Group Events G1 a

  

Individual Events

• 2

  I1 a

  6 I2 P I3 a –2 I4 a _{see the remark}

  2 IS P

  84 Q 6 b 9 b

  11 Q

  8 b 2 + 4 2

  1

  3

  c

  7 R 56 c c 462 R

  24

  4

  7

• 334

  d

  11 S 2352 d − d S _{see the remark} 1 + 2

  18 Group Events

• 2

  G1 a

  47 G2 a G3 a –10 G4 P 500 GS a _{see the remark}

  16

  b 101 b

  4.5 b Q 15 b

  1

  1

  c

  43 c 15 c 2005 R c

  12

  2

  d d

  1 d 2005 S 1 d

  9 Individual Event 1

  

I1.1 Given that there are a positive integers less than 200 and each of them has exactly three

a positive factors, find the value of . x rs r s x r s x If = , where and are positive integers, then the positive factors of may be 1, , and . r s In order to have exactly three positive factors, = = a prime number. x a Possible = 4, 9, 25, 49, 121, 169. = 6.

  I1.2 a

  2 If copies of a right-angled isosceles triangle with hypotenuse cm can be assembled to b b

  form a trapezium with perimeter equal to cm, find the least possible value of . (give the answer in surd form). or

  The perimeter = 6 + 2

  2 8.8 or 2 + 4

  2

  7.7

  ≈ ≈ b

  The least possible value of = 2 + 4

  2

  4

  2

2 I1.3 c c If sin( – 3 + 17) = , where 0 < c – 3c + 17 < 90 and c > 0, find the value of c.

  ° b

  2

  −

  4

  1

  2

  sin(c – 3c + 17) =

  ° =

  4

  2

  2

  2

• 2

  −

  2 c

• – 3c + 17 = 45

  2 c – 3c – 28 = 0

  (c – 7)(c + 4) = 0

  c = 7 or – 4 (rejected) .

  

I1.4 Given that the difference between two 3-digit numbers xyz and zyx is 700 – c, where x &gt; z

If d is the greatest value of x + z, find the value of d. xyz

• – zyx = 700 – c 100x + 10y + z – (100z + 10y + x) = 700 – 7 99x – 99z = 693

  x – z = 7

  Possible answers: x = 8, z = 1 or x = 9, z = 2

  d is the greatest value of x + z = 9 + 2 = 11

  Individual Event 2

  I2.1 In Figure 1, ABCD is a square, M is the mid-point of AD and N is the mid-point of MD. If CBN : MBA = P : 1, find the value of P.

  ∠ ∠ Let ABM = , CBM = P . Let AB = 4, AM = 2, MN = 1 = ND. ∠ θ ∠ θ

  1 tan θ =

  2 ₁

  2 ×

  2 tan θ ₂

  4 tan 2 tan P P

  θ = = = = θ , = 2 _{2 1 2}

  3 1 − tan θ 1 −

  ( ) ₂ I2.2 ABCD

  Given that is a rhombus on a Cartesian plane, and the co-ordinates of its vertices are + A (0, 0), B ( P , 1), C ( u , v ) and D (1, P ) respectively. If u v = Q , find the value of Q .

  ABCD is a rhombus, It is also a parallelogram Q ∴

  By the property of parallelogram, the diagonals bisect each other

  B D AC

  Mid point of , = mid point of

      u v

  1

  2

  1     , = , 

  2 + + + +

  2 2  

  2 2  ⇒

• u = 3, v = 3 Q = u v = 6
• I2.3 If 1 + (1 + 2) + (1 + 2 + 3) + + (1 + 2 + 3 + Q ) = R , find the value of R .

  … … R = 1 + (1 + 2) + (1 + 2 + 3) + + (1 + 2 + 3 + + 6)

  … … R = 1 + 3 + 6 + 10 + 15 + 21 = 56

  E

  I2.4 In the figure, EBC is an equilateral triangle, and A , D lie on EB

  and EC respectively. Given that AD // BC , AB = CD = R and AC

  ⊥ BD . If the area of the trapezium ABCD is S , find the value of S .

  A D ABC BCD AC BD J AC BD x x ∠ = ∠ = 60 ° , intersects at , ⊥ .

  J ACD DBA (S.A.S.)

