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2023 Canadian Team Mathematics Contest
Solutions

April 2023

© 2023 University of Waterloo

Individual Problems

  1. Working backwards, since Jin had 5 chocolates left after eating half of them, she had 2×5=10 chocolates after giving chocolates to Brian.
    Since Jin gave 8 chocolates to Brian and had 10 chocolates after doing this, Jin was given 10+8=18 chocolates by Ingrid.
    Ingrid gave one third of her chocolates to Jin, which means Ingrid started with n=3×18=54 chocolates.

    Answer: 54

  2. Since 20% of 30 is 0.2×30=6, we want to find k so that k% of 25 equals 6.
    This means k100×25=6 or k4=6 and so k=24.

    Answer: 24

  3. Note that 2023=33×60+43, and since there are 60 minutes in an hour, this means that 2023 minutes is equal to 33 hours and 43 minutes.
    For 33 hours and 43 minutes to pass, 24 hours will pass and then 9 hours and 43 minutes will pass.
    In 24 hours it will be 1:00 a.m. again.
    In 9 hours and 43 minutes after 1:00 a.m., it will be 10:43 a.m.

    Answer: 10:43 a.m.

  4. As soon as two lockers are painted blue in the top row, the other two lockers in the top row must be painted red.
    Once the top row is painted, the colours of the lockers in the bottom row are determined.
    If the lockers in the top row are numbered 1, 2, 3, and 4, then there are six possibilities for the two blue lockers.
    They are 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, and 3 and 4.

    Answer: 6

  5. Let E be on CD such that BE is perpendicular to CD as shown.

    Since AB and DE are parallel and DAB and DEB are right angles, ADE and ABE are right angles as well, and so ABED is a rectangle.
    Using AB=4 and the fact that ABED is a rectangle, we get DE=4.
    Using CD=CE+DE, DE=4, and CD=6, we get CE=64=2.
    In BEC, BEC=90° and BCE=BCD=45°, and so EBC=180°90°45°=45°.
    This means BEC is isosceles, and so BE=CE=2.
    By the Pythagorean theorem applied to right-angled BED, we have DE2+BE2=BD2, and so BD2=42+22=20.
    Since BD>0, BD=20=25.

    Answer: 25

  6. Let t be the amount of time in hours that the train will take to make the trip if it is on time and let d be the distance in kilometres between City A and City B.
    If the train is 24 minutes late, then it is 2460 hours late, which means it takes t+2460=t+25 hours to reach City B.
    Similarly, if the train is 32 minutes early, then it takes t3260=t815 hours to reach City B.
    From the given assumptions, this means 80=dt+25 and 90=dt815.

    Cross multiplying these equations, we get 80(t+25)=d and 90(t815)=d.
    Equating the values of d and solving for t, we get 80(t+25)=90(t815)80t+32=90t4880=10t8=t Substituting t=8 into d=80(t+25) gives d=80(8+25)=672.
    The speed in km/h that the train should travel in order to arrive on time is 6728=84.

    Answer: 84

  7. We are given that tanBAC=17, but BAC=BAD and BAD has a right angle at D, so tanBAD=BDAD=17. Since BD=h, AD=7BD=7h.
    Using that tanBCD=tanBCA=1, we get CD=h by similar reasoning. Therefore, AC=AD+CD=7h+h=8h.

    Applying the Pythagorean theorem to ABD, we get that AB=(7h)2+h2=50h2.
    Applying the Pythagorean theorem to BCD, we get that BC=h2+h2=2h2.
    The perimeter of ABC is AB+BC+AC=50h2+2h2+8h=2×52×h2+2h2+8h(since h>0)=5h2+h2+8h=h(8+62) Since the perimeter of ABC is given to be 24+182=3(8+62), we conclude that h=3.

    Answer: 3

  8. Suppose ABCDE is a Tim number. That is, A, B, C, D, and E are the digits of the number and since it has five digits, A0.
    Since ABCDE is a multiple of 15, it must be a multiple of both 3 and 5.
    Since ABCDE is a multiple of 5, either E=0 or E=5.
    Since ABCDE is a multiple of 3, A+B+C+D+E is a multiple of 3.
    We are also given that C=3 and D=A+B+C=A+B+3.
    Combining C=3 and D=A+B+3 with the fact that A+B+C+D+E is a multiple of 3, we have that A+B+3+A+B+3+E=2A+2B+E+6 is a multiple of 3.

