n n n n⇒ − − ( )2 . 989 500 1000 1n n n n⇒ − −

2n is a power of 2 but 1000 1n − is not. So 2n 1000 1 989 500 1000 1n n n n− ⇒ − − but

989 500 1000 1n n n− < − . This is a contradiction so 1978 1n − 1000 1n − .




SAMPLE QUESTIONS 44

SECTION 10 (Inequalities)

Note: 1

0 2x xx

> ⇒ + ≥

119. Prove that 6a b a c b c

c b a

+ + ++ + ≥ .

Solution:

2 2 2

6a c b c a b

c a c b b a

≥ ≥ ≥

+ + + + + ≥ ��

120. Find the smallest value of 3

2

c b a

a b a c b c+ + ≥

+ + + if a, b, c > 0.

Solution: Let

( )2

a b x

b c y

a c z

a b c x y z

+ = + = + + =+ + = + +

2

x y za b c

+ +⇒ + + =

( ) ( )2 2

x y z x y za a b c b c y

+ + − +⇒ = + + − + = − = , ( ) ( )

2 2

x y z x y zb a b c a c z

+ + + −= + + − + = − = ,

( ) ( )2 2

x y z x y zc a b c a b x

+ + − + += + + − + = − = .

2 2 2

x y z x y z x y zc b a

a b a c b c x y z

− + + − + + −

⇒ + + = + ++ + + 2 2 2

x y z x y z x y zT

x y z

− + + − + + −⇒ + + =

( )2 ... 2T⇒ = 1 1 1 2y z x z x y

Tx y z

+ + +⇒ − + − + − + = ( We know from the second question that

6x y x z y z

z y x

+ + ++ + ≥ )

So 3

6 3 22

T T− ≥ ⇒ ≤ . Therefore the answer is 3

2.

121. Prove that ( )( )2 21 1 1ab a b+ − − ≤ if 1 , 1a b− ≤ ≤ .

Solution:

Let sina α= and cosb β= .

( ) ( )2 2sin .cos 1 sin 1 cosα β α β⇒ + − − sin .cos cos .sinα β α β⇒ + . W.l.o.g. suppose that 0 , 1a b≤ ≤ .




SAMPLE QUESTIONS 45

( )sin .cos cos .sin cos 1α β α β α β⇒ + = − ≤

122. Prove that ( )( )2 2 21 . 1 1 1ab c c a b− + − − ≤ if 1 , , 1a b c− ≤ ≤

Solution:

Let sina α= , sinb β= and sinc γ= .

sin .sin .cos cos .cos .sin

T

α β γ α β γ⇒ +�� sin . sin cos . cos 1

we have proved before

T α β α β⇒ ≤ + ≤��

because sin

1cos

γ

γ

≤

.

123. Let’s consider 0

2

2a = ,

2

1

21 1

2n na a+ = − − , 0 1b = ,

2

1

1 1n

n

n

bb

a+

+ −= . Prove that for

any n n na b< . ( Hint: The idea is trigonometric renaming.)

Solution:

2sin sin

2n nna a

πα

+= ⇒ = , tannb α≥ .

22 2

1

11 2sin1 1 1 tan 1 1 cos 1cos 2 tan . tan

sin sin sin .cos 2 cos 22sin .cos .cos

2 2

n

n

n

bb

a

αα α α αα

α αα α α α αα+

−+ − + − −≥ ≥ = = = = ≥

(equality holds when cos 1α = )

0

0

2sin

2 4

1 tan4

a

b

π

π

= =

= =

1 tan , cos 18 4

bπ π

> ≠

1 2sin tan8 16

a bπ π

= ⇒ >2

2 2

2

0 ... 1

sin2tan sin

2 2cos

2

n

n nn n

n

b a

ππ π

π+

+ +

+

< <

⇒ > = > =

��

124. Prove that 2 2 2 2 1a b b c a c a b c+ + + + ++ + < + for , , 0a b c > .

Solution 1:

Let ( )2 , 2 , 2 , , 1a b cx y z x y z= = = > . . . 2. . . 1x y y z z x x y z⇒ + + < + . We need other steps.....