  ∆ ≅ ∆

  56

  56 AC = BD (corr. sides, ’s) ≅ ∆ y y

  ABD DCA ∠ = ∠ (corr. ∠ s, ≅ ∆ ’s)

• – JBC = 60 ABD = JCB

  ∴ ∠ ° ∠ ∠ B C JBC is a right-angled isosceles triangle.

  ∆ E

  JBC = JCB = 45 ∠ ∠ ° BJ = CJ = y , – = – AJ AC y = BD y = DJ = x

  A D JAD = JDA = 45 , ADC = 120

  ∠ ∠ ° ∠ °

• 56 x y

  J

  Apply sine formula on ACD , =

  ∆ _{o o}

  56

  56 sin 45 sin 120

  1 _{2 o}

  1 28 3 x y x y = 28 6 , area = sin + 90 = 784 6 = 2352

• 2

  ( ) ( )( )

  2 Method 2 EBC = 60 = ECB C

  ∠ ° ∠ B

  28

  28 F G AF BC DG BC BC F G Draw ⊥ and ⊥ , cutting at and .

  BF = CG = 56 cos 60 = 28 °

  AF = DG = 56 sin 60 =

  28

  3 ° AC intersects BD at J , AC BD .

  ⊥ ACD DBA (S.A.S.)

  ∆ ≅ ∆ ABD = DCA (corr. s, ’s)

  ∠ ∠ ∠ ≅ ∆ JBC = 60

• – ABD = JCB

  ∴ ∠ ° ∠ ∠ JBC is a right-angled isosceles triangle. ∆

  JBC JCB ∠ = ∠ = 45 ° CF AF

  28

  3

  = cot 45 ° =

• – FG = CF CG =

  28 3 − 1 = AD ( )

  3 −

1 ×

  2 28 ×

• 28

  2 ( )

  Area of the trapezium ABCD = S = ⋅

  28 3 = 2352 ( )

  

2

  Individual Event 3

  1

  1

  2

  ≠± ≠ =

• I3.1 Let x 1 and x –3.If a is the real root of the equation ,find the value of a

  1 x 3 x

  1

  − − ²

• x

  2

  1

  1 2 x

  1

  − ( )

• 1

  ⇒ = − = ₂ x

  3 x 1 x 1 x 3 x 1 x

  1

• − − −

  ( )( )

  1 1 x

  1

  1

  − ⇒

  = = − x

  3 x 1 x 1 x 3 x

  1

  − + + + + ( )( )

  ⇒ x +1 = –x – 3 x = –2 = a a

  1

  −

  I3.2 If b &gt; 1 ， f (b) = and g(b) = 1 + . If b satisfies the equation b b

  log log _{2 3} |f(b) – g(b)| + f(b) + g(b) = 3, find the value of b.

  Similar question: 2007 HI9 ₄

  2

  2 log 2 log

  3 2 log 2 log 3 log log _{3 b}

  12 4 b ⇒ ⇒

  1

  1

  3

  2 log log − − = = = + + + +

  log b log b log b log b log b log b

  3 b

  12

  2

  4 b log log

  = ±

  3 b

  12 ₂ 4 b

  4

  12 log log or log log

  = = ₂

  3 b

  12 3 b b

  3 b

  = 16 or b = 9 ₂

  4 b

  3

  3

  2

  3

  

2

⇒ ⇒ When b = 16, (b ) = 256 &lt; 1728 = 12 b &lt; 12 log = log &lt; 0 rejected. ₂ 3 b

  12

  4 b

  81 When b = 9, = log = log &gt; 0 accepted. log 3 b

  12

  12 Method 2

  x y x y

• −

  Remark Define the maximum function of x, y as: Max(x, y) =

  2

  x y x y − + −

  Similarly, the minimum function of x, y is: Min(x, y) =

  2

  a

  2 log 2 log

  4 1 log b log 3 log 3 b

  −

  = = =

• f (b) = , g(b) =1+ =

  log b log b log b log b log b log b log b _{2 3} The given equation is equivalent to 2 Max(f(b), g(b)) = 3

  log 4 log 3 b

  4 ⇒

  If f(b) &gt; g(b), i.e. &gt; b &lt; , then the equation is 2f(b) = 3

  b b log log

  3 ₃

    2 log

  4

  4

  4

  3

  3 ⇒ ⇒

    3 log 16 = log b b = 16 &lt; ⇒ 1 &lt;

