hongkong olympiad

Answers: (2010-11 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 25 April 2017

Group Event 3
G3.1 If a is a positive integer and a2 + 100a is a prime number, find the maximum value of a.

a(a + 100) is a prime number.
a = 1, a2 + 100a = 101 which is a prime number

G3.2 Let a, b and c be real numbers. If 1 is a root of x2 + ax + 2 = 0
and a and b be roots of x2 + 5x + c = 0, find the value of a + b + c.

1 + a + 2 = 0  a = –3

–3 + b = –5  b = –2
c = –3b = 6
a+b+c=1

G3.3 Let x and y be positive real numbers with x < y. If x  y 1, x y  10
y x 3

and x < y, find the value of y – x. Method 2
(1)2: x + y + 2 xy = 1
Let z = x , then 1  y
xy = 1  x  y   (3) y zx

2 1 10
z 3
(2): x y  10  (4) (2) becomes z  
xy 3
3z2 – 10z + 3 = 0
x y 10
Sub. (3) into (4):  3 (3z – 1)(z – 3) = 0
1x y
1
2 z = 3 or 3

6(x + y) = 10(1 – x – y)

16(x + y) = 10 x<y

x + y = 5 z= x < 1  z = 1 only
8 y 3

xy = 1 x  y = 1 1  5  = 3
2  8  16
2 y x 1 x  1 1 = 1
y  y 1 3 2
9 x 1
xy = 256 x 1
y 3

(y – x)2 = (x + y)2 – 4xy = 25  9 = 1  x  y 1  y x = 1
64 64 4 2

1   y – x = = 1
y – x = 2 y x y x 2

G3.4 Spilt the numbers 1, 2,  , 10 into two groups and let P1 be the product of the first group and

P2 the product of the second group. If P1 is a multiple of P2, find the minimum value of P1 .
P2

P1 = kP2, where k is a positive integer.

 All prime factors of P2 can divide P1.

10 = 2, 10 must be a factor of the numerator and 5 must a factor of the denominator
5

7 is a prime which must be a factor of the numerator.

Among the even numbers 2, 4, 6, 8, 10, there are 8 factors of 2.

4 factors of 2 should be put in the numerator and 4 factors should be put in the denominator.

Among the number 3, 6, 9, there are 4 factors of 3.

2 factors of 3 should be put in the numerator and 2 factors should be put in the denominator.

Minimum value of P1 = 8  7  9 10 = 7
P2 1 2  3 4  5  6

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Answers: (2010-11 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 25 April 2017

Group Event 4

G4.1 If P = 24 2007  2009  2011 2013 10  2010  2010  9  4000 , find the value of P.
Let x = 2010, 2007 = x – 3, 2009 = x – 1, 2011 = x + 1, 2013 = x + 3

   P = 24 x  3 x 1 x 1 x  3 10x2  9  4000 = 24 x2  9  x2 1 10x2  9  4000

= 24 x4 10x2  9  10x2  9  4000 = 2x – 4000 = 20

G4.2 If 9x2 + nx + 1 and 4y2 + 12y + m are squares with n > 0, find the value of n .
m
9x2 + nx + 1 = (3x + 1)2  n = 6
4y2 + 12y + m = (2y + 3)2  m = 9

n = 2
m 3

G4.3 Let n and 47  4  n  be positive integers. If r is the remainder of n divided by 15, find
5  47 141 

the value of r.

47  4  n  = 4  n = n  12 , which is an integer
5  47 141  5 15 15

n + 12 = 15k, where k is a positive integer

r=3

G4.4 In figure 1, ABCD is a rectangle, and E and F are points on

AD and DC, respectively. Also, G is the intersection of AF

and BE, H is the intersection of AF and CE, and I is the

intersection of BF and CE. If the areas of AGE, DEHF and

CIF are 2, 3 and 1, respectively, find the area of the grey

region BGHI. (Reference: 2014 FI1.1)

Let the area of EGH = x, area of BCI = z, area of BGHI = w

Area of BCE = 1 (area of ABCD) = area ADF + area BCF
2

x+w+z=3+x+2+1+z

w=6

 Area of the grey region BGHI = 6

Remark: there is a spelling mistake in the English version. old version:  gray region 

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Answers: (2010-11 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 25 April 2017

Group Spare
GS.1 Let  and  be the real roots of y2 – 6y + 5 = 0. Let m be the minimum value of

|x – | + |x – | over all real values of x. Find the value of m.