Solution 2:

( )2 1 , 2 1 , 2 1 , , 0a b cx y z x y z= + = + = + >

( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )1 1 1 1 1 1 2 1 1 1 1x y y z z x x y z⇒ + + + + + + + + < + + + +




SAMPLE QUESTIONS 46

1⇒ x+ y+ 1xy+ + y+ z+ 1yz+ + z+ x+ 2xz+ < 2x+ 2y+ 2z+ 2 2 2 1xy yz xz+ + + +

0 2xy yz xz xyz⇒ < + + + ü

125. Prove that 2 1xy yz zx xyz+ + < + if , , 1x y z > .

Solution:

Linearity y mx n= +

m = 0 ⇒ m > 0 ⇒ ( )[ ]

( ),

minx a b

f x f a∈

= and

( )[ ]

( ),

maxx a b

f x f b∈

=

m < 0 ⇒ ( )[ ]

( ),

minx a b

f x f b∈

= and ( )[ ]

( ),

maxx a b

f x f a∈

= .

So in order to see the maximum or minimum value of a linear function it is enough

to look at endpoints.

( ) ( )2 1 2 1

m n

xyz xy yz xz x yz y z yz⇒ − − − + ⇒ − − + −��

. ( ) ( )1

min 1x

mx n f≥

+ = for y, z be fixed.

2 1 2 1 1 1 1 0xyz xy yz zx⇒ − − − + > − − − + =

126. Let 1 2 3 4, , , 1x x x x ≤ . Find

( )1 2 3 4 1 2 1 3 1 4 2 3 2 4 3 4 1 2 3 1 2 4 1 3 4 2 3 4 1 2 3 4max x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x+ + + − − − − − − + + + + −

. Solution 1:

Let ( )( )( ) ( )1 2 3 4

0

1 1 1 1 1A x x x x

≥

= − − − − −��

. If 1 max 1ix A= ⇒ =

Solution 2:

( ) ( ) �1 1 2 3 4 2 3 2 4 3 4 2 3 41 .....n

m

f x x x x x x x x x x x x x x= − − − + + + − +��

( )1 1.f x m x n⇒ = + . Since ( )1f x is linear it

must have the maximum value at the endpoints.

1ix⇒ = ± . We have 42 16= possibilities.

For example 1 2 3 41, 1, 1, 1x x x x= − = + = − = − 1 8 7 1, ..., ...A⇒ = − =

127. Let [ ]0,1ix ∈ . Prove that 1 2 1 2 2 3 1 1... ...n n n n

F

x x x x x x x x x x x−+ + + − − − − − ≤��

a b[ ]

x

y y=f(x)




SAMPLE QUESTIONS 47

a) 2

n b)

2

n

Solution:

a) ( ) ( )1 1 21 ......nF x x x x= − − + .

( ) 31 2

1 2 3

1 11 1max ,... , , ,....,

0 0 0 0

n

i

n

x xx xF x

x x x x

= == = = = = =

and 2 2

n nA A

≤ ⇒ ≤

so after proving part a, part b is

obvious.

i) Let’s suppose that 1 1x =

1 2... ...

0 1 0

1 1 0 1 0 0

0 1 1 0 0 1

1 1 1 1 0 1

nx x x

GOOD→

→

→

→

. So in cyclic group we don’t have two 1’s next to each other. Therefore

1 2 2 3 1 1... 0n n nx x x x x x x x−− − − − − = 1 2 ...2

n

nx x x⇒ + + + ≤

ii) Let 1 0x =

1 2... ...

1 0 1

1 0 0 1 0 1

nx x x

→

1 2 1 2 2 3 1

0

.. ...n n

k

x x x x x x x x x⇒ + + + − − − −��

0

0

1 22

2

nn k

n

− ⇒ =

−

22

nk⇒ <

128. Prove that ( ) ( )2 2 2 2 2 23 2 3

F

x y y z z x xyz x y z+ + − + + ≤��

for [ ], , 0;1x y z∈

Solution:

( ) ( ) ( )2 2 23 3 2 ... ...F x y z yz x= + − + + ( )( ) ( ) ( )22 2 2

0

2 2 ... ...F x y z y z x

≥

⇒ = − + + + +��

I. 2 20 3 3x y z= ⇒ ≤ ü

II. ( )2 2 2 21 3 3 3 2 1 3x y z y z yz y z= ⇒ + + − + + ≤

i) 20 3 3y z= ⇒ ≤ ü

ii) 1 3y = ⇒ ( )2 23 3 2 2 3z z z z+ + − + ≤ ( )2 2 2 26 2 4 0 4 4 4 0z z z z z z z⇒ − − ≤ ⇒ − = − ≤ ü

129. Let 3 2 0ax bx cx d+ + + = has three different real roots. Prove that

2 3b ac> .




