NATIONAL MATHEMATICAL CENTRE NIGERIAN MATHEMATICS OLYMPIAD SAMPLE QUESTIONS 1 NATIONAL MATHEMATICAL CENTRE SAMPLE QUESTIONS FOR NIGERIAN MATHEMATICS OLYMPIAD PREPARIAD BY PREPARIAD BY PREPARIAD BY PREPARIAD BY RAMAZAN INCE RAMAZAN INCE RAMAZAN INCE RAMAZAN INCE (MATHEMATICS TEACHER) (MATHEMATICS TEACHER) (MATHEMATICS TEACHER) (MATHEMATICS TEACHER) BOOK 10 Downloaded from www.erudits.com.ng
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
SAMPLE QUESTIONS 1
NATIONAL MATHEMATICAL CENTRE
SAMPLE QUESTIONS FOR
NIGERIAN MATHEMATICS OLYMPIAD
PREPARIAD BYPREPARIAD BYPREPARIAD BYPREPARIAD BY RAMAZAN INCERAMAZAN INCERAMAZAN INCERAMAZAN INCE
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
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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.
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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
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≥ + >
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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
< < .
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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.
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)
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NATIONAL MATHEMATICAL CENTRE
NIGERIAN MATHEMATICS OLYMPIAD
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
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.
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NIGERIAN MATHEMATICS OLYMPIAD
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
−
− −
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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. ( )
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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
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NIGERIAN MATHEMATICS OLYMPIAD
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
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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.
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NIGERIAN MATHEMATICS OLYMPIAD
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
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NIGERIAN MATHEMATICS OLYMPIAD
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
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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.
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NIGERIAN MATHEMATICS OLYMPIAD
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
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,
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