Reference: 1994 HG1, 2001 HG9, 2004 FG4.2, 2008 HI8, 2008 FI1.3, 2010 HG6, 2012 FG2.3
 = 1,  = 5
If x < 1, |x – | + |x – | = 1 – x + 5 – x = 6 – 2x > 4
If 1  x  5, |x – | + |x – | = x – 1 + 5 – x = 4
If x > 5, |x – | + |x – | = x – 1 + x – 5 = 2x – 6 > 4
m = min. of |x – | + |x – | = 4
Method 2 Using the triangle inequality: |a| + |b|  |a + b|
|x – | + |x – |  |x – 1 + 5 – x| = 4  m = 4
Remark: there is a typing mistake in the English version.  minimum value a of 

GS.2 Let , ,  be real numbers satisfying  +  +  = 2 and  = 4. Let v be the minimum value

of || + || + ||. Find the value of v.

If at least one of , ,  = 0, then   4  , ,   0

If , ,  > 0, then

     3  (A.M.  G.M.)
3

2  3 4
3
23  274 = 108, which is a contradiction

If  < 0, in order that  = 4 > 0, WLOG let  < 0,  > 0

=2––>2

|| + || + || =  – ( + ) = 2 + 2(– – )  2 + 4    , equality holds when  = 

4 = (2 – 2)2
3 – 2 + 2 = 0
( + 1)(2 – 2 + 2) = 0
 = –1 (For the 2nd equation,  = –4 < 0, no real solution)

 = –1,  = 4

|| + || + || = 1 + 1 + 4 = 6

v = min. of || + || + || = 6

GS.3 Let y = |x + 1| – 2|x| + |x – 2| and –1  x  2. Let  be the maximum value of y. Find the value
of .
y = x + 1 – 2|x| + 2 – x = 3 – 2|x|

0  |x|  2  3  3 – 2|x|  –1
=3

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Answers: (2010-11 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 25 April 2017

GS.4 Let F be the number of integral solutions of x2 + y2 + z2 + w2 = 3(x + y + z + w).

Find the value of F.

(x, y, z, w) = (0, 0, 0, 0) is a trivial solution.
x2 + y2 + z2 + w2 – 3(x + y + z + w) = 0

 x2  3x  9    y2  3y  9    z2  3z  9    w2  3w  9   9
 4   4   4   4 

 x  3 2   y  3 2   z  3 2   w  3 2  9
 2   2   2   2 

(2x – 3)2 + (2y – 3)2 + (2z – 3)2 + (2w – 3)2 = 36
Let a = 2x – 3, b = 2y – 3, c = 2z – 3, d = 2w – 3, the equation becomes a2 + b2 + c2 + d2 = 36

For integral solutions of (x, y, z, w), (a, d, c, d) must be odd integers.

In addition, the permutation of (a, b, c, d) is also a solution. (e.g. (b, d, c, a) is a solution)

 a, b, c, d are odd integers and a2 + b2 + c2 + d2  0

If one of the four unknowns, say, a > 6, then L.H.S. > 36, so L.H.S.  R.H.S.
 a , b, c, d = 1, 3, 5
When a = 5, then 25 + b2 + c2 + d2 = 36  b2 + c2 + d2 = 11
The only integral solution to this equation is b = 3, c = 1 = d or its permutations.
When the largest (in magnitude) of the 4 unknowns, say, a is 3, then 9 + b2 + c2 + d2 = 36
 b2 + c2 + d2 = 27, the only solution is b = 3, c = 3, d = 3 or its permutations.

 The integral solutions are (a, b, c, d) = (5, 3, 1, 1) and its permutations  (1)  P24 = 12

(3, 3, 3, 3)  (2)  1
If (a, b, c, d) is a solution, then (a, b, c, d) are also solutions.
There are 16 solutions with different signs for (a, b, c, d).
 F = (12 + 1)16

= 208

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Individual Events

SI P 30 I1 A 12 I2 P 5 I3 α 45 I4 A 22 IS P 95
*Q 120 108 Q 12 β 56
B 280 R3 γ 23 B 12 Q 329
see the remark 69 S 17 δ 671
*C Group Events C 12 *R 6
R 11 D7
*S 72 see the remark see the remark

see the remark D S 198

SG q 3 G1 tens digit 1 G2 2 G3 z BD 2 GS *m 4
6 G4 see the remark 6
k1 *P 1031 K2 *r
CE 1v
see the remark see the remark 540 Q

w 25 k 21 l 45 D 998 R1 α3

3 *S∆BCD 32 see the remark 1 *F2012(7) 1 x5 5 F 208
p 4 see the remark
see the remark
2

Sample Individual Event (2009 Final Individual Event 1)
SI.1 Let a, b, c and d be the roots of the equation x4 – 15x2 + 56 = 0.