SAMPLE QUESTIONS 48

Solution:

and

( )3 2 23 2 0ax bx cx d ax bx c′+ + + = + + = 2 24 4.3 0 3b ac b ac⇒ ∆ = − > ⇒ >

130. Let 4 3 2 0ax bx cx dx e+ + + + = has four different real roots. Prove that

2 9.

4c bd> .

Solution: Rolle’s Theorem.

( ) ( )4 3 2 3 24 3 2f x ax bx cx dx e f x ax bx cx d′= + + + + ⇒ = + + + has three distinct real roots. We habe

eliminated e but in order to eliminate a we need additional ideas.

Let 1

tx

= 3 2

3 2

2 34 0 2 3 4 0

d c ba dt ct bt a

x x x⇒ + + + = ⇒ + + + = it has also three distinct real roots t’s.

( ) 23 4 3 0f t dt ct b′⇒ = + + = 2 2 36 90 16 36 0 .

16 4

bdc bd c bd⇒ ∆ > ⇒ − > ⇒ > = .




SAMPLE QUESTIONS 49

SECTION 11 (Fillings and Colourings)

dominoe triminoe angle triminoe

131. Is it possible to fit a chess board by ? (8x8 chess board) Solution:

8 8 64× = 3 . So it is impossible.

132. Let’s cut off two neighbour corner of a chess board. Is it possible to fill the remaining part by dominoes?

Solution: Since we cut off 1 white and 1 black square the number of black squares and white squares is 31 and total number of squares is 62 which is divisible by 2. Therefore the answer is YES.

133. Let’s cut off two opposite corner of a chess board. Is it possible to fill the remaining part by dominoes?

Solution:

62 2 but the answer is NO.

The figure contains 30 black and 32 white and each dominoe covers 1 black and 1 white.

134. Let’s suppose thatt we cut off twoo arbitrary squares of different colours from a chess board. Prove that it is possible to fill the remaining part with dominoes.

Solution: If we fill the cyclic pattern with dominoes there is no empty space because there are always even number of squares in this pattern.

135. Is it possible to fill out 5x9 rectangular board using angle dominoes without remaining part?

Solution:

136. After cutting only one square from 5x5 square we can fill out the remaining part using linear triminoes. Determine all the possible squares that are cut off.




SAMPLE QUESTIONS 50

Solution: Since each triminoe covers each number we must remove one of

2’s. So we have 9 alternatives but we can eliminate by symmetry. Therefore

If we combine these two figures we get

So the center is the only case.

137. After cutting only one square from 8x8 square we can fill out the remaining part using linear

triminoes. Determine all the possible squares that are cut off. Solution:

1 – 21, 2 – 22, 3 – 22. By the same idea, after rotating

Tetraminoes

138. Prove that a chess board can be filled by any kind of tetraminoes except z-tetraminoes. Solution:

So a 4x4 board can be covered by our tetraminoes.

Theorem: Any board of the form 4k x 4k can be filled by tetraminoes except z-tetraminoes. Theorem: There exist no rectangular board that can be filled by z-tetraminoes.

1 8

2 9

3 8

−

− −




SAMPLE QUESTIONS 51

They cannot be filled by z.

139. Prove that it is impossible to fill out 10x10 board by T-triminoes.( In general (4k+2)x(4k+2) is impossible. )

Solution 1:

B – 50 W – 50

1 , 3x B W→ 1 , 3y W B→

100 4 25 x y= = +

( ){

252 25

3 50

x yy y

x y B

+ == ⇒ ∉

+ =�

Solution 2:

25

* * * ... * 100+ + + + =�� , { }* 1,3∈ but the sum of 25 odd numbers must be odd. (100 2 1k≠ + )

140. Prove that a 10x10 board cannot be filled out by L-triminoes. ( ) Solution 1: B – 50 3B, 1W 3W, 1B W – 50 100 : 4 = 25 figures.

252 25

3 50

x yx x

x y

+ =⇒ = ⇒ ∉

+ =�

Solution 2: One kind of same type of figure will be more. So it is impossible.