    Since 6 is a multiple of 3, this means 2A+2B+E is a multiple of 3.
    We are looking for triples (A,B,E) with the property that E=0 or E=5 and 2A+2B+E is a multiple of 3. However, we must keep in mind that A+B+3=D is a digit, which means A+B+39 or A+B6.
    We will consider two cases: E=0 and E=5. We note that once we choose values for A and B, the value of D will be determined.

    E=0: In this case, we need 2A+2B to be a multiple of 3, so A+B is a multiple of 3.
    Combined with the requirement that A+B6, this means A+B=3 or A+B=6. We also need to remember that A0, which means (A,B) is one of (1,2), (2,1), (3,0), (1,5), (2,4), (3,3), (4,2), (5,1), and (6,0) for a total of nine possibilities.

    E=5: In this case, we need 2A+2B+5 to be a multiple of 3. For the same reason as in the previous case, A+B6.
    If A+B=1, then 2A+2B+5=7 which is not a multiple of 3.
    If A+B=2, then 2A+2B+5=9 which is a multiple of 3.
    If A+B=3, then 2A+2B+5=11 which is not a multiple of 3.
    If A+B=4, then 2A+2B+5=13 which is not a multiple of 3.
    If A+B=5, then 2A+2B+5=15 which is a multiple of 3.
    If A+B=6, then 2A+2B+5=17 which is not a multiple of 3.

    Therefore, we must have that A+B=2 or A+B=5. Keeping in mind that A0, we get that (A,B) can be (1,1), (2,0), (1,4), (2,3), (3,2), (4,1), or (5,0) for a total of seven possibilities.
    This gives a total of 9+7=16 Tim numbers, which are listed below. One can check that all 16 of these integers are indeed Tim numbers. 12360,15390,21360,24390,30360,33390,42390,51390,60390,11355,14385,20355,23385,32385,41385,50385

    Answer: 16

  9. Solution 1

    Multiplying 4x+7y+z=11 by 2 gives 8x+14y+2z=22 and multiplying 3x+y+5z=15 by 3 gives 9x+3y+15z=45.
    Adding 8x+14y+2z=22 and 9x+3y+15z=45 gives 17x+17y+17z=67, and after dividing through by 17, we get x+y+z=6717.
    The fraction 6717 is in lowest terms, which means p=67 and q=17, so pq=6717=50.

    Solution 2

    We will try to solve the system of equations 4x+7y+z=113x+y+5z=15x+y+z=pq for x, y, and z in terms of p and q and see what happens. For this particular system, it is impossible to express x, y, and z in terms of only p and q. However, a relationship between p and q will be revealed by the process of attempting to solve the system.

    Subtracting the second equation from the first, we get x+6y4z=4, and subtracting x+y+z=pq from this equation gives 5y5z=4pq.
    After dividing by 5, we get yz=45p5q.
    Next, we subtract 4 times the second equation from 3 times the first equation to get 3(4x+7y+z)4(3x+y+5z)=3(11)4(15)12x+21y+3z12x4y20z=336017y17z=27yz=2717

    We now have two expressions for yz, and if we set them equal, we get an equation involving p and q. We can manipulate this equation as follows. 45p5q=271745+2717=p5q4+5×2717=pq68+13517=pq6717=pq and since 6717 is in lowest terms, p=67 and q=17, so pq=6717=50.

    Answer: 50

  10. Fix a positive integer k and let n=2k (we are interested in even n).
    The two-element subsets of S2k consisting of two integers that have a sum of 2k+1 are {1,2k}, {2,2k1}, {3,2k2}, and so on up to {k,k+1}. There are k subsets in the list.
    A subset X of S2k contains two integers with a sum of 2k+1 exactly when it contains both of the integers from at least one of the k subsets above.

    We will now count the number of subsets X of S2k with the property that no two integers in the subset have a sum of 2k+1.
    We will do this by counting the number of subsets X of S2k that contain at most one of the two integers from each of the k two-element subsets listed earlier.
    Suppose {a,b} is one of these k subsets. The set X could contain neither a nor b, it could contain a and not b, and it could contain b and not a.
    This gives 3 possibilities for each of the k subsets.
    Every integer in S2k appears in exactly one of these k subsets, so a choice of one of these three possibilities for each of the k subsets {a,b} completely determines X.
    These choices are independent, which means the number of subsets X with the desired property is exactly 3k.