If P = a2 + b2 + c2 + d2, find the value of P.
x4 – 15x2 + 56 = 0 ⇒ (x2 – 7)(x2 – 8) = 0

a= 7 , b=− 7 , c= 8, d=− 8
P = a2 + b2 + c2 + d2 = 7 + 7 + 8 + 8 = 30

SI.2 In Figure 1, AB = AC and AB // ED.

If ∠ABC = P° and ∠ADE = Q°, find the value of Q.

∠ABC = 30° = ∠ACB (base ∠s isos. ∆)

∠BAC = 120° (∠s sum of ∆)

∠ADE = 120° (alt. ∠s, AB // ED)

Q = 120

Remark: Original question … AB // DE ….

It is better for AB and ED to be oriented in the same direction. Figure 1

SI.3 Let F = 1 + 2 + 22 + 23 + … + 2Q and R = log(1 + F ) , find the value of R.

log 2

F = 1 + 2 + 22 + 23 + … + 2120 = 2121 −1 = 2121 – 1
2 −1

R = log(1 + F ) = log 2121 = 11

log 2 log 2

SI.4 Let f (x) be a function such that f (n) = (n – 1) f (n – 1) and f (1) ≠ 0.
(R)
If S = (R − f f (R − 3) , find the value of S.

1)

f (n) = (n – 1) f (n – 1) = (n – 1)(n – 2)f (n – 2) = ……
(11) (8)
S = f (11 − = 10 ×9 × 8 ×f = 9×8 = 72
f 10 × f
(11) 3) (8)

Remark: Original question:

Let f (x) be a function such that f (n) = (n – 1) f (n – 1). If S = (R − f (R) − 3) , find the value of S.
f (R
1)

Note that S is undefined when f (n) = 0 for some integers n.

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Individual Event 1
I1.1 If A is the sum of the squares of the roots of x4 + 6x3 + 12x2 + 9x + 2, find the value of A.

Let f (x) = x4 + 6x3 + 12x2 + 9x + 2 By division,

f (–1) = 1 – 6 + 12 – 9 + 2 = 0 x2 +3x +1

f (–2) = 16 – 48 + 48 – 18 + 2 = 0 x2 + 3x + 2 x4 + 6x3 + 12x2 + 9x + 2
∴ By factor theorem, (x2 + 3x + 2) is a factor of f (x). x4 − 3x3 + 2x2
x4 + 6x3 + 12x2 + 9x + 2 = (x + 1)(x + 2)(x2 + 3x + 1)
3x3 + 10x2 + 9x
The roots are –1, –2 and α, β; where α + β = –3, αβ = 1
3x3 + 9x2 + 6x
A = (–1)2 + (–2)2 + α2 + β2 = 5 + (α + β)2 – 2αβ

=5+9–2 x2 + 3x + 2
A = 12 x2 + 3x + 2

Method 2 By the change of subject, let y = x2, then the equation becomes

x4 + 12x2 + 2 = –x(6x2 + 9) ⇒ y2 +12 y + 2 = m y (6 y + 9)

(y2 + 12y + 2)2 – y(6y + 9)2 = 0
Coefficient of y4 = 1, coefficient of y2 = 24 – 36 = –12
If α, β, δ and γ are the roots of x, then α2, β2, δ2 and γ2 are the roots of y

α2 + β2 + δ2 + γ2 =− coefficient of y3 = 12
coefficient of y4

Method 3
Let α, β, δ and γ are the roots of x, then by the relation between roots and coefficients,

α + β + δ + γ = –6 LL (1)

αβ + αδ + αγ + βδ + βγ + δγ = 12 LL (2)
α2 + β2 + δ2 + γ2 = (α + β + δ + γ)2 – 2(αβ + αδ + αγ + βδ + βγ + δγ)

= (–6)2 – 2(12) = 12

I1.2 Let x, y, z, w be four consecutive vertices of a regular A-gon. If the length of the line segment

xy is 2 and the area of the quadrilateral xyzw is a + b , find the value of B = 2a⋅3b.

Let O be the centre of the regular dodecagon. x2 y
Let Ox = r = Oy = Oz = Ow

∠xOy = ∠yOz = ∠zOw = 360o = 30° (∠s at a point) z
12

In ∆xOy, r2 + r2 – 2r2 cos 30° = 22 (cosine rule)

( )(2 – 3 )r2 = 4 ⇒ r2 = 4 = 4 2 + 3 w
2− 3 O

Area of xyzw = area of Oxyzw – area of ∆Oxw

( )=3× 1 r2 1 r2 =  3 1 
2 ⋅ sin 30o − 2 sin 90o  4 − 2  ⋅ 4 2 + 3 =2+ 3

a = 2, b = 3, B =22⋅33 = 4×27 = 108

I1.3 If C is the sum of all positive factors of B, including 1 and B itself, find the value of C.
108 = 22⋅33
C = (1 + 2 + 22)⋅(1 + 3 + 32 + 33) = 7×40 = 280

Remark: Original version: 若C是B的所有因子之和… If C is the sum of all factors L

Note that if negative factors are also included, then the answer will be different.