141. Prove that a 10x10 board cannot be filled out by linear triminoes. ( )




SAMPLE QUESTIONS 52

Solution 1:

13 2 2 B→ × − , 12 2 2 W→ × −

52 – B, 48 – W OR 52 – W, 48 – B. We have 100 : 4 = 25 figyres. Each different tetraminoe has two white and two black squares. So after filling the number of whites and blacks must be same. CONTRADICTION! Note: ( 4k+1)x( 4k+1), ( 4k+3)x( 4k+3) can be proved by the area of board

considering mod 4. Solution 2:

1 – 25, 2 – 26, 3 – 25, 4 – 24. Every linear tetraminoe has each color. Therefore the number of colors must be same. CONTRADICTION! Solution 3:

24 – B, 76 – W. Any tetraminoe in any position has exactly one black

square. So we need 25 – B squares to fill out all board. CONTRADICTION!

142. Let’s consider a 6x6x6 cubic box. Determine (with proof) the maximum number of bricks of

the form 1x1x4 that can be put into box. (Hint: It is impossible to fill the box without any remainder and any extra.)

Solution:

2x2x2 – B, 14x8 = 112 small 1x1x1 2x2x2 –W, 13x8 = 104 small 1x1x1 Every brick has exactly 2 – B and 2 – W small cubes so the number of B and W must be same. CONTRADICTION!

143. We add 1 next two parts or two opposite parts of the circle given at

right. Is it possible to get all parts a) equal b) even c) divisible by 3 after finite number of steps? Solution:

a) 6x→∑ IMPOSSIBLE




SAMPLE QUESTIONS 53

So we have added �6 1

initial

x − but 6 1x − 2 . Since we added 2 in each step.

b) 2∑ but 2 1k + 2 . IMPOSSIBLE.

c)

1

B

W

B W

a

b

=

− =

− =

∑∑

∑ ∑

invariant because we add 1 to one – W and one – B in

each step.

1 3 5

2 4 6

3

3

1 3

B

W

B W

a a a

a a a

= + +

− = + +

− =

∑∑

∑ ∑

CONTRADICTION!

144. In each step we add 1 to both points on the vertice of a side. Is it possible to get all points

a) equal b) even c) divisible by 3 ? Solution:

a) Suppose that they are all equal so 8x=∑ . It means we

have added 8x – 1 but this is not divisible by 2. b) Let’s suppose that all of them are even and we made n moves.

2 1n⇒ = +∑ but 2 1n + 2 .

c) 1B W− =∑ ∑ (initial)

For each move we add 1 to B and 1 to W. So 1B W− =∑ ∑ will not change but 1 3 .

145. Rectangular even area was covered by two different covers. ( , ). A bad boy has

broken the figure and replaced one of by . Prove that it is impossible to cover again by the new set of figures.

2

2

2

2

2

2




SAMPLE QUESTIONS 54

Solution:

, (1) a b (2) a – 1 b + 1

If we color the rectangular area like that, then

4B – 0W OR 2B – 2W

3B – 1W Since he has changed the parity of B it is impossible to cover again. Let x be the number of B initiallly i) x was even ( new set has x – 1 which is odd.) ii) x was odd ( new set has x – 1 which is even.)

146. Is it possible to divide the rectangle of the form 34x35 into the small

recatangles of the form 5x7? Solution: 35 = 5x + 7y 34 = 2.7 + 4.5 YES.

147. The rectangle of the form AxB can be covered by the rectangles of the

form axb if and only if

1 1

2 2

,i i

A a or B a

A b or B b

A x a y bx y

B x a y b

= + ∈ = +

�

Solution: (ïïïï)

I. A a

B b

II.

A b

B a

III.

A a

A b

IV.

B b

B a

I. II.

III. IV. Change only sides A&B on figure.




SAMPLE QUESTIONS 55

(ðððð)

3 3

2 2

B a b

A a b

= +

= +

So A and B must be a linear combination of a and b. Now we must prove

that A a or B a

A b or B b

.

Lemma: Let AxB be filled with 1x n. Then A n or B n .

A B× →

A B× →

Proof of Lemma:

Let A 1 1 1. , 0n A n q r r n

B

= + < <

2 2 2. , 0n B n q r r n

= + < <.

Now we need n colors.

Each 1x n recatangles has all n colors. If AxB is filled by 1x n then for any color we have the same number of 1x1 squares. However this

impossible if there is a remainder of sides. (A&B, A n , B n )

a

b

a

b




SAMPLE QUESTIONS 56

If 1 20 ,r r n< < then the number of colors will not be same.

k < n ( k < k+1 colors )

148. Prove that the length of closed broken line is even. ( In figure we have 44 ) Solution: If we move an odd number of unit squares the color will change. (Even move will not change)

149. If every vetex has integer coordinates and length of any segment is natural, then the length of closed line is even.