  = b  

  27 log

  3 ⇒ 27 &lt; 4, which is a contradiction; rejected

  ∴

  If f(b) g(b), the equation is equivalent to 2 g(b) = 3

  ≤

  2 log 3 b

  2

  3

  2

  3

  i.e. = 3 ⇒ log 9b = log b ⇒ 9b = b ⇒ b = 9 log b ₂

  2 x

  I3.3 Given that x satisfies the equation x – 5x + (b – 8) = 0. If c = , find the value of c. _{4 2} x x

  1

  2

  2

  4

  2

  2

  4

  2

  2 x – 5x + 1 = 0, x + 1 = 5x, x + 2x + 1 = 25x , x + x + 1 = 24x _{2 2} x x

  1

  c = = = _{4 2 2} x x

  1 24x

  24

  2 I3.4 If –2 and 216c are the roots of the equation px + dx = 1, find the value of d.

  2

• –2 and 9 are roots of px + dx – 1 = 0

  1

  1

  ⇒ Product of roots = = –2

  9 p =

  − × p

  18

  d

  7

  ⇒ ⇒

  Sum of roots = = –2+9 –18d = 7 d =

  − − p

  18 Individual Event 4

  I4.1 Let a be a real number. _{x x}

• 1

  If a satisfies the equation log _{x x x} 2 (4 + 4) = x + log 2 (2 – 3), find the value of a.

• 1

  log _{x x x +1} 2 (4 + 4) = log

  2

  2 + log

  2 (2 – 3)

  4 + 4 = 2 (2 – 3) _x ⋅

  2 x 2 x

  (2 ) + 4 = 2 (2 ) – 3

  2 _x ⋅ ⋅

  2 x

  0 = (2 ) – 3 2 – 4 _{x x} ⋅ (2 – 4)(2 + 1) = 0 _x 2 = 4, x = 2 = a

  I4.2 Given that n is a natural number.

  3

  2 (Reference: 2011 FI3.3) If b = n – 4an – 12n + 144 is a prime number, find the value of b.

  3

2 Let f (n) = n – 8n – 12n + 144

  3

  2 f (6) = 6 – 8

  6 – 12 6 + 144 = 216 – 288 – 72 + 144 = 0

  ⋅ ⋅ f (6) is a factor

  ∴

  By division, f (n) = (n – 6)(n – 6)(n + 4)

  3

  2 ⇒ b = n – 8n – 12n + 144 , it is a prime n – 6 = 1, n = 7, b =11

  I4.3 In Figure 1, S 1 and S 2 are two different inscribed squares of the right-angled triangle ABC.

  If the area of S

  1 is 40b + 1, the area of S 2 is 40b and AC + CB = c, find the value of c.

  Reference: American Invitation Mathematics Examination 1987 Q15 A

A

R F E y c Q

b

b c h

  440

  

S

x

  441

  B B C C D P

a a

  Add the label D, E, F, P, Q, R, S as shown. CDEF, PQRS are squares.

  2 a b Let DE = x, SR = y, then x = 441 = 21, y = 440 . Let BC = a, AC = b, AB = c = .

• 2

  

Let the height of the triangle drawn from C onto AB be h, then ab = ch = 2 area of .....(*)

∆ b x x ab

  − AFE ~ ACB : ⇒ x = = 21 .................(1)

  ∆ ∆ =

• b a a b

  height of CSP from C h h y h ch

  ∆ − ⇒ ⇒

  CSP ~ CAB :

  y = = 440

  ∆ ∆ = =

• SP c y c c h
²

^{2ab ab a b}

  ⇒

  By (*), = 440 440 ......(2)

  = ₂

₂

ab a ab b

• c c

  From (1) ab = 21(a + b) ............ (3), sub. (3) into (2):

  2

  2

  2

  21

  2

  21

  42

  21

  42 ( ) ( ) − ( ) ( ) − ( ) ( ) − ( )

+ + + + + +

• a b a b ab a b a b a b a b a b

  440 = = =

  2

  21 a b ab a b 21 a b −

• 2
• a b

  − − ( )