    Since there are 22k subsets in total, we have p(2k)=3k22k=(34)k We want to find values of k that satisfy p(2k)<14. Using the formula above, this is equivalent to (34)k<14, which is equivalent to 3k<4k1.
    The table below has positive integer values of k in the left column, the value of 3k in the middle column, and 4k1 in the right column.

    k 3k 4k1
    1 3 1
    2 9 4
    3 27 16
    4 81 64
    5 243 256

    We see that k=5 is the smallest even positive integer for which 3k<4k1, so the answer is n=2×5=10.

    Answer: 10

Team Problems

  1. The cod accounts for 100%40%40%=20% of the total pieces of fish sold.
    Therefore, the number of pieces of cod sold was 0.2×220=44.

    Answer: 44

  2. Rearranging the given equation, we get x2=14 which implies that x=28=22×7.
    Therefore, 7x=7×22×7=22×72=2×7=14.

    Answer: 14

  3. The slope of AB is 43(23)33(13)=2020=1.

    Since AB is neither horizontal nor vertical, the slope of a line perpendicular to it is the negative

    of the reciprocal of 1, or 11=1.

    Answer: 1

  4. Factoring 1192172 as a difference of squares, we have 1192172=(11917)(119+17).
    Then 119217211917102=(11917)(119+17)11917102=119+17100=36=6

    Answer: 6

  5. Solution 1

    Substituting p=2q into p+q+r=70 gives 2q+q+r=70 or 3q+r=70.
    Substituting q=3r into 3q+r=70 gives 3(3r)+r=70 or 10r=70 which means r=7.
    Substituting r=7 into q=3r gives q=21, and substituting q=21 into p=2q gives p=42.

    Solution 2

    The equation p=2q is equivalent to q=12p.

    The equation q=3r is equivalent to r=13q.

    Substituting q=12p into r=13q gives r=16p.

    Substituting these values of q and r into p+q+r=70 gives p+12p+16p=70.
    Therefore, p=701+12+16=70106=42

    Answer: 42

  6. If we let d=ba, then we have b=a+d, and since cb=ba=d, we have c=b+d=a+2d. Thus, (a,b,c)=(a,a+d,a+2d).
    Since b>a, d>0, and so we can enumerate the triples by considering possible values of d starting at d=1.
    When d=1, we get the triples (1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), and (8,9,10) for a total of 8 triples.
    When d=2, we get the triples (1,3,5), (2,4,6), (3,5,7), (4,6,8), (5,7,9), and (6,8,10) for a total of 6 triples.
    When d=3, we get the triples (1,4,7), (2,5,8), (3,6,9), and (4,7,10) for a total of 4 triples. When d=4, we get the triples (1,5,9) and (2,6,10) for a total of 2 triples.
    If d5, then since a is at least 1, c=a+2d1+2×5=11. This means d cannot be any larger than 4, so we have found all triples.
    In total, there are 8+6+4+2=20 triples.

    Answer: 20

  7. The positive integer 4446 is even and the sum of its digits is 4+4+4+6=18 which is a multiple of 9, so 4446 itself is divisible by both 2 and 9. (Similar to the well-known divisibility rule for 3, a positive integer is divisible by 9 exactly when the sum of its digits is divisible by 9.)
    Factoring, we have 4446=2×32×247, and it can be checked that 247=13×19.
    Therefore, 4446=2×32×13×19, so the distinct prime factors of 4446 are 2, 3, 13, and 19, the sum of which is 37.

    Answer: 37

  8. Since every line segment must be used exactly once, the integer will have seven digits with each digit from 1 to 7 used exactly once.
    Observe that the path that goes from E to B to C to E to D to B to A to C creates the integer 7645123.

