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  • Maths Lab: Wordings

    Lfs Maths Lab, SMO Senior 1

  • Maths Lab: Elements of solutionSingapore Mathematics OlympiadSenior Section 2010Duration: 2 h30.

    The number of indicates the relativ difficulty of the question.

    Multiple Choice Questions

    1 Answer (A).

    Find the value of H123L + H246L + H369L + + H3356701005L

    H136L + H2612L + H3918L + + H33510052010L .We have

    H123L + H246L + H369L + + H3356701005LH136L + H2612L + H3918L + + H33510052010L =

    H123LA1 + 23 + 33 + + 3353EH136LA1 + 23 + 33 + + 3353E then

    H123L + H246L + H369L + + H3356701005LH136L + H2612L + H3918L + + H33510052010L =

    1

    3.

    2 Answer (E).

    Lets a, b, c and d be real numbers such that b + c + d

    a=

    a + c + d

    b=

    a + b + d

    c=

    a + b + c

    d= r .

    As r + 1 = 1 +b + c + d

    a=

    a + b + c + d

    a so aHr + 1L = a + b + c + d .

    Identically, bHr + 1L = cHr + 1L = d Hr + 1L = a + b + c + d so Hr + 1L Ha + b + c + d L = 4 Ha + b + c + d L.Thus we obtain Hr - 3L Ha + b + c + d L = 0.Then r = 3 or a + b + c + d = 0.

    Yet a + b + c + d = 0 gives -a = b + c + d so b + c + d

    a=-a

    a= -1 and identically -b = a + c + d , -c = a + b + d and

    -d = a + b + c so a + c + d

    b=

    a + b + d

    c=

    a + b + c

    d= -1.

    Finally, r = 3 or r = -1.

    3 Answer (E).

    If 0 < x 0.As 4907 = 7701 with 701 prime number.

    (In fact as 252 < 701 < 302, we know that if 701 wasnt a prime number it will be divisible by 2,3,5,7,11,13,17,19,23 or29 and thats not the case.)Therefore we obtain:

    k - m = 1 and k + m = 4907, then k = 2454 and n = 2453 k - m = 4907 and k + m = 1, then k = 2454 and n = -2453 k - m = 7 and k + m = 701, then k = 354 and n = 347 k - m = 701 and k + m = 7, then k = 354 and n = -347 .

    As y = n2, we deduce y = 24532 or y = 3472.Moreover

    or x - 90 = 2454 so x = 2544 or 90 - x = 2454 so x = -2364: we have 2 ordered pairs I-2364, 24532M andI2544, 24532M,

    or x - 90 = 354 so x = 444 or 90 - x = 354 so x = -264: we have 2 ordered pairs I-264, 3472M andI354, 3472M .

    7 Answer (D).

    Lets describe the cases of non-empty subsets verifying the condition Good: the singletons: only composed by an even number so 5 subsets;

    the pairs: two even numbers or 1 even and 1 odd, so 52

    +51

    51

    = 10 + 25 = 35 subsets;

    the 3lists: two even and one odd or three even, so 52

    51

    +53

    = 105 + 10 = 60 subsets;

    the 4lists: 2 even and 2 odd, 3 even and 1 odd or 4 even, so 52

    52

    +53

    51

    +54

    = 100 + 50 + 5 = 155

    subsets;

    the 5lists: 3 even and 2 odd, 4 even and 1 odd or 5 even, so 53

    52

    +54

    51

    +55

    = 100 + 25 + 1 = 126

    subsets; the 6lists: 3 even and 3 odd, 4 even and 2 odd or 5 even and 1 odd, so

    53

    53

    +54

    52

    +55

    51

    = 100 + 50 + 5 = 155 subsets;

    the 7lists: 4 even and 3 odd or 5 even and 2 odd, so 54

    53

    +55

    52

    = 50 + 10 = 60 subsets;

    the 8lists: 4 even and 4 odd or 5 even and 3 odd, so 54

    54

    +55

    53

    = 25 + 10 = 35 subsets;

    the 9lists: 5 even and 4 odd so 55

    54

    = 5 subsets;

    the 10lists: 5 even and 5 odd so 1 subset.the overall number of subsets is then 5 + 35 + 60 + 155 + 126 + 155 + 60 + 35 + 5 + 1 = 637 subsets.

