Answers: (1994-95 HKMO Heat Events) Created by: Mr.

Francis Hung Last updated: 29 October 2016

2 11 4
1 1111111 2 3 0, 2 4 5 
94-95 3 450 3
Individual 1
6 5 7 1 8 7 9 10 12
6

45
1 1 2 132 3 4 45 5 24
94-95 2
Group 2
6 5130 7 2 3 3 8 8 9 124 10
3
Individual Events
I1 Find the square root of 1234567654321.
Observe the pattern 112 = 121; 1112 = 12321, 11112 = 1234321, 
1234567654321 = 11111112
 1234567654321 = 1111111 (7 digits)
1 x
I2 Given that f    , find the value of f (2).
 x  1 x
2

1 1
2
f (2) = f  1  = 2 2 =
 2  1  2  3
1

I3 Solve 32x + 9 = 10(3x).

Let y = 3x, then y2 = 32x
y2 + 9 = 10y
y2 – 10y + 9 = 0
(y – 1)(y – 9) = 0
y = 1 or y = 9
3x = 1 or 3x = 9
x = 0 or 2
I4 A three-digit number is selected at random. Find the probability that the number selected is a
perfect square.
Reference: 1997 FG1.4
The three-digit numbers consists of {100, 101,  , 999}, altogether 900 numbers.
Favourable outcomes = {100, 121,  , 961} = {102, 112,  , 312}, 22 outcomes
22 11
P(perfect squares) = =
900 450

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Answers: (1994-95 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 29 October 2016

1
I5 Given that sin x + cos x = and 0  x  , find tan x.
5
Reference: 1992 HI20, 1993 G10, 2007 HI7, 2007 FI1.4, 2014 HG3
1
(sin  + cos )2 =
25
1
sin2  + 2 sin  cos  + cos2  =
25
1
1 + 2 sin  cos  =
25
12
sin  cos  = 
25
25 sin  cos  = –12(sin2  + cos2 )
12sin2  + 25 sin  cos  + 12cos2  = 0
(3 sin  + 4 cos )(4 sin  + 3 cos ) = 0
4 3
tan  =  or 
3 4
4 4 3
Check when tan  =  , then sin  = , cos  = 
3 5 5
4  3 1
LHS = sin  + cos  = +    = = RHS
5  5 5
3 3 4
When tan  =  , then sin  = , cos  = 
4 5 5
3 4 1
LHS = sin  + cos  = – = –  RHS
5 5 5
4
 B = tan  = 
3
I6 How many pairs of positive integers x, y are there satisfying xy – 3x – 2y = 10?
xy – 3x – 2y + 6 = 10 + 6
(x – 3)(y – 2) = 16
x–3 y–2 16 x y
1 16 4 18
2 8 5 10
4 4 7 6
8 2 11 4
16 1 19 3
 There are 5 pairs of positive integers.
I7 x, y are positive integers and 3x + 5y = 123. Find the least value of |x – y|.
x = 41, y = 0 is a particular solution of the equation.
The general solution is x = 41 – 5t, y = 3t, where t is any integer.
|x – y| = |41 – 5t – 3t| = |41 – 8t|
The least value is |41 – 85| = 1.

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Answers: (1994-95 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 29 October 2016

I8 Find the remainder when 1997913 is divided by 10.

Note that 71 = 7, 72 = 49, 73 = 343, 74 = 2401.
Also, 74n+1  7 (mod 10), 74n+2  9 (mod 10), 74n+3  3 (mod 10), 74n  1 (mod 10)
1997913  7913 (mod 10)  7912 +1  74(228)+1  7 (mod 10)
The remainder is 7.
Area of ADE
I9 In figure 1, if BC = 3DE, find the value of r where r = .
Area of BDC
ADE ~ ABC

Area of ADE  1  1
2

 =  =
Area of ABC  3  9

Area of ADE 1 1
= =  (1)
Area of BCED 9  1 8
AE : AC = DE : BC = 1 : 3 (ratio of sides, ~)
AE : EC = 1 : 2
ADE and CDE have the same height with base ratio 1 : 2
Area of ADE AE 1
= =  (2)
Area of CDE CE 2
Area of ADE 1 1
r= = = by (1) and (2)
Area of BDC 8  2 6
I10 A, B, C, D are points on the sides of the right-angled triangle PQR as
shown in figure 2. If ABCD is a square, QA = 8 and BR = 18, find AB.
Let BRC = , then DQA = 90 –  (s sum of )
DAQ = 90 ( of a square), QDA =  (s sum of )
BC = BR tan  = 18 tan  = AD (opp. sides of square)
QA = 8 = AD tan  = 18 tan2 
2
 tan  =
3
2
AB = BC = 18 tan  = 18 = 12
3
Method 2
It is easy to show that PDC ~ AQD ~ BCR (equiangular)
Let AB = AD = BC = CD = x
PD : PC : x = 8 : x : QD = x : 18 : CR (cor. sides, ~s)
x2 = 818
AB = x = 12

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Answers: (1994-95 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 29 October 2016