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

I1.4 If C! = 10Dk, where D and k are integers such that k is not divisible by 10, find the value of D.

Reference: 1990 HG6, 1994 FG7.1, 1996 HI3, 2004 FG1.1, 2011 HG7, 2012 FG1.3

Method 1 Method 2

When each factor of 5 is multiplied by 2, a trailing We can find the total number of factors of 5 by

zero will appear in n!. division as follow:

The number of factors of 2 is clearly more than 5 2 8 0 ∴Total no. of factors of 5 is

the number of factors of 5 in 280! 5 56 56 + 11 + 2 = 69

It is sufficient to find the number of factors of 5. 5 1 1 … 1 D = 69
5, 10, 15, … , 280; altogether 56 numbers, each 2…1
have at least one factor of 5.

25, 50, 75, … , 275; altogether 11 numbers, each

have at least two factors of 5.

125, 250; altogether 2 numbers, each have at least

three factors of 5.

∴ Total number of factors of 5 is 56 + 11 + 2 = 69

D = 69

Individual Event 2

I2.1 If the product of the real roots of the equation x2 + 9x + 13 = 2 x2 + 9x + 21 is P, find the

value of P.

Let y = x2 + 9x, then the equation becomes y + 13 = 2 y + 21
(y + 13)2 = 4y + 84
y2 + 26y + 169 – 4y – 84 = 0
y2 + 22y + 85 = 0

(y + 17)(y + 5) = 0

y = –17 or y = –5

Check put y = –17 into the original equation: –17 + 13 = 2 −17 + 21

LHS < 0, RHS > 0, rejected

Put y = –5 into the original equation: LHS = –5 + 13 = 2 − 5 + 21 = RHS, accepted
x2 + 9x = –5
x 2 + 9x + 5 = 0

Product of real roots = 5

Method 2

Let y = x2 + 9x + 21 ≥ 0
Then the equation becomes y2 – 8 = 2y ⇒ y2 – 2y – 8 = 0
(y – 4)(y + 2) = 0 ⇒ y = 4 or –2 (rejected)
⇒ x2 + 9x + 21 = 16
x2 + 9x + 5 = 0
∆ = 92 – 4(5) > 0
Product of real roots = 5

I2.2 25x and Q = f  1  + f  2  + L + f  24  , find the value of Q.
If f (x) = 25x + P  25   25   25 

Reference: 2004 FG4.1, 2011 HG5

f (x) + f (1 − x)= 25x 5 + 251− x = 25 + 5⋅ 25x + 25 + 5⋅ 251−x = 1
25x + 251−x + 5 25 + 5⋅ 251−x + 5⋅ 25x + 25

Q = f  1  + f  2  +L+ f  24 
 25   25   25 

= f  1  + f  24  + f  2  + f  23  + L + f  12  + f  13  = 12
 25   25   25   25   25   25 

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

I2.3 If X = (100)(102)(103)(105) + (Q − 3) is an integer and R is the units digit of X, find the value

of R.

Reference: 1993 HG6, 1995 FI4.4, 1996 FG10.1, 2000 FG3.1, 2004 FG3.1

Let y = 102.5, then

(100)(102)(103)(105) + (12 – 3)

= (y – 2.5)(y – 0.5)(y + 0.5)(y + 2.5) + 9
= (y2 – 6.25)(y2 – 0.25) + 9

= y4 – 6.5y2 + 25 + 9 = y4 – 6.5y2 + 169
16 16

=  y2 − 13 2 = 102.52 − 13 2 =  2052 − 13 2
 4  4 4 4

X = 420252 −13 = 10503
4

R = the units digit of X = 3

Method 2 X = (100)(102)(103)(105) + 9 = (100)(100 + 5)(100 + 2)(100 + 3) + 9

( )( ) ( ) ( )= 1002 + 500 1002 + 500 + 6 + 9 = 1002 + 500 2 + 6 1002 + 500 + 9 = (1002 + 500) + 3

R = the units digit of X = 3

I2.4 If S is the sum of the last 3 digits (hundreds, tens, units) of the product of the positive roots of

2012 ⋅ xlog2012 x = xR , find the value of S.