    8 Answer (B).

    As f Hr L = k and f HsL = k, then r and s are symetric regarding the x-coordinate of the summit of the quadratic function,it is to say -

    b

    2 a.

    We then deduce s + r = 2 -b

    2 a= -

    b

    a .

    Thus f Hr + sL = a - ba

    2

    + b -b

    a+ c =

    b2

    a-

    b2

    a+ c = c .

    4 Lfs Maths Lab, SMO Senior

  • 9 Answer (D).

    Find the number of positive integers k < 100 such that 2 H36 nL + kH23 n+1L - 1 is divisible by 7 for any positive integer n.Note that 2 H36 nL + kH23 n+1L - 1 = 2 H36Ln + 2 k H23Ln - 1 = 2 HH36Ln + k H23LnL - 1.We have 23 = 8 = 7 + 1 so the remainder of 23 in the division by 7 is 1. Thus the remainder of 23 n is 1 too.

    Identically, 36 = H32L3 so the remainder of 36 in the division by 7 is 1, thus the remainder of 36 n is 1.Finally the remainder of HH36Ln + k H23LnL is 1 + k1 = k + 1.We have to determine the number of positive integers k < 100 such that 2 Hk + 1L - 1 = 2 k + 1 is divisible by 7.So it must exist integer p such that 2 k + 1 = 7 p , 2 k = 7 p - 1 = 7 H p - 1L + 6 and we deduce that k = 7 p ' + 3.As 0 7 p ' + 3 < 100, 0 p ' < 13 .We obtain 14 possibilities.

    10 Answer (C).

    Lets consider point E, intersection of the lines AD et CM.The angles A M E and B M E are opposite angles they are congruent.As lines AD and CB are parallel and line CM is a secant then the angles A E M and B C M are alternate angles andcongruent.Moreover A M = M E as M is the midpoint of AB .Theregore the triangles AEM and CBM are congruent.We deduce A E = C B and M E = C M so A E + A D + D C = 17 and C E = 13 and clearly the area of the trapeziumis equal to the area of triangle CDE.The triangle EDC is rectangle at point D, according to the Pythagoreas theorem, C E2 = C D2 + D E2.

    So 132 = C D2 + H17 - D CL2 so D C is solution of the equation 2 x2 - 34 x + 120 = 0.The roots of this quadratic equation are 5 and 12.

    (The discriminant is D = 196 > 0 so the roots are 34 - 196

    22= 5 and

    34 + 196

    22= 12).

    Finally D E = 12 or D E = 5 so the area of the triangle is in both cases 512

    2= 30.

    Remark:Lets note S the area of trapezium ABCD.

    Then S =1

    2D ED C as previously explained.

    Now HD E + D CL2 = D E2 + 2 D ED C + D C2 = 4 S + C E2. In fact, Pythagoreas Theorem in triangle DEC right-angled at D, shows that C E2 = D E2 + D C2.

    Finally as D E + D C = A D + C D + C B = 17, we deduce 4 S + 132 = 172 and so S = 30 .

    Short Questions

    11 Answer: x = 132 .

    Denote L et w the dimensions of the original rectangle.Then the area is given by A = Lw .We then know that HL + 6L Hw - 2L = Lw so we obtain -2 L + 6 w - 12 = 0, L - 3 w + 6 = 0.We also know that HL - 12L Hw + 6L = Lw so we also obtain 6 L - 12 w - 72 = 0 , L - 2 w - 12 = 0.So the dimensions of the original rectangle are the couple solution of the system

    L - 3 w + 6 = 0L - 2 w - 12 = 0

    .

    Thus L = 48 and w = 18 so x = 2 HL + wL = 132.

    Lfs Maths Lab, SMO Senior 5

  • Denote L et w the dimensions of the original rectangle.Then the area is given by A = Lw .We then know that HL + 6L Hw - 2L = Lw so we obtain -2 L + 6 w - 12 = 0, L - 3 w + 6 = 0.We also know that HL - 12L Hw + 6L = Lw so we also obtain 6 L - 12 w - 72 = 0 , L - 2 w - 12 = 0.So the dimensions of the original rectangle are the couple solution of the system

    L - 3 w + 6 = 0L - 2 w - 12 = 0

    .

