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2018 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 18, 2018
(in North America and South America)

Thursday, November 19, 2018
(outside of North American and South America)

©2018 University of Waterloo


PART A

  1. Since each box contains 12 trays, then 6 boxes contain 12×6=72 trays.
    Since Paul has 4 extra trays, then he has 72+4=76 trays.
    Since each tray can hold 8 apples, then the trays can hold 76×8=608 apples.

    Answer: 608 apples

  2. Since the rabbit runs 3 times as quickly as the skunk, then the rabbit takes 13 of the time that the skunk takes, or 2 minutes.
    Since the rabbit runs 5 times as quickly as the turtle, then the turtle takes 5 times as long as the rabbit, or 10 minutes, to finish the race.

    Answer: 10 minutes

  3. Solution 1
    With 6 crayons in the box, there are 6×52=15 pairs of crayons that can be removed. (This is because there are 6 crayons that can be chosen first and 5 crayons that can chosen second; since there are 2 orders in which any specific pair of crayons can be chosen, we divide 6×5 by 2.)
    Using similar reasoning, with 3 red crayons in the box, there are 3×22=3 pairs of red crayons that can be removed.
    Therefore, the probability that Jakob removes 2 red crayons is 315 or 15.

    Solution 2
    We label the crayons in the box R1, R2, R3, B1, B2, G1.
    The possible pairs that Jakob can choose are

    R1R2, R1R3, R1B1, R1B2, R1G1, R2R3, R2B1, R2B2, R2G1,
    R3B1, R3B2, R3G1, B1B2, B1G1, B2G1

    for a total of 15 pairs.
    Of these 15 pairs, three pairs (R1R2, R1R3, R2R3) consist of two red crayons.
    Therefore, the probability that Jakob removes 2 red crayons is 315 or 15.

    Solution 3
    We remove the two crayons one after the other. In order for both crayons to be red, both the first and second crayons removed must be red.
    When the first crayon is removed, the probability that it is red is 36, since there are 6 crayons 3, of which are red.
    After 1 red crayon is removed, there are 5 crayons remaining, 2 of which are red.
    Therefore, the probability that the second crayon removed is red is 25.
    Thus, the probability that both crayons are red is 3625=15.

    Answer: 15

  4. Since x21=(x+1)(x1), then setting x=10n gives 102n1=(10n+1)(10n1).
    Therefore, a=102n13(10n+1)=(10n+1)(10n1)3(10n+1)=10n13 because 10n+10.
    Since 10n is the integer consisting of a 1 followed by n 0s, then 10n1 is the integer consisting of n 9s.
    Thus, a=10n13 is the integer consisting of n 3s. This is because each digit 9 in 10n1 can be divided by 3 to obtain the new digit 3 without any re-grouping (borrowing).
    Therefore, the sum of the digits of the integer a is 3n.
    Since the sum of the digits of a is given to be 567, then 3n=567 or n=189.

    Answer: n=189

  5. When we translate the parabola and hexagon horizontally, the distance between the x-intercepts of the parabola does not change.
    This means that we can position the hexagon and parabola in a horizontal position that is more convenient.
    Thus, we position the hexagon and parabola so that the y-axis passes through the midpoint of FE and so the hexagon and parabola are symmetric across the y-axis.
    Since FE=2, then the midpoint of FE is 1 unit from each of F and E.
    In other words, the coordinates of F and E are (1,0) and (1,0), respectively.
    Next, we drop a perpendicular from D to P on the x-axis.

    Since ABCDEF is a regular hexagon, then each of its interior angles measures 120. This is because the sum of the interior angles is 4180 or 720 and so each angle is 16720 which equals 120.
    Thus, DEP=180120=60, which means that DEP is a 30-60-90 triangle.
    Since DE=2, then EP=1 and DP=3.
    Since the coordinates of E are (1,0), then the coordinates of D are (1+1,0+3) or (2,3).
    Similarly, C is one unit to the left of D and 3 units above, which means that the coordinates of C are (1,23). This is because a right-angled triangle can be drawn with hypotenuse CD=2, legs parallel to the axes, and one angle equal to 60 by reflecting EPD through the horizontal line through D.
    Since the hexagon is symmetric across the y-axis, then the coordinates of A and B are (2,3) and (1,23), respectively.
    Suppose that the parabola has equation y=ax2+bx+c.
    Since the points (2,3), (2,3) and (1,23) lie on the parabola, then 3=4a2b+c3=4a+2b+c23=a+b+c Subtracting the second equation from the first, we obtain 4b=0 and so b=0. (We could also have determined this using a symmetry argument.)
    Substituting into the second and third equations, we obtain 4a+c=3 and a+c=23. Subtracting these equations, we obtain 3a=3 and so a=33.
    Substituting a=33 into a+c=23, we obtain c=23a=23+33=733.
    Therefore, the equation of the parabola is y=33x2+733.
    We find the x-intercepts of the parabola by setting y=0 to obtain 0=33x2+733 or x2=7 and so x=±7.
    Thus, the parabola crosses the x-axis at the points (7,0) and (7,0).
    The distance between these two points is 7(7)=27.

