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2021 Canadian Intermediate
Mathematics Contest
Solutions

Wednesday, November 17, 2021
(in North America and South America)

Thursday, November 18, 2021
(outside of North American and South America)

©2021 University of Waterloo


PART A

  1. The area of a rectangle is equal to its length multiplied by its width, and so width equals area divided by length.
    The width of Rectangle A is equal to 36 cm26 cm=6 cm.

    The width of Rectangle B is equal to 36 cm212 cm=3 cm.

    The width of Rectangle C is equal to 36 cm29 cm=4 cm.

    The smallest of these widths is 3 cm, and so x=3.
    (Note that since the three rectangles have the same area, the smallest width would be the one with the largest length.)

    Answer: x=3

  2. The prime numbers that are greater than 10 and less than 20 are 11, 13, 17, 19.
    (Since 12=2×6 and 14=2×7 and 15=3×5 and 16=2×8 and 18=2×9, none of these integers is a prime number.)
    The sum of these prime numbers is 11+13+17+19=(11+19)+(13+17)=30+30=60

    Answer: 60

  3. From the 22nd floor to the nth floor, Taya and Jenna each go up n22 floors.
    Since Taya goes from each floor to the next in 15 seconds, this takes her 15×(n22) seconds.
    Since Jenna waits for 2 minutes, she waits for 120 seconds. Since Jenna goes from each floor to the next in 3 seconds, it takes a total of 120+3×(n22) seconds to reach the nth floor.
    Since their travel times are equal, then 15×(n22)=120+3×(n22)15×(n22)3×(n2)=12012×(n22)=120n22=10 and so n=32.

    Answer: n=32

  4. Since DQC is a straight angle, RQP=180RQDPQC=180wx.
    Since BPC is a straight angle, RPQ=180RPBQPC=180zy.
    The sum of the measures of the angles in PQR is 180, and so RQP+RPQ+PRQ=180(180wx)+(180zy)+30=180390wxyz=180210=w+x+y+z and so w+x+y+z=210.

    Answer: 210

  5. Using the given rules, the first 7 numbers in the list are 3,4,53,(5/3)+14=8/34=23,(2/3)+15/3=5/35/3=1,1+12/3=22/3=3,3+11=4 Since the 6th number equals the 1st number, the 7th number equals the 2nd number, and each number in the list depends on the previous two numbers, then the 8th number equals the 3rd number, which will mean that the 9th number will equal the 4th number, and so on.
    In other words, the list will repeat every 5 numbers since each group of 5 numbers is produced in exactly the same way as the previous 5 numbers.

    The sum of the first 5 numbers is 3+4+53+23+1=8+73=1013.
    Since 2021÷1013195.6, then 195 groups of 5 numbers will still have a sum less than 2021.
    In fact, since 195×1013=1950+65=2015 and 195×5=975, then the sum of the first 975 numbers in the list is 2015.
    We make a chart of the next terms and the sum to that point until we reach a sum that is greater than 2021 and is an odd integer:

    Term # 976 977 978 979 980 981 982
    Term 3 4 53 23 1 3 4
    Sum to this point 2018 2022 202323 202413 202513 202813 203213
    Term # 983 984 985 986 987 988 989
    Term 53 23 1 3 4 53 23
    Sum to this point 2034 203423 203523 203823 204223 204413 2045

    Therefore, the smallest positive integer N for which the sum of the first N numbers is equal to an odd integer greater than 2021 is N=989 when the sum is 2045.

    Answer: 989

  6. Solution 1

    Suppose that Dragomir removes the socks one at a time. There are 12 socks that he can remove 1st, then 11 socks 2nd, then 10 socks 3rd, and finally 9 socks 4th.
    This means that there are 12×11×10×9 ways in which he can remove 4 socks one at a time.
    Suppose that there is exactly one pair among the 4 socks removed. There are 6 different possibilities for how this pair was formed: 1st and 2nd socks match, 1st and 3rd match, 1st and 4th match, 2nd and 3rd match, 2nd and 4th match, 3rd and 4th match.
    We determine the number of ways in which each of these possibilities can arise.

    Case 1: Matching pair 1st and 2nd
    There are 12 socks that he can pick 1st (no restrictions).
    There is 1 sock that he can pick 2nd (pair of 1st).
    There are 10 socks that he can pick 3rd (any of the 10 remaining socks).
    There are 8 socks that he can pick 4th (any of the 9 remaining socks other than pair of 3rd).
    In this case, there are 12×1×10×8 ways.