  ( ) ( ) ( )

  Cross multiplying and squaring both sides:

  2

  2

  440[(a + b) – 42(a + b) + 441] = 441[(a + b) – 42(a + b)]

  2

  (a + b) – 42(a + b) – 440 441 = 0

  ×

  (a + b – 462)(a + b + 420) = 0

  AC + CB = a + b = 462

  2 I4.4 Given that 241c + 214 = d , find the positive value of d.

  Method 3

  2 d = 241 462 + 214 = 111556

  ×

  Observe the number

  d = 111556

  patterns:

  2 昌爸工作坊 昌爸工作坊圖解直式開平方 圖解直式開平方

  Reference: 昌爸工作坊 昌爸工作坊 圖解直式開平方 圖解直式開平方

  34 = 1156

  2 Divide 111556 into 3 groups of numbers 11, 15, 56.

  334 = 111556

  2

  2 ⇒

  Find the maximum integer p such that p 11 p = 3

  ≤ 3334 = 11115556

  2

  11 – p = 2 ................................ 3 + 3 = 6 d = 334

  ∴

  2 Find the maximum integer q such that (60 + q)q 215 ≤ Also, 33 = 1089

  2 ⇒

  q = 3 333 = 110889

  2

  215 – 63 3 = 26

  ×

  3333 = 11108889 60 + q + q = 66 ..................................... Find the maximum integer r such that (660 + r)r 2656

  ≤

  3 4 = 12

  × ⇒

  r = 4 33 34 = 1122

  × d = 334

  333 334 = 111222

  ×

  2 Method 2 : d = 241 462 + 214 = 111556 ×

  3333 3334 = 11112222

  ×

  2

  2 ⇒

  300 = 90000 &lt; 111556 &lt; 160000 = 400 300 &lt; d &lt; 400 .....................................

  2

  2

⇒

  330 = 108900 &lt; 111556 &lt; 115600 = 340 330 &lt; d &lt; 340

  2 ⇒

  The unit digit of d is 6 the unit digit of d is 4 or 6

  2 ⇒

  335 = 112225 d = 334 or 111556 is not divisible by 3, but 336 is divisible by 3

  ∴ d = 334 ⇒

  Remark: Original question: , find the value of d. d = 334 LL ±

  Individual Event (Spare)

  IS.1 In figure 1, ABC is an acute triangle, AB = 15, AC = 13, and its

  ∆ altitude AD = 12. If the area of the ABC is P, find the value of P. _{2 2} ∆ BD =

  15 12 = 9 ₂ − ₂ CD =

  13 12 = 5 −

  1

  9

  5 12 = 84

  ∆ × ( )

• P = area of =

  2

  

4

  4 IS.2 Given that x and y are positive integers. If x y is divided by x y , the quotient is P + 13

• and the remainder is Q , find the value of Q .

  4

  4

• x y = 97( x y ) + Q , 0 Q &lt; x y

  ≤

  Without loss of generality, assume x y ,

  ≥

  4

  4

  4

  &lt; + + + x x y = 97( x y ) + Q &lt; 98( x y ) 98(2 x ) = 196 x

  ≤

  3 x &lt; 196 x

  5

  ≤

  4

  4

  3

  3

  3

• On the other hand, + + x y = 97( x y ) + Q = x ( x y ) – – y ( x y )

  3

  3

  3 ⇒

  ( x – 97)( + – x y ) = y ( x y ) + Q

  3 ⇒ ⇒

  RHS LHS x

  97

  ≥ ≥ ≥ x

  ≥

  5

  x = 5 ∴

  4

  4

  3

  3

  3

• Next, = 97( – + + x y x y ) + Q = y ( x y ) + x ( x y )

  3

  3

  3 ⇒

  (97 – y )( + – x y ) = x ( x y ) – Q

  3

  3

  3 ⇒

  (97 – )( + ) &gt; + – y x y x ( x y ) – ( x y )

  3 ⇒

  (98 – y )( + x y ) &gt; 0

  3 ⇒

  98 &gt; y

  ⇒ y

  4 ≥ ........... (1)

  4

  4 ⇒

• Q x y – 97( x y )

  ≤ ≤

  4 ⇒

  97(5 + y ) 625 + y

  ≤

  4 ⇒ y y

  97 140 + ............... (2)

  ≤

  By trial and error, y = 1 and y = 4 satisfies (2) So there are two possible pairs ( x , y ) = (5, 1) or (5,4).