    We will now argue that this is the largest integer that can be created.
    Since every digit from 1 through 7 is used exactly once, if we start with any digits other than 7 and 6, in that order, the integer formed must be smaller than 7645123.
    This means that the path forming the largest possible integer must start with E, then go to B, and then C.
    From C, the options are to go to A or E. If the path continues to A, then the integer would start with 763, which is guaranteed to be smaller than 7645123, regardless of how the path is completed.
    The path must start with E, B, C, E, so the first three digits of the largest inter are 764.
    The line segments connecting E to B and E to C have already been used, which means that the path must continue to D since otherwise we would have to reuse a line segment. We now have shown that the first four digits of the largest integer are 7645.
    Since the path arrived at D from E, there is nowhere to go from D but to B, and then by similar reasoning, it must continue to A and finally to C, so the largest integer must be 7645123.

    Answer: 7645123

  9. Solution 1

    Let V equal the volume of the pool in litres and let x be the rate at which each hose outputs water in litres per hour.
    With three hoses, the pool will fill in 12 hours at a rate of 3x litres per hour. This means 3x=V12 or x=V36.

    One of the hoses stops working after 5 of the 12 hours, which means that the hose stops working when there are 512V litres in the pool, or 712V more litres to be added to the pool.

    Starting at 11:00 a.m., the new rate at which water is entering the pool is 2x, and 712V litres need to be added to the pool.
    If we let t be the amount of time in hours remaining to fill the pool with two hoses, then we have 2x=712Vt and so t=7V24x.
    Substituting x=V36, we get t=7V24V36=212=10.5 Therefore, another 10.5 hours are required, so the pool will be full 10.5 hours after 11:00 a.m., or at 9:30 p.m.

    Solution 2

    With three hoses working, the pool will fill in 12 hours, which means that one hose could fill the pool in 36 hours. We will say that the pool takes 36 hose-hours to fill.

    At the time when one of the hoses stops working, three hoses have been working for 5 hours. In other words, which is the same as if one hose had been working for 3×5=15 hours. In other words, 15 hose hours have been spent putting water in the pool.
    This means 3615=21 hose hours are still required. There are two hoses still working, so it will take and additional 212=10.5 hours to fill the pool.

    Since 10.5 hours after 11:00 a.m. is 9:30 p.m., the pool will be filled at 9:30 pm.

    Answer: 9:30 p.m.

  10. Substituting x=2 and y=3 into the two equations gives 2(a2+1)2b(3)=42(1a)+b(3)=9 which can be simplified to get

    2a2+6b=22a+3b=7 Dividing the first equation by 2 gives a2+3b=1 from which we can subtract 2a+3b=7 to get a22a=8.
    Rearranging gives a22a8=0 which can be factored to get (a4)(a+2)=0.
    Therefore, there are two possible values of a, which are a=4 and a=2. Substituting these values into 2a+3b=7 and solving for b gives b=5 and b=1, respectively.
    The ordered pairs (a,b)=(4,5) and (a,b)=(2,1) are the two possibilities. It can be checked that the two systems of equations

    17x+10y=45x+2y=43x5y=93xy=9 both have the unique solution (x,y)=(2,3).

    Answer: (4,5) and (2,1)

  11. If we set x=2023, then 20234(2022)(2024)(1+20232)=x4(x1)(x+1)(1+x2) Expanding, we have (x1)(x+1)(1+x2)=(x21)(x2+1)=x41.
    This means x4(x1)(x+1)(1+x2)=x4(x41)=1, so the answer is 1.

    Answer: 1

  12. For the square root in the numerator to be defined, we need 75x0 or x75, and for the square root in the denominator to be defined, we need x250 or 25x. Therefore, for the expression to be an integer, we must have 25x75.

    Suppose 75xx25 is equal to some integer n. Then 75x=nx25.
    Squaring both sides, we get 75x=n2(x25) or 75x=xn225n2. This equation can be rearranged to get 25n2+75=x(n2+1) and then solved for x to get x=25n2+75n2+1.
    Rearranging, we have x=25n2+75n2+1=25n2+25+50n2+1=25(n2+1)n2+1+50n2+1=25+50n2+1 Observe that n2+1>0 for all n, so the cancellation above is justified.
    Since x and 25 are both integers, 50n2+1 must also be an integer, so 50 is a multiple of n2+1.
    Since n2+1 is a divisor of 50 and it must be positive, we have that n2+1 must be one of 1, 2, 5, 10, 25, or 50.