    Answer: 27

  6. Solution 1
    When 0<A<90 and 0<B<90, we know that tanA and tanB are defined with tanA>0 and tanB>0.
    For all real numbers x and y, we know that (xy)20 with equality exactly when x=y.
    Rearranging, we obtain x22xy+y20 and x2+y22xy with equality exactly when x=y.
    Setting x=2 and y=tanA, we obtain 4+tan2A2(2)(tanA).
    Thus, 4+tan2A4tanA with equality exactly when tanA=2.
    Setting x=5 and y=tanB, we obtain 5+tan2B2(5)(tanB).
    Thus, 5+tan2B25tanB with equality exactly when tanB=5.
    Since 4+tan2A4tanA and 5+tan2B25tanB, then (4+tan2A)(5+tan2B)(4tanA)(25tanB) since all quantities are positive.
    For equality to occur in this inequality, we need equality to occur in both of the component inequalities. (If one inequality is actually “>”, then the product becomes “>”.) In other words, (4+tan2A)(5+tan2B)=(4tanA)(25tanB) exactly when tanA=2 and tanB=5.
    We note that (4tanA)(25tanB)=85tanAtanB=645tanAtanB=320tanAtanB Therefore, (4+tan2A)(5+tan2B)=320tanAtanB exactly when tanA=2 and tanB=5.
    We now need to determine the value of cosAsinB.
    Since 0<A<90 and 0<B<90, then cosA,sinA,cosB,sinB are all positive.
    When tanA=2, we obtain sinAcosA=2 and so 1cos2AcosA=2 which gives 1cos2A=4cos2A or 5cos2A=1.
    Since cosA>0, then cosA=15.
    When tanB=5, we obtain sinBcosB=5 and so sinB1sin2B=5.
    This gives sin2B=55sin2B or 6sin2B=5.
    Since sinB>0, then sinB=56.
    Finally, this gives cosAsinB=1556=16.

    Solution 2
    Let X=tanA and Y=tanB.
    The given equation becomes successively: (4+X2)(5+Y2)=320XYX2Y2+5X2+4Y2+20=85XYX2Y285XY+5X2+4Y2+20=0X2Y22(25)XY+20+5X2+4Y245XY=0X2Y22(25)XY+(25)2+5X2+4Y245XY=0(XY25)2+5X245XY+4Y2=0(XY25)2+(5X)22(5X)(2Y)+(2Y)2=0(XY25)2+(5X2Y)2=0 For the sum of squares to be 0, each part must equal 0.
    Therefore, XY=25 and 5X=2Y.
    From the first equation, X(2Y)=45 and so X(5X)=45 which gives X2=4 or X=±2.
    Since 0<A<90, then X=tanA>0 and so X=tanA=2.
    Since 2Y=5X, then Y=tanB=5.
    We can then proceed as in Solution 1 to obtain cosAsinB=16.

    Answer: 16

PART B

    1. To find the x-intercept of the line with equation y=3x+6, we set y=0 to obtain 3x+6=0 or 3x=6 and so x=2.
      Therefore, the x-intercept is 2.

    2. Solution 1
      To find the equation of the line that is symmetric across the y-axis with the line with the equation y=3x+6, we reverse the sign of the slope and keep the same y-intercept to obtain y=3x+6.

      Solution 2
      The line with equation y=3x+6 has x-intercept 2 (from (a)) and y-intercept 6 (from the equation of the line).
      Since the letter A is symmetric about the y-axis, the right side of the letter A will lie along the line with x-intercept 2 (that is, (2)) and y-intercept 6.
      The equation of this line is thus of the form y=mx+6 for some m.
      Since it passes through the point (2,0), then 0=2m+6 and so 2m=6 or m=3.
      Thus, the equation of this line is y=3x+6.