Solution:

2 2 2a b c+ = ( ) ( ) ( )2 22 22 mod 2c a b ab c a b⇒ − + = − ⇒ ≡ +

( )mod 2c a b⇒ ≡ + . So c and a+b are of the same parity. For any non-vertical

and non-horizontal line we can choose hor. & ver. Lines. Then the solution is O.K. because of the previous question.

150. There is a frog on integer lattice. It can jump only on diagonals. It made several jumps and come to the same point. Prove that the number of jumps was even.

1 2 32

33

. ... .

.. ...

.

n

n

n

nn

k

k

k

k

k

k+1

k+1

k+1

k+1

k+1

.k k+1

. nr1

r2

n




SAMPLE QUESTIONS 57

Solution:

# 0 1 2 3 4 ....

....

of jumps even

colour B W B W B B

151. Let’s consider our frog’s step on the digonal . If it made a trip and return the same point, prove that the number of steps was even.

Solution: As seen at right in each step it will change its color. So the number of steps must be even.

152. Let m and n be two fixed natural numbers. This time our frog jumps on diagonals ,

. If it start from an initial point and come the same point, prove that the number of steps must be even.

Solution: i) If m+n is odd, then it will change its color. So in order to come the same point the

number of steps must be even. ii) If m and n are odd, then it will change the color.

iii)Let m and n are even, ( ) 1 1, ,m n

d m n m nd d

= ⇒ = = . Then

1 1

1 1

1 1

,

,

,

m n odd

m odd n even

n odd m even

⇒

it

follows by the same procedure.




SAMPLE QUESTIONS 58

NIGERIAN TEAM FOR PAMO 2005, ALGERIA 1. SEGUN ARIYIBI, NIGERIAN TURKISH

INTERNATIONAL COLLEGES, LAGOS

HONORABLE MENTION 2. SAMUEL CHUKUMWA, NIGERIAN TURKISH

INTERNATIONAL COLLEGES, ABUJA

3. JACINTE EBELE, LOYALA JESUIT COLLEGE,

ABUJA 4. AHIGBE METTHEUS, LOYALA JESUIT COLLEGE,

ABUJA

NIGERIAN TEAM FOR PAMO 2006, SENEGAL 1. UCHENDU NDUBISI, NIGERIAN TURKISH


SILVER MEDAL 2. MUAZZAM IDRIS, NIGERIAN TURKISH


3. SEGUN ARIYIBI, NIGERIAN TURKISH


4. EKWUE WINNER, CHRIST THE KING COLLEGE,

ANAMBRA

NIGERIAN TEAM FOR IMO 2006, SLOVENIA 1. UCHENDU NDUBISI, NIGERIAN TURKISH


2. MUAZZAM IDRIS, NIGERIAN TURKISH


3. SEGUN ARIYIBI, NIGERIAN TURKISH


4. EKWUE WINNER, CHRIST THE KING COLLEGE,

ANAMBRA 5. OLUWAKAYODE JOHN, NIGERIAN TURKISH


6. OLURKOYO F. ADETUNJI, REGIANA P.G.S.S., ABUJA

NIGERIAN TEAM FOR PAMO 2007, NIGERIA 1. UCHENDU NDUBISI, NIGERIAN TURKISH


GOLD MEDAL 2. EFUNBAJO OYEWOLE NIGERIAN TURKISH


BRONZ MEDAL 3. EMMANUEL AUNDE COMMERCIAL COLLEGE

GBOKO, BENUE BRONZ MEDAL

4. MUAZZAM IDRIS, NIGERIAN TURKISH INTERNATIONAL COLLEGES, ABUJA

HONORABLE MENTION

USEFUL WEBSITES FOR


1ST, 2ND AND 3RD ROUND (NMO)

http://science.up.ac.za/samo/questions.html

PAN AFRICAN MATHEMATICAL OLYMPIAD (PAMO)

http://www.saasta.ac.za/pamo/ http://www.pamo2007.com

INTERNATIONAL MATHEMATICAL OLYMPIAD (IMO)

http://imo.math.ca/ http://www.kalva.demon.co.uk

WE ARE TRYING OUR BEST FOR NIGERIA

NMC




SAMPLE QUESTIONS 59


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