    Case 2: Matching pair 1st and 3rd
    There are 12 socks that he can pick 1st (no restrictions).
    There are 10 socks that he can pick 2nd (all but pair of 1st).
    There is 1 sock that he can pick 3rd (pair of 1st).
    There are 8 socks that he can pick 4th (any other than pair of 2nd).
    In this case, there are 12×10×1×8 ways.

    Case 3: Matching pair 1st and 4th
    There are 12 socks that he can pick 1st (no restrictions).
    There are 10 socks that he can pick 2nd (all but pair of 1st).
    There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2nd).
    There is 1 sock that he can pick 4th (pair of 1st).
    In this case, there are 12×10×8×1 ways.

    Case 4: Matching pair 2nd and 3rd
    There are 12 socks that he can pick 1st (no restrictions).
    There are 10 socks that he can pick 2nd (all but pair of 1st).
    There is 1 sock that he can pick 3rd (pair of 2nd).
    There are 8 socks that he can pick 4th (all but pair of 1st).
    In this case, there are 12×10×1×8 ways.

    Case 5: Matching pair 2nd and 4th
    There are 12 socks that he can pick 1st (no restrictions).
    There are 10 socks that he can pick 2nd (all but pair of 1st).
    There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2nd).
    There is 1 sock that he can pick 4th (pair of 2nd).
    In this case, there are 12×10×8×1 ways.

    Case 6: Matching pair 3rd and 4th
    There are 12 socks that he can pick 1st (no restrictions).
    There are 10 socks that he can pick 2nd (all but pair of 1st).
    There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2nd).
    There is 1 sock that he can pick 4th (pair of 3rd).
    In this case, there are 12×10×8×1 ways.

    In each of the six cases, there are 12×10×8×1 ways in which the socks can be chosen.
    Therefore, the overall probability is 6×12×10×8×112×11×10×9=6×811×9=2×811×3=1633.

    Solution 2

    Since there are 12 socks in the drawer, there are (124)=12×11×10×94×3×2×1=11×5×9=495 ways of choosing 4 socks.
    Next, we count the number of ways in which the 4 chosen socks can include exactly 1 matching pair.
    There are 6 possible matching pairs that can be chosen.
    The remaining 2 socks must come from 2 different pairs of the remaining 5 pairs.
    If we label the remaining pairs A, B, C, D, E, there are 10 possible two-pair combinations from which these 2 socks can be chosen: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. (We could also use a binomial coefficient to calculate this total.)
    For each of the 2 pairs from which these 2 socks will be chosen, there are in fact 2 socks to choose.
    Thus, the total number of ways in which the 4 socks can be chosen that satisfy the given conditions is 6×10×2×2=240.

    This means that the probability is 240495=16×1533×15=1633.

    Solution 3

    Since there are 12 socks in the drawer, there are (124)=12×11×10×94×3×2×1=11×5×9=495 ways of choosing 4 socks.
    These 4 socks can include 0 matching pairs, 1 matching pair, or 2 matching pairs.
    We count the number of ways in which the 4 chosen socks can include 0 matching pairs or 2 matching pairs and subtract this from the total number of ways of choosing 4 socks to obtain the number of ways of choosing 1 matching pair.

    To choose 0 matching pairs, we choose 4 of the 6 pairs and choose 1 of the 2 socks in each pair.
    There are (64)=6×5×4×34×3×2×1=15 ways of choosing 4 of 6 pairs, and 24=16 ways of choosing 1 of the 2 socks in each of 4 pairs.
    Thus, there are 15×16=240 ways of choosing exactly 0 matching pairs.

    To choose 2 matching pairs, we choose 2 of the 6 pairs and choose both socks in each pair.
    There are (62)=6×52×1=15 ways of choosing 2 of 6 pairs, and 1 way to choose both socks in each pair.
    Thus, there are 15×1=15 ways of choosing exactly 2 matching pairs.

    This means that there are 49524015=240 ways of choosing exactly 1 matching pair.

    This means that the probability is 240495=16×1533×15=1633.
    Answer: 1633

PART B

    1. Since Karol uses 28 cups of Gluze, and the recipe calls for 4 cups of Gluze, he must have repeated the recipe 28÷4=7 times.
      Since the recipe calls for 3 cups of Blurpos, and the recipe was repeated 7 times, he must have used 7×3=21 cups of Blurpos.