    3. Since the x-intercepts of the lines that form the sides are 2 and 2, then the base of the triangle has length 2(2)=4.
      Since both lines pass through (0,6), then the height of the triangle is 6.
      Therefore, the area of the triangle is 1246=12.

    4. Solution 1
      We note first that c<6.
      Since the top vertex of the shaded triangle lies at y=6 and the base of the triangle lies along y=c, then the height of the shaded triangle is 6c.
      The bottom right vertex of the shaded triangle lies at the point of intersection of the line with equation y=3x+6 and the line with equation y=c.
      At this point, c=3x+6 and so 3x=6c and so x=6c3.
      By symmetry, the x-coordinate of the bottom left vertex is 6c3.
      Therefore, the base of the shaded triangle has length 6c3(6c3)=2(6c)3.
      Since the area of the shaded triangle is 49 of the area of the original triangle, then the area of the shaded triangle is 4912=163.
      Putting all of this together and using the area obtain in (c), we have 122(6c)3(6c)=16313(6c)2=163(6c)2=166c=±4 Since c<6, then 6c>0 and so 6c=4 which gives c=2.

      Solution 2
      Since the base of the shaded triangle is parallel to the base of the larger triangle, then their base angles are equal and so the triangles are similar.
      Since the area of the shaded triangle is 49 of the area of the larger triangle, then the side lengths of the shaded triangle are 49=23 of the corresponding side lengths of the larger triangle.
      Furthermore, the height of the shaded triangle is 23 that of the larger triangle.
      Since the height of the larger triangle is 6, then the height of the shaded triangle is 236 which equals 4.
      Since the top vertex of the shaded triangle lies at y=6 and the base of the triangle lies along y=c, then the height of the shaded triangle is 6c.
      Thus, 6c=4 and so c=2.

    1. Re-arranging and noting that x0, we obtain the following equivalent equations: 141x=161416=1x312212=1x112=1x and so x=12.

    2. Factoring, we obtain the following equivalent equations: abb+a1=4b(a1)+(a1)=4(a1)(b+1)=4 Since a and b are integers, then a1 and b+1 are integers. Therefore, b+1 and a1 are a divisor pair of 4.
      Since b1, then b+12.
      Since b+1>0 and 4>0 and (a1)(b+1)=4, then a1>0.
      This means that b+1 and a1 must form a positive divisor pair of 4 with b+12.
      There are two possibilities:

      • b+1=2 and a1=2; this gives (a,b)=(3,1).

      • b+1=4 and a1=1; this gives (a,b)=(2,3).

      Therefore, the pairs of positive integers that solve the equation are (3,1) and (2,3).

    3. Since y0 and z0, we multiply the given equation by 12yz and manipulate to obtain the following equivalent equations: 1y1z=11212z12y=yz0=yz+12y12z0=y(z+12)12z144=y(z+12)12z144144=y(z+12)12(z+12)144=(y12)(z+12) Since y and z are integers, then y12 and z+12 are integers.
      Since z>0, then z+12>0.
      Since 144<0 and z+12>0 and (y12)(z+12)=144, then y12<0.
      Since y is a positive integer, then 1y11. This makes y12>12.
      Since (y12)(z+12)=144 and y12<0, then y12 and z+12 are a divisor pair of 144 with 12<y12<0.
      We make a table to determine the possible values:

      y12 z+12 y z
      1 144 11 132
      2 72 10 60
      3 48 9 36
      4 36 8 24
      6 24 6 12
      8 18 4 6
      9 16 3 4
      Therefore, there are 7 pairs of positive integers (y,z) with 1y1z=112, namely (y,z)=(11,132),(10,60),(9,36),(8,24),(6,12),(4,6),(3,4)