( )log2012 2012 ⋅ xlog2012 x = log2012 x3

( )1 + 2 = 3log2012 x

2
log2012 x

Let y = log2012 x, then 2y2 – 6y + 1 = 0

y = log2012 x = 3 ± 7
2

3+ 7 3− 7

⇒ x = 2012 2 or 2012 2

3+ 7 3− 7

Product of positive roots = 2012 2 × 2012 2
= 20123

≡ 123 (mod 1000)

= 1728 (mod 1000)

S = sum of the last 3 digits = 7 + 2 + 8 = 17

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Individual Event 3

I3.1 In Figure 1, a rectangle is sub-divided into 3

identical squares of side length 1.

If α° = ∠ABD + ∠ACD, find the value of α.

Method 1 (compound angle)

tan ∠ABD = 1 , tan ∠ACD = 1
32

0° < ∠ABD, ∠ACD < 45° Figure 1

∴ 0° < ∠ABD + ∠ACD < 90°

tan α° = tan ∠ABD + tan ∠ACD = +1 1 =1>0

32

1 − tan ∠ABD ⋅ tan ∠ACD 1 − 1 ⋅ 1
3 2

α = 45

Method 2 (congruent triangles) L KJ A

Draw 3 more identical squares BCFE, CIGH, IDHG

of length 1 as shown in the figure. ALBD, DBEH are

identical rectangles. Join BG, AG.

BE = GH = AD (sides of a square) B CI D
EG = HA = CD (sides of 2 squares)

∠BEG = 90° = ∠GHA = ∠ADC (angle of a square)

∆BEG ≅ ∆GHA ≅ ∆ADC (SAS) 1 1
F G 1H
Let ∠BGE = θ = ∠GAH (corr. ∠s ≅ ∆'s) E
∠AGH = 90° – θ (∠s sum of ∆)

∠AGB = 180° – ∠AGH – ∠BGE (adj. ∠s on st. line)

= 180° – θ – (90° – θ) = 90°

BG = AG (corr. sides ≅ ∆'s)

∠ABG = ∠BAG (base ∠s isos. ∆)

180o − 90o = 45° (∠s sum of ∆)
=
2

α° = ∠ABD + ∠ACD

= ∠ABD + ∠GBI (corr. ∠s, ≅∆’s)
= 45° ⇒ α = 45

Method 3 (similar triangles) L KJ A
Join AI.

AI = 2 (Pythagoras’ theorem)

BI = 2 = 2, AI = 2 2
AI 2 =
CI 1
B CI D
∠AIB = ∠CIA (common ∠) L

∆AIB ~ ∆CIA (2 sides proportional, included ∠)

∴ ∠ACD = ∠BAI (corr. ∠s, ~∆’s)

α° = ∠ABD + ∠ACD = ∠ABI + ∠BAI

= ∠AID (ext. ∠ of ∆)
= 45° (diagonal of a square) ⇒ α = 45

Method 4 (vector dot product) KJ A

Define a rectangular system with BC = i, BL = j. j
AB = –3i – j, AC = –2i – j, AI = –i – j.

AB ⋅ AI = | AB || AI | cos ∠BAI
(− 3)(−1) + (−1)(−1)
cos∠BAI = =2 B iC I D

(− 3)2 + (−1)2 ⋅ (−1)2 + (−1)2 5

cos ∠ACD = 2 ⇒ ∠BAI = ∠ACD
5

α° = ∠ABD + ∠ACD = ∠ABI + ∠BAI = ∠AID (ext. ∠ of ∆) ⇒ α = 45

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Method 5 (complex number) L KJ A

Define the Argand diagram with B as the origin, BD

as the real axis, BL as the imaginary axis. z1
Let the complex numbers represented by AB and AC z2
be z1 and z2 respectively.

z1 = 3 + i, z2 = 2 + i BC I D

z1⋅z2 = (3 + i)(2 + i) = 6 – 1 + (2 + 3)i = 5 + 5i

Arg(z1z2) = Arg(z1) + Arg(z2)

α° = ∠ABD + ∠ACD = tan–1 5 = 45°
5

α = 45

I3.2 Let ABC be a triangle. If sin A = 36 , sin B = 12 and sin C = β , find the value of β, where β
α 13 y

and y are in the lowest terms.

sin A = 36 = 4 > 0 ⇒ 0° < A < 90° and cos A = 1 −  4 2 = 3
45 5 5 5

sin B = 12 > 0 ⇒ 0° < B < 90° and cos B = 1 −  12 2 =5
13  13  13

sin C = sin(180° – (A + B)) = sin (A + B) = sin A cos B + cos A sin B C
= 4 ⋅ 5 + 3 ⋅ 12 = 56 O γ°
5 13 5 13 65 D (2β -1)°

β = 56 AB
I3.3 In Figure 2, a circle at centre O has three points on its

circumference, A, B and C. There are line segments OA, OB, AC

and BC, where OA is parallel to BC. If D is the intersection of

OB and AC with ∠BDC = (2β – 1)° and ∠ACB = γ°, find the
value of γ.
∠AOB = 2γ° (∠ at centre twice ∠ at circumference)
∠OBC = 2γ° (alt. ∠, OA // CB)
γ° + 2γ° + (2β – 1)° = 180° (∠s sum of ∆)
3γ + 111 = 180
γ = 23

Figure 2
I3.4 In the expansion of (ax + b)2012, where a and b are relatively prime positive integers.