    2. Suppose that Karol uses 48 cups of Gluze.
      Since 48÷4=12, he uses his recipe 12 times and so uses 12×3=36 cups of Blurpos.
      Suppose instead that Karol uses 48 cups of Blurpos.
      Since 48÷3=16, he uses his recipe 16 times and so uses 16×4=64 cups of Gluze.
      Therefore, the two possible values of N are 36 and 64.

    3. Karol starts with 64 cups of Gluze and 42 cups of Blurpos. One of these ingredients will be the “limiting ingredient”; that is, when Karol repeats his recipe, he will probably run out of one of the ingredients before the other.
      If Karol used all 64 cups of Gluze, then he must have used his recipe 64÷4=16 times.
      If Karol used all 42 cups of Blurpos, then he must have used his recipe 42÷3=14 times.
      Therefore, he can make the recipe at most 14 times since after the 14th time, he would run out of Blurpos.
      Each time Karol uses his recipe, he makes 60 Zippies.
      When he uses his recipe 14 times, he makes 14×60=840 Zippies.

    4. We start by assuming that Karol makes 1 recipe worth of Zippies.
      In other words, suppose that Karol makes 60 Zippies using 4 cups of Gluze and 3 cups of Blurpos.
      He sells each Zippie for $0.50, which earns him 60×$0.50=$30.00.
      His profit on each Zippie is $0.30, which means that his total profit is 60×$0.30=$18.00.
      Since he earns $30.00 and his profit is $18.00, his costs must be $30.00$18.00=$12.00.
      Since Karol pays $1.80 for each cup of Gluze, he pays a total of 4×$1.80=$7.20 for Gluze.
      This means that he pays $12.00$7.20=$4.80 for Blurpos.
      Since he uses 3 cups of Blurpos, Karol pays $4.80÷3=$1.60 per cup of Blurpos.

    1. Since ABCD is a rectangle and AB=3, then DC=3.
      Thus, DE=DC+CE=3+6=9.
      Now, the area of trapezoid ABED is 48.
      Since ABCD is a rectangle, then AD is perpendicular to DE which means that AD is a height of trapezoid ABED.
      The area of trapezoid ABED equals 12(AB+DE)×AD.
      (We could instead note that trapezoid ABED is made up of a rectangle with base 3 and an unknown height, and a right-angled triangle with base 6 and the same unknown height.)
      Therefore, 12(3+9)×AD=48 and so 6×AD=48 which gives AD=8.

      Since ABCD is a rectangle, then BC=AD=8.
      Finally, by the Pythagorean Theorem, BE=BC2+CE2=82+62=100=10.

    2. Draw a perpendicular from Q to F on PS.

      Quadrilateral FQTS has three right angles and so it must be a rectangle.
      Therefore, FS=QT. Since QT is a radius, then FS=QT=16.
      Since PS is a radius, then PS=25.
      Therefore, PF=PSFS=2516=9.
      We note that PQ=PX+XQ=25+16=41 (they are both radii).
      By the Pythagorean Theorem in PFQ, which is right-angled at F, FQ=PQ2PF2=41292=168181=40 Since FQTS is a rectangle, ST=FQ=40.
      Finally, the area of trapezoid PQTS equals 12(PS+QT)×ST=12(25+16)×40=41×20=820

    3. Suppose that the centre of the circle of radius r is O and that this circle touches the line at point V.
      Join P to O and Q to O.
      Join V to O. Since the circle is tangent to at V, OV is perpendicular to . Also, draw perpendiculars from O to G on PS and from O to H on QT.

      As in (b), GOVS is a rectangle with GS=OV=r and SV=GO.
      Since PS=25, then PG=PSGS=25r.
      Since PO joins the centres of two tangent circles, then the length of PO equals the sum of the radii of the two circles. In other words, PO=25+r.
      By the Pythagorean Theorem in POG, GO2=PO2PG2=(25+r)2(25r)2=(625+50r+r2)(62550r+r2)=100r Since GO>0, then GO=100r=100×r=10r.
      Thus, SV=GO=10r.
      Using a similar analysis, HO2=QO2QH2=(16+r)2(16r)2=(256+32r+r2)(25632r+r2)=64r Since HO>0, then HO=64r=64×r=8r.
      Thus, TV=HO=8r.
      From (b), ST=40 and so SV+TV=40.
      Therefore, 10r+8r=40 which gives 18r=40 and so r=4018=209.
      Finally, r=(209)2=40081.