    4. Since r0 and s0 and p2, we multiply the given equation by p2rs and manipulate to obtain the following equivalent equations: 1r1s=1p2p2sp2r=rs0=rs+p2rp2s0=r(s+p2)p2sp4=r(s+p2)p2sp4p4=r(s+p2)p2(s+p2)p4=(rp2)(s+p2) Since r and s are integers, then rp2 and s+p2 are integers.
      Since s>0, then s+p2>0.
      Since p4<0 and s+p2>0 and (rp2)(s+p2)=p4, then rp2<0.
      Since r is a positive integer, then 1r<p2.
      Since (rp2)(s+p2)=p4 and rp2<0, then rp2 and s+p2 are a divisor pair of p4 with p2<rp2<0.
      We consider the possibilities that rp2=1 and rp2=p.
      In these cases, s+p2=p4 and s+p2=p3, respectively.
      These give the pairs (r,s)=(p21,p4p2) and (r,s)=(p2p,p3p2).
      Since p is prime, then p2, which means that these pairs are distinct and that each component is a positive integer.
      Therefore, there are at least two pairs (r,s) of positive integers for which 1r1s=1p2.

    1. If AB and BA both occur as substrings, they either overlap as ABA or BAB or they do not overlap.
      If they do not overlap, then such a string must be ABBA or BAAB. Since the strings have length 4, there are no other possibilities.
      A 4 character string including the substring ABA can have this substring start in either the 1st or 2nd position, which gives the possibilities: ABAAABABABACAABABABACABA A 4 character string including the substring BAB can have this substring start in either the 1st or 2nd position, which gives the possibilities: BABABABBBABCABABBBABCBAB Removing duplicate strings, we obtain the list ABBABAABABAAABABABACAABA BABACABABABBBABCBBABCBAB (There are 12 such strings.)

    2. Solution 1
      Since there are 3 choices for each character in such a string, then there are 37=2187 strings of length 7 with characters from the set {A,B,C} and no further restrictions.
      We count the number of such strings that do not include the substring CC and subtract this total from 2187 to obtain our desired answer.
      Let tn be the number of strings of length n with characters from {A,B,C} that do not contain the substring CC.
      Let an be the number of strings of length n with characters from {A,B,C} that do not contain the substring CC and whose nth character (i.e. rightmost character) is A.
      Let bn be the number of strings of length n with characters from {A,B,C} that do not contain the substring CC and whose nth character (i.e. rightmost character) is B.
      Let cn be the number of strings of length n with characters from {A,B,C} that do not contain the substring CC and whose nth character (i.e. rightmost character) is C.
      Note that tn=an+bn+cn since each string ends in A, B or C.
      Also, a1=b1=c1=1 (the strings A, B and C), which means that t1=1+1+1=3.
      Suppose that n2.
      Consider a string of length n that ends with A and does not contain CC. The second last character can be any of A, B or C. This means that the first n1 characters are a string of length n1 that ends in A, B or C, and so an=an1+bn1+cn1.
      Consider a string of length n that ends with B and does not contain CC. The second last character can be any of A, B or C. This means that the first n1 characters are a string of length n1 that ends in A, B or C, and so bn=an1+bn1+cn1.
      Consider a string of length n that ends with C and does not contain CC. The second last character can be either A or B. (It cannot be C otherwise, there would be an occurrence of the substring CC.) This means that the first n1 characters are a string of length n1 that ends in A or B, and so cn=an1+bn1.
      Therefore, a2=a1+b1+c1=1+1+1=3b2=a1+b1+c1=1+1+1=3c2=a1+b1=1+1=2 and a3=a2+b2+c2=3+3+2=8b3=a2+b2+c2=3+3+2=8c3=a2+b2=3+3=6 Continuing in this way, we can build a table:

      n an bn cn
      1 1 1 1
      2 3 3 2
      3 8 8 6
      4 22 22 16
      5 60 60 44
      6 164 164 120
      7 448 448 328
      Therefore, t7=a7+b7+c7=448+448+328=1224.
      Finally, we obtain that the total number of strings of length 7 that do include CC is 21871224=963.