If the coefficients of xγ and xγ+1 are equal, find the value of δ = a + b.

Coefficient of x23 = C 2012 ⋅ a 23 ⋅ b1989 ; coefficient of x24 = C 2012 ⋅ a 24 ⋅ b1988
23 24

C 2012 ⋅ a 23 ⋅ b1989 = C 2012 ⋅ a 24 ⋅ b1988
23 24

b = C 2012 ⋅a
24

C 2012
23

b = 2012 − 24 + 1 ⋅ a
24

24b = 1989a

8b = 663a

Q a and b are relatively prime integers
∴ a = 8, b = 663
δ = 8 + 663 = 671

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Individual Event 4

I4.1 If A is a positive integer such that 1 + 1 + L + (A 1 3) = 12 , find the value of A.
1× × 25
3 3 5 + 1)(A +

(n + 1 + 3) = 1  n 1 − n 1 3  for n ≥ 0
2  +1 + 
1)(n

1 1 − 1 + 1 − 1 +L+ 1 − 1  = 12
2 3 3 5 A+1 A+ 3 25

1− 1 = 24
A+ 3 25

1 =1
A + 3 25
A = 22

I4.2 If x and y be positive integers such that x > y > 1 and xy = x + y + A.

Let B = x , find the value of B.
y

Reference: 1987 FG10.4, 2002 HG9

xy = x + y + 22

xy – x – y + 1 = 23

(x – 1)(y – 1) = 23

Q 23 is a prime number and x > y > 1

∴ x – 1 = 23, y – 1 = 1
x = 24 and y = 2

B = 24 = 12
2

I4.3 Let f be a function satisfying the following conditions:
(i) f (n) is an integer for every positive integer n；

(ii) f (2) = 2;

(iii) f (mn) = f (m)f (n) for all positive integers m and n and

(iv) f (m) > f (n) if m > n.
If C = f (B), find the value of C.

Reference: 2003 HI1
2 = f (2) > f (1) > 0 ⇒ f (1) = 1

f (4) = f (2×2) = f (2)f (2) = 4

4 = f (4) > f (3) > f (2) = 2
⇒ f (3) = 3

C = f (12) = f (4×3) = f (4)f (3) = 4×3 = 12
I4.4 Let D be the sum of the last three digits of 2401×7C (in the denary system).

Find the value of D.
2401×7C = 74×712 = 716 = (72)8 = 498 = (50 – 1)8

= 508 – 8×507 + … – 56×503 + 28×502 – 8×50 + 1

≡ 28×2500 – 400 + 1 (mod 1000)

≡ –399 ≡ 601 (mod 1000)

D=6+0+1=7
Method 2 2401×7C = 74×712 = 716
74 = 2401
78 = (2400 + 1)2 = 5760000 + 4800 + 1 ≡ 4801 (mod 1000)
716 ≡ (4800 + 1)2 ≡ 482×10000 + 9600 + 1 ≡ 9601 (mod 1000)

D=6+0+1=7

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Individual Spare (2011 Final Group Spare Event)
IS.1 Let P be the number of triangles whose side lengths are integers less than or equal to 9. Find

the value of P.
The sides must satisfy triangle inequality. i.e. a + b > c.
Possible order triples are (1, 1, 1), (2, 2, 2), … , (9, 9, 9),
(2, 2, 1), (2, 2, 3), (3, 3, 1), (3, 3, 2), (3, 3, 4), (3, 3, 5),

(4, 4, 1), (4, 4, 2), (4, 4, 3), (4, 4, 5), (4, 4, 6), (4, 4, 7),

(5, 5, 1), … , (5, 5, 4), (5, 5, 6), (5, 5, 7), (5, 5, 8), (5, 5, 9),
(6, 6, 1), … , (6, 6, 9) (except (6, 6, 6))
(7, 7, 1), … , (7, 7, 9) (except (7, 7, 7))
(8, 8, 1), … , (8, 8, 9) (except (8, 8, 8))
(9, 9, 1), … , (9, 9, 8)
(2, 3, 4), (2, 4, 5), (2, 5, 6), (2, 6, 7), (2, 7, 8), (2, 8, 9),

(3, 4, 5), (3, 4, 6), (3, 5, 6), (3, 5, 7), (3, 6, 7), (3, 6, 8), (3, 7, 8), (3, 7, 9), (3, 8, 9),
(4, 5, 6), (4, 5, 7), (4, 5, 8), (4, 6, 7), (4, 6, 8), (4, 6, 9), (4, 7, 8), (4, 7, 9), (4, 8, 9),

(5, 6, 7), (5, 6, 8), (5, 6, 9), (5, 7, 8), (5, 7, 9), (5, 8, 9), (6, 7, 8), (6, 7, 9), (6, 8, 9), (7, 8, 9).