    Thus L = 48 and w = 18 so x = 2 HL + wL = 132.

    12 Answer: 2575.

    For all integer r, u r = 1 + 2 + + r =r Hr + 1L

    2.

    Thus for all integer j 1, i=1

    j 1

    u i=i=1

    j 2

    iHi + 1L .

    Lets remind that 1

    i-

    1

    i + 1=

    i + 1 - i

    iHi + 1L =1

    iHi + 1L , so i=1j 1

    u i= 2

    i=1

    j 1

    i-

    1

    i + 1= 2 1 -

    1

    j + 1=

    2 j

    j + 1.

    Then j

    i=1j 1u i=

    j

    2 jj+1

    =j + 1

    2, so

    1

    1

    u 1

    +2

    1

    u 1+

    1

    u 2

    +3

    1

    u 1+

    1

    u 2+

    1

    u 3

    + +100

    1

    u 1+

    1

    u 2+ +

    1

    u 100

    =1

    2i=1

    100 H j + 1L.

    Yet, i=1

    100 H j + 1L =i=2

    101

    j =H2 + 101L100

    2= 5150 so the value of

    1

    1

    u 1

    +2

    1

    u 1+

    1

    u 2

    +3

    1

    u 1+

    1

    u 2+

    1

    u 3

    + +100

    1

    u 1+

    1

    u 2+ +

    1

    u 100

    is 2575 .

    13 Answer: k = 501 .

    As 2010 ! = 122010 and as 10 = 25, so it is enough to count how many 5s in 2010 ! .We have 2010 = 5402 so there are 402 multiples of 5 in the set 81; 2; ; 2010< and then at least 402 factors 5s in theproduct 2010 ! . Identically, 2010 = 5280 + 10 so there are 80 multiples of 52 in the set 81; 2; ; 2010 1 and 1

    log aHbL +1

    log bHaL = 1229 , find the value of 1

    log a bHbL -1

    log a bHaL .As a > b > 1,

    1

    log a bHbL -1

    log a bHaL > 0.Then

    1

    log a bHbL -1

    log a bHaL =1

    lnHbLlnHa bL

    -1

    lnHaLlnHa bL

    =lnHa bLlnHbL -

    lnHa bLlnHaL =

    lnHaL + lnHbLlnHbL -

    lnHaL + lnHbLlnHaL =

    lnHaLlnHbL -

    lnHbLlnHaL .

    As lnHaLlnHbL -

    lnHbLlnHaL > 0,

    1

    log a bHbL -1

    log a bHaL =lnHaLlnHbL -

    lnHbLlnHaL

    2

    .

    Lets yet note 1

    log aHbL +1

    logbHaL =1

    lnHbLlnHaL

    +1

    lnHaLlnHbL

    =lnHaLlnHbL +

    lnHbLlnHaL so

    lnHaLlnHbL -

    lnHbLlnHaL

    2

    =lnHaLlnHbL +

    lnHbLlnHaL

    2

    - 4lnHaLlnHbL

    lnHbLlnHaL = 1229 - 4 = 1225.

    Finally 1

    log a bHbL -1

    log a bHaL = 1225 = 35.

    6 Lfs Maths Lab, SMO Senior

  • If a > b > 1 and 1

    log aHbL +1

    log bHaL = 1229 , find the value of 1

    log a bHbL -1

    log a bHaL .As a > b > 1,

    1

    log a bHbL -1

    log a bHaL > 0.Then

    1

    log a bHbL -1

    log a bHaL =1

    lnHbLlnHa bL

    -1

    lnHaLlnHa bL

    =lnHa bLlnHbL -

    lnHa bLlnHaL =

    lnHaL + lnHbLlnHbL -

    lnHaL + lnHbLlnHaL =

    lnHaLlnHbL -

    lnHbLlnHaL .

    As lnHaLlnHbL -

    lnHbLlnHaL > 0,