    1. The key move in Beryl’s strategy is to choose 2 on turn #2 (her first turn).
      On turn #1, Alphonse must choose 1, for a running total of 1.
      On turn #2, Beryl chooses 2, for a running total of 3.
      On turn #3, Alphonse can choose 1, 2 or 3, for a running total of 4, 5 or 6, respectively.
      On turn #4, Beryl can choose 1, 2, 3, or 4, and so chooses 4, 3 or 2 (corresponding to Alphonse’s choices of 1, 2 or 3, respectively) to achieve a running total of 8 in each case.
      Therefore, if Beryl chooses 2 on turn #2 and then chooses appropriately on turn #4, she is guaranteed to win.

    2. Alphonse has a winning strategy when N=17.
      To show this, we use the fact that 1+4+5+7=17.

      On turn #1, Alphonse must choose 1.

      On turn #2, Beryl can choose 1 or 2; on turn #3, Alphonse can choose 1, 2 or 3.
      No matter what Beryl chooses on turn #2, Alphonse can choose a number to make the sum of their numbers on those two turns equal to 4.
      (If Beryl chooses 1, Alphonse chooses 3. If Beryl chooses 2, Alphonse chooses 2.)
      Thus, Alphonse can ensure that the running total after 3 turns is 1+4=5.

      On turn #4, Beryl can choose 1, 2, 3, or 4; on turn #5, Alphonse can choose 1, 2, 3, 4, or 5.
      No matter what Beryl chooses on turn #4, Alphonse can choose a number to make the sum of their numbers on those two turns equal to 5.
      (If Beryl chooses 1, 2, 3, or 4, Alphonse chooses 4, 3, 2, or 1, respectively.)
      Thus, Alphonse can ensure that the running total after 5 turns is 5+5=10.

      On turn #6, Beryl can choose 1, 2, 3, 4, 5, or 6; on turn #7, Alphonse can choose 1, 2, 3, 4, 5, 6, or 7.
      No matter what Beryl chooses on turn #6, Alphonse can choose a number to make the sum of their numbers on those two turns equal to 7.
      (If Beryl chooses x with 1x6, Alphonse chooses 7x. Note that 17x6.)
      Thus, Alphonse can ensure that the running total after 7 turns is 10+7=17.
      Note as well that, following this method, the largest that Beryl can make the running total after turn #6 is 10+6=16 and so she cannot win “early” and thus break Alphonse’s strategy.

      Therefore, Alphonse has a winning strategy when N=17.

    3. When N=1, Alphonse wins automatically.
      When N=2 or N=3, Beryl has a winning strategy by choosing 1 or 2, respectively, on her first turn.
      When N=4 or N=5, Alphonse has a winning strategy by choosing 2 or 3 if Beryl chooses 1 (making the previous running total 2) or by choosing 1 or 2 if Beryl chooses 2 (making the previous running total 3).
      When N=6, Beryl has a winning strategy by choosing 1 on turn #2, which forces Alphonse to make the running total after turn #3 equal to 3, 4 or 5, after which Beryl can win by choosing 3, 2 or 1 on turn #4.
      When N=7, Beryl has a winning strategy by choosing 1 on turn #2, which forces Alphonse to make the running total after turn #3 equal to 3, 4 or 5, after which Beryl can win by choosing 4, 3 or 2 on turn #4.
      We saw what happens when N=8 in (a).
      We can continue these types of argument to show that Alphonse has a winning strategy when N=9, N=10, and N=11, and Beryl has a winning strategy when N=12, N=13, N=14, and N=15.
      So far, we have:

      Winning strategy for Alphonse Winning strategy for Beryl
      N=1 N=2,3
      N=4,5 N=6,7,8
      N=9,10,11 N=12,13,14,15

      These small values of N suggest that Alphonse has a winning strategy for k values of N starting with N=k2 (that is, for N=k2 up to and including N=k2+k1) and Beryl has a winning strategy for k+1 values of N leading up to N=(k+1)21 (that is, for N=k2+k up to and including N=k2+2k).

      We note that 452=2025 and 462=2116.
      Based on our hypotheses, our guess is that Beryl has a winning strategy for N=1980,1981,1982,,2023,2024 while Alphonse has a winning strategy for N=2025,2026,2027,,2068,2069 and Beryl has a winning strategy for N=2070,2071,2072,,2114,2115 If this is true, the smallest positive integer m with m>2021 for which Alphonse has a winning strategy when N=m and Beryl has a winning strategy when N=m+1 is m=2069.
      In order to prove that this is true, we need to show that none of m=2022 through m=2068 satisfy the given conditions and that m=2069 does. To do this, we prove that Beryl has a winning strategy for N=2022 and N=2023, Alphonse has a winning strategy for N=2025 through N=2069, and Beryl has a winning strategy for N=2070. (Can you explain why we do not need to consider N=2024?)