Total number of triangles = 9 + 6 + 6 + 8×5 + 6 + 9 + 9 + 6 + 4 = 95
Method 2 First we find the number of order triples.
Case 1 All numbers are the same: (1, 1, 1), … , (9, 9, 9).
Case 2 Two of them are the same, the third is different: (1, 1, 2), … , (9, 9, 1)

There are C19 × C18 = 72 possible triples.

Case 3 All numbers are different. There are C39 = 84 possible triples.

∴ Total 9 + 72 + 84 = 165 possible triples.
Next we find the number of triples which cannot form a triangle, i.e. a + b ≤ c.
Possible triples are (1, 1, 2), … (1, 1, 9) (8 triples)
(1, 2, 3), … , (1, 2, 9) (7 triples)
(1, 3, 4), … , (1, 3, 9) (6 triples)
(1, 4, 5), … , (1, 4, 9) (5 triples)
(1, 5, 6), … , (1, 5, 9) (4 triples)
(1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 7, 8), (1, 7, 9), (1, 8, 9),

(2, 2, 4), … , (2, 2, 9) (6 triples)

(2, 3, 5), … , (2, 3, 9) (5 triples)
(2, 4, 6), … , (2, 4, 9) (4 triples)
(2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 8), (2, 6, 9), (2, 7, 9),

(3, 3, 6), … , (3, 3, 9) (4 triples)
(3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 8), (3, 5, 9), (3, 6, 9), (4, 4, 8), (4, 4, 9), (4, 5, 9).

Total number of triples which cannot form a triangle

= (8 + 7 + … + 1) + (6 + 5 + … + 1) + (4 + 3 + 2 + 1) + (2 + 1) = 36 + 21 + 10 + 3 = 70

∴ Number of triangles = 165 – 70 = 95

IS.2 Let Q = log128 23 + log128 25 + log128 27 + … + log128 2P. Find the value of Q.

Q = 3 log128 2 + 5 log128 2 + 7 log128 2 + … + 95 log128 2

= (3 + 5 + … + 95) log128 2 = 3 + 95 ×47× log128 2 = log128 22303 = log128 (27)329 = 329
2

IS.3 Consider the line 12x – 4y + (Q – 305) = 0. If the area of the triangle formed by the x-axis, the

y-axis and this line is R square units, what is the value of R?

12x – 4y + 24 = 0 ⇒ Height = 6, base = 2; area R = 1 ⋅6⋅2 = 6
2

Remark: the original question is …12x – 4y + Q = 0….
The answer is very difficult to carry forward to next question.

IS.4 If x + 1 = R and x3 + 1 = S, find the value of S.

x x3

S =  x + 1  x2 −1+ 1  = R x + 1 2 −  = R3 – 3R = 216 – 3(6) = 198
x  x2   x  3
 

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Sample Group Event (2009 Final Group Event 1)

SG.1 Given some triangles with side lengths a cm, 2 cm and b cm, where a and b are integers and

a ≤ 2 ≤ b. If there are q non-congruent classes of triangles satisfying the above conditions,

find the value of q.

When a = 1, possible b = 2

When a = 2, possible b = 2 or 3

∴q=3

SG.2 Given that the equation x − 4 = 3 x has k distinct real root(s), find the value of k.

xx

When x > 0: x2 – 4 = 3x ⇒ x2 – 3x – 4 = 0 ⇒ (x + 1)(x – 4) = 0 ⇒ x = 4
When x < 0: –x2 – 4 = –3x ⇒ x2 – 3x + 4 = 0; D = 9 – 16 < 0 ⇒ no real roots.

k = 1 (There is only one real root.)