  4

  4 5 + 1 = 626 = (5 + 1) 104 + 2, the quotient is not 97.

  ×

  4

4 Q

  5 + 4 = 881 = (5 + 4) 97 + 8, = 8

  ×

  12 IS.3 Given that the perimeter of an equilateral triangle equals to that of a circle with radius cm.

  Q

  2

2 If the area of the triangle is R cm , find the value of R .

  π

  12

  ⇒

  Radius of circle = cm = 1.5 cm circumference = 2 1.5 cm = 3 cm

  ×π× π

  8 Side of the equilateral triangle = cm

  π

  1 ² _o

  2

  3 ²

  2

  3

  ⇒

  Area of the triangle = sin 60 cm = cm R =

  π π

  2

  4

  4

  1

  3 IS.4 Let W = , S = W + , find the value of S.

  1

  2 R

• W

  1

• W
• W

  L

  1

  2

  1 ±

  2

  ⇒ W = 2, S = 2 + S – 2S – 1 = 0, S = , S &gt; 0, S = 1 + 2 only

  ∴ S

  2

  Group Event 1 _a G1.1 Given that a is an integer. If 50! is divisible by 2 , find the largest possible value of a.

  2, 4, 6, 8, , 50 are divisible by 2, there are 25 even integers.

  …

Reference: 1990 HG6, 1994 FG7.1, 1996 HI3, 2004 FG1.1, 2011 HG7, 2012 FI1.4, 2012 FG1.3

4, 8, , 48 are divisible by 4, there are 12 multiples of 4.

  … 8, , 48 are divisible by 8, there are 6 multiples of 8. … 16, , 48 are divisible by 16, there are 3 multiples of 16.

  … 32 is the only multiple of 32. a = 25 + 12 + 6 + 3 + 1 = 47 G1.2 Let [x] be the largest integer not greater than x. For example, [2.5] = 2.

   

  11

  77

  12

  78

  13

  79

  14

  80

  × × × × + + + If b = 100 , find the value of b . ×

  

  11

  76

  12

  77

  13

  78

  14 79 

  × × × × + + +

  11

  77

  12

  78

  13

  79

  14

  80

  × × × × + + +

  100

  ×

  11

  76

  12

  77

  13

  78

  14

  79

  × × × × + + +

  11 ×

  76 12 ×

  77 13 ×

  78 14 ×

  79

  11

  12

  13

  14 = 100

  ×

  11

  76

  12

  77

  13

  78

  14

  79

  × × × × + + +

   

  11

  12

  13

  14

  = 100

  1

  ×

•  

  

  11

  76

  12

  77

  13

  78

  14 79 

  × × × × + + +

  11 100 12 100 13 100 14 100

  × × × × + + +

  = 100 +

  76 12 ×

  77

  13

  78

  14

  79

  × × × + +

• 11

  11 × 76+12 × 77+13 × 78+14 × 79 &lt;11 × 100+12 × 100+13 × 100+14 × 100&lt;2(11 × 76+12 × 77+13 × 78+14 × 79)

  11

  77

  12

  78

  13

  79

  14

  80

  × × × × + + +

  1

  2

  &lt; &lt;

  11

  76

  12

  77

  13

  78

  14

  79

  × × × × + + +

  11

  77

  12

  78

  13

  79

  14

  80

  × × × × + + +

  101 100 102 , b = 101

  &lt; × &lt;

  11

  76

  12

  77

  13

  78

  14

  79

  × × × × + + + G1.3 If there are c multiples of 7 between 200 and 500, find the value of c .