    Setting n2+1 equal to each of these values and solving gives n2+1=1 n=0n2+1=2 n=±1n2+1=5 n=±2n2+1=10n=±3n2+1=25n=±24n2+1=50n=±7 Note that n is the ratio of two square roots, which means it must be nonnegative. As well, 24 is not an integer, so this means n=0, n=1, n=2, n=3, or n=7.
    Substituting these values of n into x=25n2+75n2+1, we get that the possible values of x are 75, 50, 35, 30, and 26, of which there are 5.

    Answer: 5

  13. The two-digit perfect squares are 16, 25, 36, 49, 64, and 81.
    Notice that no two of these perfect squares have a common first digit, which means that in a mystical integer, either a digit is the rightmost digit or there is only one possibility for the digit to its right.
    This means, for each possible first digit, the largest mystical integer with that first digit is the mystical integer with as many digits as possible.
    To make a mystical integer starting with 1 have as many digits as possible, we must follow the 1 by a 6, which must be followed by 4 and then 9. No two-digit perfect square starts with 9, so 1649 is the longest, and hence, the largest mystical integer with its first digit equal to 1.
    The only mystical integer with its first digit equal to 2 is 25 since no two-digit perfect square has a first digit of 5.
    The longest mystical integer starting with 3 is 3649.
    The longest mystical integer starting with 4 is 49.
    There are no mystical integers starting with 5, 7, or 9.
    The longest mystical integer with a first digit of 6 is 649.
    The longest mystical integer starting with 8 is 81649. This is the only five-digit mystical integer, so it must be the largest.

    Answer: 81649

  14. Suppose a, b, c, and d are the digits 1 through 4 in some order and are placed in the top-left subgrid as shown below.

    A 3 by 3 grid. The first two entries in the top row of the grid are a and b. The first two entries in the middle row are c and d. All other squares are empty.

    In order for the bottom-left subgrid to contain the integers from 1 through 4, the leftmost two cells in the bottom row must contain c and d in some order. Similarly, in order for the top-right subgrid to contain the digits from 1 through 4, the third cells in the first two rows must contain a and c in some order.
    Therefore, there are four possible ways to fill in the rest of the cells, excluding the bottom-right cell. They are shown below.

    In the rightmost column, the top entry is a and the middle entry is c. In the bottom row, the leftmost entry is a and the middle entry is b.     In the rightmost column, the top entry is c and the middle entry is a. In the bottom row, the leftmost entry is a and the middle entry is b.     In the rightmost column, the top entry is a and the middle entry is c. In the bottom row, the leftmost entry is b and the middle entry is a.     In the rightmost column, the top entry is c and the middle entry is a. In the bottom row, the leftmost entry is b and the middle entry is a.

    In each of these four grids, the top-left, top-right, and bottom-left subgrids all satisfy the condition that each digit occurs exactly once in it.
    In the first three of four, the three digits already in the bottom-right subgrid are different, so there is exactly one way to place a digit in the bottom right cell to make the grid satisfy the condition.
    For the first of the four grids above, we need to place a in the bottom-right cell. For the second, we need to place c. For the third, we need to place b.
    The fourth of these grids lacks b and c in the bottom-right subgrid. With only one cell left to fill, there is no way to place a digit in it so that every digit occurs in the subgrid.

    There are 24 ways to order the digits from 1 through 4 as a, b, c, and d. We have shown that there are three ways to fill the grid for each of these 24 orderings, so the answer is 3×24=72.

    Answer: 72

  15. Two numbers are reciprocals if their product is 1. Therefore, we are looking for real numbers x for which (x5x)(x4)=1.
    Multiplying through by x gives (x25)(x4)=x. Expanding gives x34x25x+20=x which can be rearranged to get x34x26x+20=0.
    Given that the problem states that there are exactly three real numbers with the given condition, these three real numbers must be the solutions to the cubic equation above. A cubic equation has at most three real roots, which means this cubic must have exactly three real roots. The sum of the roots of a cubic is the negative of the coefficient of x2, which means the answer is 4.
    However, in the interest of presenting a solution that does not assume the question “works”, we will factor this cubic to find the three real numbers.