Group Events
G1 Find the number of positive integral solutions of the equation x3 + (x + 1)3 + (x + 2)3 = (x + 3)3
Expand: x3 + x3 + 3x2 + 3x + 1 + x3 + 6x2 + 12x + 8 = x3 + 9x2 + 27x + 27
2x3 – 12x – 18 = 0
x3 – 6x – 9 = 0; let f (x) = x3 – 6x – 9
f (3) = 27 – 18 – 9 = 0  x – 3 is a factor.
By division, (x – 3)(x2 + 3x + 3) = 0
3 3
x = 3 or (rejected)
2
 There is one positive integral solution x = 3.
G2 In figure 1, ABCD is a quadrilateral whose diagonals intersect at O.
If AOB = 30, AC = 24 and BD = 22,
find the area of the quadrilateral ABCD.
1
The area = 24  22  sin 30 
2
= 132
1 2 3 n 1 n 1
G3 Given that      = ,
n n n n 2
1 1 2 1 2 3 1 9
find the value of +    +     ++      . Reference: 1996 FG9.4
2 3 3  4 4 4  10 10 
1 1 2 1 2 3 1 9
         
2 3 3 4 4 4  10 10 
1 2 3 4 9
=    
2 2 2 2 2
1  2    9 45
= 
2 2
G4 Suppose x and y are positive integers such that x2 = y2 + 2000, find the least value of x.
Reference: 1993 HI7, 1997 HI1
x2 – y2 = 2000 = 12000
= 21000
= 4500
= 5400
= 8250
= 10200
= 16125
= 20100
= 2580
= 4050
(x + y)(x – y) = 2000
 x and y are positive integers
 x + y and x – y are also positive integers
x>y
x is the least when y is the largest
 The difference between x and y is the largest
x + y = 50, x – y = 40
Solving, x = 45

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Answers: (1994-95 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 29 October 2016

G5 Given that 37100 is a 157-digit number, and 3715 is an n-digit number. Find n.
Reference: 2003 FI2.1
Let y = 37100, then log y = log 37100 = 156 + a, where 0  a < 1
100 log 37 = 156 + a
15 log 37 =
15
156  a 
100
log 3715 = 23.4 + 0.15a
23 < log 3715 < 24
3715 is a 24 digit number.
n = 24.

G6 Given that 12 + 22 + 32 +  + n2 =
n
n  12n  1 ,
6
find the value of 1921 + 1822 + 1723 +  + 139.
1921 + 1822 + 1723 +  + 139
= (20 – 1)(20 + 1) + (20 – 2)(20 + 2) + (20 – 3)(20 + 3) +  + (20 – 19)(20 + 19)
= (202 – 12) + (202 – 22) + (202 – 32) +  + (202 – 192)
= 202 +  + 202 (19 times) – (12 + 22 + 32 + + 192)

= 19400 –
19
2039
6
= 7600 – 2470
= 5130
G7 In figure 2, ABCD is a square where AB = 1 and CPQ is an equilateral
triangle. Find the area of CPQ.
Reference: 2008 FI4.4
Let AQ = AP = x.
Then BQ = DP = (1 – x)
By Pythagoras’ Theorem,
CP = CQ  1 + (1 – x)2 = x2 + x2
2 – 2x + x2 = 2x2
x2 + 2x – 2 = 0  x2 = 2 – 2x
x =  1 3
Area of CPQ = Area of square – area of APQ – 2 area of CDP
x2 1 1  x  x2 2  2x
=1–  2 =x =x = 2x – 1
2 2 2 2

 
= 2  1  3  1= 2 3  3

1 1
Method 2 Area of CPQ = PQ 2 sin 60   x 2  x 2 
2 2

2
3

=
3x 2


3 1 3  2 3  
 2 3 3 
2 2

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Answers: (1994-95 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 29 October 2016

G8 The number of ways to pay a sum of $17 by using $1 coins, $2 coins and $5 coins is n. Find n.
(Assume that all types of coins must be used each time.)
Suppose we used x + 1 $1 coins, y + 1 $2 coins, z + 1 $5 coins, where x, y, z are non-negative
integers. Then (x + 1) + 2(y + 1) + 5(z + 1) = 17
x + 2y + 5z = 9
(x, y, z) = (9, 0, 0), (7, 1, 0), (5, 2, 0), (3, 3, 0), (1, 4, 0), (4, 0, 1), (2, 1, 1), (0, 2, 5).
Altogether 8 ways.
G9 In figure 3, find the total number of triangles in the 33 square.
Reference: 1998 HG9
There are 36 smallest triangles with length = 1
There are 36 triangles with length = 2

There are 24 triangles with length = 2

There are 16 triangles with length = 2 2

There are 8 triangles with length = 3

There are 4 triangles with length = 3 2

Altogether 124 triangles.

G10 In figure 4, the radius of the quadrant and the diameter of the large
semi-circle is 2. Find the radius of the small semi-circle.
Let the radius of the smaller semi-circle be r cm.
Let A, D, E be the centres of the quadrant, the larger and the smaller
semi-circles respectively.
BAC = 90
DE intersects the two semicircles at F. B
AE = EC = 1 cm r cm

BD = DF = r cm D

AC = AB = 2 cm F
AD = (2 – r) cm, DE = (1 + r) cm (2 - r) cm
AD2 + AE2 = DE2 (Pythagoras’ theorem) A
12 + (2 – r)2 = (1 + r)2 C 1 cm E 1 cm

1 + 4 – 4r + r2 = 1 + 2r + r2
2
r=
3

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