      Solution 2
      Since there are 3 choices for each character in such a string, then there are 37=2187 strings of length 7 with characters from the set {A,B,C} and no further restrictions.
      We count the number of such strings that do not include the substring CC and subtract this total from 2187 to obtain our desired answer.
      A string of length 7 that does not include the substring CC can include 0, 1, 2, 3 or 4 C’s. (If a string of length 7 includes 5 or more C’s, two of them must be adjacent.)
      Case 1: 0 C’s
      There are 2 choices (A or B) for each of the 7 characters and so there are 27=128 strings in this case.
      Case 2: 1 C
      There are 7 positions in which the C can go and 2 choices for each of the remaining 6 characters.
      In total, there are 726=448 strings in this case.
      Case 3: 2 C’s
      There are 15 pairs of positions in which the two C’s can go and not be adjacent: (1,3),(1,4),(1,5),(1,6),(1,7),(2,4),(2,5),(2,6),(2,7) (3,5),(3,6),(3,7),(4,6),(4,7),(5,7) (We use, for example, the notation “(1,3)” to mean C in each of the 1st and 3rd positions.) To find these we start with (1,3), move the second C along position by position until the end, then start again with (2,4), and so on.
      There are 2 choices for each of the remaining 5 characters.
      In total, there are 1525=480 strings in this case.
      Case 4: 3 C’s
      There are 10 triplets of positions in which the three C’s can go and not be adjacent: (1,3,5),(1,3,6),(1,3,7),(1,4,6),(1,4,7),(1,5,7) (2,4,6),(2,4,7),(2,5,7),(3,5,7) To find these we start with (1,3,5), move the third C along position by position until the end, then start again with (1,4,6), and so on, eventually moving the first C to positions 2 and 3.
      There are 2 choices for each of the remaining 4 characters.
      In total, there are 1024=160 strings in this case.
      Case 5: 4 C’s
      The C’s must go in positions (1,3,5,7) in order for them not to be adjacent.
      There are 2 choices for each of the remaining 3 characters.
      In total, there are 123=8 strings in this case.
      Combining totals, there are 128+448+480+160+8=1224 strings that do not include the substring CC.
      Finally, we obtain that the total number of strings of length 7 that do include CC is 21871224=963.

    3. First, we calculate f(2097).
      Every string counted by f(2097) has a first occurrence of the substring CC.
      Starting from the left, this first occurrence can begin in any of the positions 1,2,3,,2096. (It cannot begin in the last position.)
      Suppose that k is a positive integer with 1k2096.
      We determine the number of strings counted by f(2097) whose first occurrence of CC begins in position k.
      In such a string, positions k and k+1 are CC.
      This leaves k1 positions to the left of the CC and 2097(k+1)=2096k positions to the right.
      Based on the restrictions for the string, there are no restrictions for the characters in each of the 2096k positions to the right of CC.
      Therefore, there are 3 choices for each of these positions and so 32096k ways to fill these 2096k positions.
      For the k1 positions to the left of CC, there are two restrictions: no substring AB or BA can occur (from the given restrictions) and no substring CC can occur (since the CC in positions k and k+1 is the first such substring).
      Since position k is C, then position k1 cannot be C and so must be A or B. Thus, there are 2 choices for position k1.
      If position k1 were A, then position k2 could be A or C, but not B (as this would create a substring BA). In this case, there are 2 choices for position k2.
      If position k1 were B, then position k2 could be B or C, but not A (as this would create a substring AB). In this case, there are 2 choices for position k2.
      In general, if we consider the character at position j for some j with 2jk, then

      • if position j is A, position j1 must be A or C,

      • if position j is B, position j1 must be B or C, and

      • if position j is C, position j1 must be A or B.

      In other words, for each position from k1 back to 1, there are 2 choices for the character, giving 2k1 choices overall for these characters.
      Note that this formula is valid when k=1 since this gives “1 choice” for the non-existent preceding characters which does not affect the formula that follows.
      Therefore, there are 2k132096k strings that obey the given restrictions whose first occurrence of CC begins in position k.
      Adding from k=1 to k=2096, we obtain f(2097)=2032095+2132094+2232093++2209332+2209431+2209530=32095(1+23+2232++2209332093+2209432094+2209532095)=32095(1+(23)1+(23)2++(23)2093+(23)2094+(23)2095)=32095(1(1(23)2096)123)=32095(1(23)2096)13)=32096(1(23)2096)=3209622096 To show that f(2097)=3209622096 is divisible by 97, we use the fact that, for all real numbers x and y, xnyn=(xy)(xn1+xn2y+xn3y2++x2yn3+xyn2+yn1) Setting x=38 and y=28 and n=262, we get f(2097)=(38)262(28)262=x262y262=(xy)(x261+x260y++xy260+y261)=(3828)((38)261+(38)26028++38(28)260+(28)261)=(34+24)(3424)((38)261+(38)26028++38(28)260+(28)261) factoring a difference of squares in the last step.
      Since 34+24=81+16=97 and the remaining two factors are integers, then f(2097) is a multiple of 97, as required.