    A bit of guessing and checking reveals that 234(22)6(2)+20=0, which means x2 is a factor of the cubic. We can factor to get (x2)(x22x10)=0 By the quadratic formula, the other two roots of the cubic are 2±224(10)2=1±11 The three real numbers are 2, 111, and 1+11 which have a sum of 2+(111)+(1+11)=4

    Answer: 4

  16. From the similarity of PBA and ABC, we get PBAB=ABCB.
    Substituting, we get PB8=811 which can be rearranged to get PB=6411.
    From the similarity of QAC and ABC, we get QCAC=ACBC.
    Substituting, we get QC6=611 which can be rearranged to get QC=3611.
    We now know the lengths of PB, QC, and BC. Since PQ=BCPBQC, we get PQ=1164113611=121643611=2111

    Answer: 2111

  17. Let t be the x-coordinate of B and note that t>0. The parabola has reflective symmetry across the y-axis, which implies ABCD has reflective symmetry across the y-axis. Thus, the x-coordinate of A is t.
    Therefore, AB=2t, and so the area of ABCD is (2t)2=4t2.
    The point C lies on the parabola with equation y=x24 and has the same x-coordinate as B. Therefore, the coordinates of C are (t,t24).
    Since BC is vertical, its length is the difference between the y-coordinates of B and C.
    It is given that C is below the x-axis, so its y-coordinate is negative. The y-coordinate of B is 0, and so we have that BC=0(t24)=4t2.
    Since ABCD is a square, AB=BC, and so 2t=4t2.
    Rearranging this equation gives t2+2t4=0, and using the quadratic formula, we get t=2±224(4)2=1±5 The quantity 15 is negative, and we are assuming t is positive, which means t=1+5.
    We know that the area of ABCD is 4t2, so the area is 4(1+5)2=4(625)=2485.

    Answer: 2485

  18. In this solution, “log” (without explicitly indicating the base) means “log10”.
    Using logarithm rules, we have ab=(log49)(108log38)=108log9log4×log8log3=108log32log22×log23log3=108×2×3×(log3)×(log2)2×(log2)×(log3)=108×3=324 Since ab=324=182 and ab>0, we have ab=18.

    Answer: 18

  19. The multiples of 3 from 1 through 25 are 3, 6, 9, 12, 15, 18, 21, and 24.
    The only multiple of 5 in this list is 15.

    Jolene chooses the red ball numbered 15 with probability 15 since there are exactly five red balls.
    If Jolene chooses the red ball numbered with 15, then there will be 7 green balls remaining with a multiple of 3 on them. The probability that Tia chooses a ball numbered with a multiple of 3 in this situation is 724.

    The probability that Jolene and Tia win the game if Jolene chooses the ball numbered with 15 is 15×724=7120.

    Jolene chooses a red ball other than the one numbered with 15 with probability 45.
    In this situation, there are 8 green balls numbered with a multiple of 3, so the probability that Tia chooses a green ball numbered with a multiple of 3 is 824.
    The probability that Jolene and Tia win the game if Jolene does not choose the red ball numbered by 15 is 45×824=32120.

    Since exactly one of the events “Jolene chooses the ball numbered with 15” and “Jolene does not choose the ball numbered with 15” must occur, the probability that Jolene and Tia win the game is equal to the sum of the probabilities that they win in each of the two situations.
    Therefore, the probability that they win the game is 7120+32120=39120=1340

    Answer: 1340

  20. The remainder when 468 is divided by d is r. This means that 0r<d and there is a unique integer q1 such that 468=q1d+r.
    Similarly, there are unique integers q2 and q3 such that 636=q2d+r and 867=q3d+r.
    This gives three equations 468=q1d+r636=q2d+r867=q3d+r Subtracting the first equation from the second, the first from the third, and the second from the third gives the three equations 168=(q2q1)d399=(q3q1)d231=(q3q2)d Since q1, q2, and q3 are integers, so are q2q1, q3q1, and q3q2. Therefore, d must be a divisor of 168, 399, and 231.

    Since d is a divisor of 168 and 231, it is a factor of 231168=63.
    Since d is a divisor of 63 and 168, it is a factor of 1682×63=42.
    Since d is a divisor of 63 and 42, it is a factor of 6342=21.
    Now note that 168=8×21, 399=19×21, and 231=11×21, so 21 is a divisor of all three integers.
    We have shown that d must be a divisor of 21 and that 21 is a divisor of all three integers. This implies that the possible values of d are the positive divisors of 21, which are 1, 3, 7, and 21.
    If d=1, then the remainder is 0 when each of 468, 636, and 867 is divided by d. In this case, d+r=1+0=1.
    If d=3, then the remainder is 0 when each of 468, 636, and 867 is divided by d. In this case, d+r=3+0=3.
    If d=7, then the remainder is 6 when each of 468, 636, and 867 is divided by d. In this case, d+r=7+6=13.
    If d=21, then the remainder is 6 when each of 468, 636, and 867 is divided by d. In this case, d+r=21+6=27.