SG.3 Given that x and y are non-zero real numbers satisfying the equations x − y = 7 and
y x 12
x – y = 7. If w = x + y, find the value of w.
The first equation is equivalent to x − y = 7 ⇒ xy = 12 ⇒ xy = 144

xy 12

Sub. y = 144 into x – y = 7: x − 144 = 7 ⇒ x2 – 7x – 144 = 0 ⇒ (x + 9)(x – 16) = 0
xx

x = –9 or 16; when x = –9, y = –16 (rejected Q x is undefined); when x = 16; y = 9
w = 16 + 9 = 25
Method 2 The first equation is equivalent to x − y = 7 ⇒ xy = 12 ⇒ xy = 144 …… (1)

xy 12

Q x – y = 7 and x + y = w

∴ x = w+7 , y = w−7
22

Sub. these equations into (1):  w + 7   w − 7  = 144
 2  2 

w2 – 49 = 576 ⇒ w =±25

Q From the given equation x − y = 7 , we know that both x > 0 and y > 0
y x 12

∴ w = x + y = 25 only

SG.4 Given that x and y are real numbers and x − 1 + y2 −1 = 0 . Let p = |x| + |y|, find the value
2

of p.

Reference: 2006 FI4.2 … y2 + 4y + 4 + x + y + k = 0. If r = |xy|, …

Both x − 1 and y2 −1 are non-negative numbers.
2

The sum of two non-negative numbers = 0 means each of them is zero
x = 1 , y = ±1; p = 1 + 1 = 3

2 22

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

Group Event 1
G1.1 Calculate the tens digit of 20112011.

20112011 ≡ (10 + 1)2011 mod 10
≡ 102011 + … + 2011×10 + 1 ≡ 11 mod 10

The tens digit is 1.

G1.2 Let a1, a2, a3, … be an arithmetic sequence with common difference 1 and

a1 + a2 + a3 + … + a100 = 2012. If P = a2 + a4 + a6 + … + a100, find the value of P.
100(2a + 99) = 2012
Let a1 = a then
2

2a + 99 = 1006
25

2a = 1006 − 99 × 25 = − 1469
25 25

P = a2 + a4 + a6 + … + a100 = 50(a + 1+ a + 99) = 25×(2a + 100) = 25 × 100 − 1469 
2 25 

P = 2500 – 1469 = 1031

Method 2 P = a2 + a4 + a6 + … + a100
Q = a1 + a3 + a3 + … + a99

P – Q = 1 + 1 + 1 + … + 1 (50 terms) = 50
But since P + Q = a1 + a2 + a3 + … + a100 = 2012
∴ P = 2012 + 50 = 1031

2
Remark: the original question …等差級數…, …arithmetic progression…

The phrases are changed to …等差數列…and …arithmetic sequence… according to the

mathematics syllabus since 1999.

G1.3 If 90! is divisible by 10k, where k is a positive integer, find the greatest possible value of k.

Reference: 1990 HG6, 1994 FG7.1, 1996 HI3, 2004 FG1.1, 2011 HG7, 2012 FI1.4

Method 1 Method 2

When each factor of 5 is multiplied by 2, a trailing zero We can find the total number of factors

will appear in n!. of 5 by division as follow:

The number of factors of 2 is clearly more than the number 5 9 0 No. of factors of 5 is 18+3

of factors of 5 in 280! 5 1 8 k = 21

It is sufficient to find the number of factors of 5. 3…3

5, 10, 15, … , 90; altogether 18 numbers, each have at

least one factor of 5.

25, 50, 75, altogether 3 numbers, each have at least two

factors of 5.

∴ Total number of factors of 5 is 18 + 3 = 21

k = 21

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Answers: (2011-12 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 11 April 2016

G1.4 In Figure 1, ∆ABC is a right-angled triangle with AB ⊥ BC. A
If AB = BC, D is a point such that AD ⊥ BD with AD = 5 and
BD = 8, find the value of the area of ∆BCD. D

AB = BC = 52 + 82 = 89 (Pythagoras’ theorem)

Let ∠ABD = θ, then cos θ = 8
89

∠CBD = 90° – θ, sin∠CBD = 8 BC
89
Figure 1
S∆BCD = 1 BD⋅BC sin∠CBD = 1 ⋅8⋅ 89 ⋅ 8 = 32
2 2 89

Method 2 A

Rotate ∆ABD about the centre at B in clockwise direction 5

through 90° to ∆CBE. D
Then ∆ABD ≅ ∆CBE

∠CEB = ∠ADB = 90° (corr. ∠s ≅ ∆s)

CE = 5, BE = 8 (corr. sides ≅ ∆s) 8

∠DBE = ∠DBC + ∠CBE

= ∠DBC + ∠ABD (corr. ∠s ≅ ∆s) BC
= 90°

∠DBE + ∠CEB = 180° 85

BD // CE (int. ∠s supp.)

∴ BDCE is a right-angled trapezium. E

Area of ∆BCD = area of trapezium BDCE – area of ∆BCE

= 1 (8 + 5)×8 – 1 ⋅8×5
22

= 32

Remark: the original question “… right-angle triangle…”

It should be changed to right-angled triangle. Furthermore, the condition AD ⊥ BD is not

specified.

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