      Consider N=2025.
      Note that 1+3+5++85+87+89=45+(1+89)+(3+87)++(41+49)+(43+47)=45+2290=2025 The following table summarizes Alphonse’s strategy, based on Beryl’s moves:

      Turn(s) Details Condition
      #1 A must choose 1
      #2 and #3 B can choose 1 or 2
      A chooses 2 or 1, respectively
      Sum over two turns is 3
      #4 and #5 B can choose 1, 2, 3, or 4
      A chooses 4, 3, 2, or 1, respectively
      Sum over two turns is 5
      #(2q) and #(2q+1)
      (1q44)
      B can choose 1,2,3,,2q1, or 2q
      A chooses 2q,2q1,,2,1, respectively
      Sum over two turns is 2q+1

      Therefore, after 89 turns, using this strategy, Alphonse can ensure that the running total is 1+3+5++87+89=2025 and so Alphonse has a winning strategy.

      Consider the case where N equals one of 2026,2027,,2068,2069.
      Let k=N2025. Note that 1k44.
      Alphonse’s winning strategy for these values of N is to reproduce the strategy above for N=2025 with the change that for k of his turns from turn #3, turn #5, , turn #87, turn #89 (there are 44 turns in this list), he makes the combined sum 1 larger than the corresponding combined sum in strategy for N=2025.
      By doing this, after turn #89, Alphonse can ensure that the running total is 2025+k=N, and so Alphonse has a winning strategy.
      (Note that in the N=2025 case, by choosing one of 2q,2q1,,2,1 on turn 2q+1, Alphonse could ensure that the sum of the numbers chosen on turns #(2q) and #(2q+1) is 2q+1. Since Alphonse can choose up to 2q+1 on turn 2q+1, then he can make his choice one larger on turn 2q+1 than he did in the N=2025 case and so make the sum of the numbers on turns 2q and 2q+1 equal to (2q+1)+1.)

      Consider N=2070.
      Note that 2+4+6++86+88+90=90+(2+88)+(4+86)+(42+48)+(44+46)=90+2290=2070 The following table summarizes Beryl’s strategy, based on Alphonse’s moves:

      Turn(s) Details Condition
      #1 and #2 A must choose 1
      B chooses 1
      Sum over two turns is 2
      #3 and #4 A can choose 1, 2 or 3
      B chooses 3, 2 or 1, respectively
      Sum over two turns is 4
      #5 and #6 A can choose 1, 2, 3, 4, or 5
      B chooses 5, 4, 3, 2, or 1, respectively
      Sum over two turns is 6
      #(2q1) and #(2q)
      (1q45)
      A can choose 1,2,3,,2q2, or 2q1
      B chooses 2q1,2q2,,2,1, respectively
      Sum over two turns is 2q

      Therefore, after 90 turns, using this strategy, Beryl can ensure that the running total is 2+4+6++86+88+90=2070 and so Beryl has a winning strategy.

      Consider N=2022 and N=2023.
      Note that 2+4+7+9++87+89=6+(7+89)+(9+87)++(47+49)=6+21×96=2022 and 2+5+7+9++87+89=7+(7+89)+(9+87)++(47+49)=7+21×96=2023 (The first sum starts with 2+4 and continues with the sum of all of the odd numbers from 7 to 89, inclusive. The second sum starts with 2 and continues with the sum of all of the odd numbers from 5 to 89, inclusive.)
      By following the idea of the cases above, in particular making the sum of the numbers on turn #1 and turn #2 equal to 2, the sum of the numbers on turn #3 and turn #4 equal to 4 or 5, and the sum of the numbers on the pairs of turns #5 and #6 through #87 and #88 equal to 7,9,,87,89, respectively, Beryl can guarantee that the running total after turn #88 is 2022 or 2023. (For example, on turns #5 and #6, if Alphonse chooses 1, 2, 3, 4, or 5 on turn #5, Beryl then chooses 6, 5, 4, 3, or 2, respectively, on turn #6.)

      In summary, we have shown that Beryl has a winning strategy for N=2022 and N=2023, Alphonse has a winning strategy for N=2025 through N=2069, and Beryl has a winning strategy for N=2070, which shows that the value of m in question is m=2069.