  200 = 28.6, the least multiple of 7 is 7 29 = 203

  ×

  7 500

  = 71.4, the greatest multiple of 7 is 7 71 = 497

  ×

  7 497 = a + ( c – 1) d = 203 + ( n – 1) 7; c = 43

  ⋅ x

  π

  Given that 0 ≤ ≤ and satisfies the equation sin 1 − 1 − sin = sin . If d = tan x ,

• G1.4 x x x x

  2 2 find the value of d. _{2 2}

   

  x

  1 1 sin x sin

•   sin x

  − − = ( )

   

  2 _{2 −} 1 cos x 1 + sin x + 1 – sin x – 2 1 sin x

  − =

  2 2(2 – 2 cos x) = 1 – cos x cos x = 1

  x = 0 d

  = tan x = 0

  Group Event 2 G2.1 If the tens digit of ^{5 5} 5 is a, find the value of a. _{5 5} 5 = LL125, the tens digit = a = 2

  = × b

  The prime are: {2, 3, 5, 7, 13, 17, 23, 37, 43, 47, 53, 67, 73, 83, 97} c = 15 G2.4 If the lines y = x + d and x = –y + d intersect at the point (d – 1, d), find the value of d. x

  = × ⋅

G2.3 Given that there are c prime numbers less than 100 such that their unit digits are not square

numbers, find the values of c.

  1

  2

  5

  9

  5

  9

  2

  =

  3

  Remark Original question: If the tenth-place digit LL, this is the position of the first digit to the right of the decimal point.

  3

  5

  9

  5

  ∆ ABE from A = AB cos B =

  3 Height of

  5

  2 , find the value of b. cos B =

  ∆ ABE is b cm

  ∆ ABC is a right-angled triangle, AB = 3cm, AC = 4 cm and BC = 5 cm. If BCDE is a square and the area of

  G2.2 In Figure 1,

  = –(x + d) + d x = 0 = d – 1 d = 1

  Group Event 3 G3.1 If a is the smallest real root of the equation x x 1 x 2 x

  3

  1 71 , find the value of a.

• =

  ( )( )( ) Reference: 1993 HG6, 1995 FI4.4, 1996 FG10.1, 2000 FG3.1, 2012 FI2.3 Let t = x + 1.5, then the equation becomes t 1 . 5 t . 5 t . 5 t 1 .

  5

  1

  71 − − = + + +

  ( )( )( )( )    

  9

  1 ^{2 2} +     t t

  1

  71 − − =

     

  4

  4 ₄

  5 ₂

  9 t t

  1

  71 − = + +

  2

  16

  5

  25 ^{4 2} + ⇒ t t

  71 − =

  2 ₂

  16

    ₂

  5

  ⇒

   

  t

  71

  − =

   

  4 ₂ 5 289 ₂

  17

  17 ⇒

  t

  71 t ⇒ t = or t = –

  − = =

  4

  4

  2

  2

  17

  3

  x = t – 1.5 = = 7 or –10 ±

• – 2

  2

  a = the smallest root = –10 G3.2 Given that p and q are prime numbers satisfying the equation 18p + 30q = 186. p If log = b 0, find the value of b.

  ≥ ⁸ +

  3 q

  1 18p + 30q = 186 ⇒ 3p + 5q = 31 Note that the number “2” is the only prime number which is even.

  3p + 5q = 31 = odd number ⇒ either p = 2 or q = 2

  p

  2

  1 If p = 2, then q = 5; b = log = log log &lt; 0 (rejected) _{8 8} = ₈ 3 q

  1

  3

  5

  1

  16

  × + + p

  7 If q = 2, then p = 7; b = log = log _{8 8} = 3 q

  1

  3

  2

  1

• ×

  G3.3 Given that for any real numbers x, y and z, is an operation satisfying

⊕

x

  (i) ⊕ 0 = 1, and (ii) (x y ) z = (z xy ) + z.

  ⊕ ⊕ ⊕ If 1 2004 = c, find the value of c. ⊕ c

  = 1 2004

  ⊕

  = (1 0) 2004

  ⊕ ⊕

  = (2004 0)+ 2004

  ⊕

  = 1 + 2004 = 2005

  4 3 2004 G3.4 f

  Given that f (x) = (x + 2x + 4x – 5) + 2004. If 3 − 1 = d, find the value of d.

  ( ) x

  3

  1

  = − ₂

  2

  (x + 1) =

  3

  ( )

  2 x + 2x – 2 = 0

  4

  3

  2

  2 By division, x + 2x + 4x – 5 = (x + 2x – 2)(x + 2) – 1 = –1 d = f

  3

  1

  − ( )

  = f(x)

  4 3 2004

  = (x + 2x + 4x – 5) + 2004

  2004

  = (–1) + 2004 = 2005

  Group Event 4 _x

       

  4

  1 2 1000       G4.1 If f x and P = f f f , find the value of P.

  = L

+ + +

( ) ^x +

       

  4 2 1001 1001 1001 Reference: 2011 HG5, 2012 FI2.2 _x

  1 − x x 1 − x

  4

  4

  4

  4

  2

  4

  2

  4

• ⋅ ⋅

  f x f x

  1

  1

  − = = = + + ( ) ( ) _x

  1 − x x 1 − x

  4

  2

  4

  2

  4

  4

  2

  4

  2

  4

• ⋅ ⋅

       

  1 2 1000

  P f   f   f  

  =

  L       1001 1001 1001             1 1000

  2 999 500 501   

  f f  f   f   f   f  

  = = 500

  L

              1001 1001 1001 1001 1001 1001

  G4.2 Let f (x) = |x – a| + |x – 15| + |x – a – 15|, where a x 15 and 0 &lt; a &lt; 15.

  ≤ ≤ If Q is the smallest value of f (x), find the value of Q.

  Reference: 1994 HG1, 2001 HG9, 2008 HI8, 2008 FI1.3, 2010 HG6, 2011 FGS.1, 2012 Final G2.3 f (x) = x – a + 15 – x + 15 – x + a = 30 – x 30 – 15 = 15 = Q

  ≥

  1

  1

• G4.3 If 2 = 3 = 36 and R = , find the value of R. ^{m n}

  m n Reference: 2001 HI1, 2003 FG2.2, 2004 FG4.3, 2005 HI9, 2006 FG4.3 _{m n}

  log 2 = log 3 = log 36

  m log 2 = n log 3 = log 36

  log 36 log

  36

  m = ; n =

  log 2 log

  3

  1 1 log 2 log 3 log 6 log

  6

  1 = = = =

  m n log

  36 log 36 log

  36 2 log

  6

  2 G4.4 Let [x] be the largest integer not greater than x, for example, [2.5] = 2.

  1

  1

  1 If a 1 and S = [a], find the value of a.

  = L _{2 2} + + + + ₂

  2 3 2004

  1

  1

  1

  1

  1

  1 1 &lt; a 1 &lt;

  1

  = L L _{2 2 2} + + + + + + + +

  2 3 2004

  1

  2

  2 3 2003 2004

  × × ×

  1

  1

  1

  1

  1 =

  1

  1

  − − L − + + + +

  2

  2 3 2003 2004

  1 = 2 &lt; 2

  −

  2004

  S = [a] = 1

  Group Event (Spare) GS.1 For all integers n, F n is defined by F n = F n + F n , F = 0 and F = 1.

• –1 –2

  1 If a = F + F + F + F , find the value of a. +

• –5 –4 …

  4

  5 F 2 = 0 + 1 = 1, F 3 = 1 + 1 = 2, F 4 = 1 + 2 = 3, F 5 = 2 + 3 = 5 F

• –1 + 0 = 1 ⇒ F –1 = 1, F –2 + 1 = 0 ⇒ F –2 = –1, F –3 + (–1) = 1 ⇒ F –3 = 2,

  F + 2 = –1 ⇒ F = –3, F + (–3) = 2 ⇒ F = 5

• –4 –4 –5 –5
• a = F –5 + F –4 + F 4 + F

  5 = 5 – 3 + 2 – 1 + 1 + 0 + 1 + 1 + 2 + 3 + 5 = 16 …

  2

  4

  3

  2 GS.2 Given that x satisfies the equation x + x + 2 = 0. If b = x + 2x + 3x + 2x +1, find the value of b.

  4

  3

2 By division, b = x + 2x + 3x + 2x +1

  2

  2 = (x + x + 2)(x + x ) + 1 = 1

  GS.3 Figure 1 shows a tile. If C is the minimum number of tiles required to tile a square, find the value of C.

  C = 12

GS.4 If the line 5x + 2y – 100 = 0 has d points whose x and y coordinates are both positive integers,

find the value of d.

  (x, y) = (18, 5), (16, 10), (14, 15), (12, 20), (10, 25), (8, 30), (6, 35), (4, 40), (2, 45) d = 9