    Answer: 27

  21. Solution 1

    Let X be the foot of the altitude of PQO from O to PQ, and let r=OQ.

    It is given that RBQ is isosceles and since QBR=90°, we must have that BQR=45°. As well, ROQ is equilateral so RQO=60°.
    Points X, Q, and B are on a line, so XQO=180°60°45°=75°.
    By very similar reasoning, XPO=75°, which means OPQ is isosceles with OP=OQ.
    Since OQX=75°, we have sin75°=OXOQ=OXr and so OX=rsin75°.
    Similarly, QX=rcos75°, and since OPQ is isosceles, OX bisects PQ, which means PQ=2rcos75°.

    From the double-angle formula, sin150°=2sin75°cos75°. Using this fact and the fact that sin150°=12, we can compute the area of OPQ as 12×PQ×OX=12(2rcos75°)(rsin75°)=r22(2sin75°cos75°)=r22sin150°=r22×12=r24

    Since ROQ is equilateral, QR=OQ=r, and since RQB=45°, cos45°=BQQR=BQr.
    Rearranging and using that cos45°=12, we get BQ=r2.
    Since BQ=BR, we get BR=r2 as well.
    Therefore, the area of BRQ is 12×BQ×BR=12×r2×r2=r24 The areas of PQO and BRQ are equal, so the ratio is 1:1.

    Solution 2

    Let OQ=r, which implies QR=r as well since ROQ is equilateral.
    Since BRQ has a right angle at B and is isosceles, the Pythagorean theorem implies that BQ2+BR2=r2 or 2BQ2=r2, so BQ=r2.
    Therefore, the area of BRQ is 12×r2×r2=r24.

    Since BRQ is isosceles and right angled at B, RQB=45°. As well, ROQ is equilateral, which means ROQ=60°. Using that P, Q, and B lie on the same line, we can compute PQO=180°60°45°=75°.
    By similar reasoning, QPO=75°, which implies that PQO is isosceles with OP=OQ=r.
    Using the two known angles in PQO, we get POQ as POQ=180°75°75°=30°.
    This means the area of OPQ is 12×OP×OQ×sin30°=r24.

    The areas of PQO and BRQ are equal, so the ratio is 1:1.

    Answer: 1:1

  22. Since a1,a2,a3, has a1=1 and a common difference of 3, we have that an=1+3(n1) for all n1. By similar reasoning, we get that bm=2+10(m1) for all m1.

    For fixed m, the term bm appears in both sequences exactly when there is some n such that an=bm.
    Using the closed forms established earlier, the term bm appears in both sequences exactly when there exists n such that 1+3(n1)=2+10(m1), which is equivalent to n=10m63.
    In other words, bm is in both sequences exactly when the expression 10m63 is a positive integer, which happens exactly when 10m6 is a positive multiple of 3.

    Note that m is a positive integer, so 10m6 is at least 4, which means 10m6 is positive for all m.
    Since 6 is a multiple of 3, 10m6 is a multiple of 3 exactly when 10m is a multiple of 3, but 3 and 10 have no prime factors in common, 10m is a multiple of 3 exactly when m is a multiple of 3.
    We have now shown that bm is in both sequences exactly when m is a multiple of 3.

    Let m=3k for some k1. Then bm=2+10(3k1)=30k8 and we wish to find the smallest possible k for which 30k8>2023.
    This inequality is equivalent to 30k>2031 which is equivalent to k>203130=67+2130. The smallest positive integer k with this property is k=68. Therefore, b3×68=b204=2+10(2041)=2+10(203)=2032 is the smallest integer that is larger than 2023 and is in both sequences.

    Answer: 2032

  23. For every real number x, 1sinx1, which means that sin3A and sinC are both at most 1.
    Therefore, 3sinC has a maximum possible value of 3, so sin3A+3sinC4.
    The only way that sin3A+3sinC=4 is if sin3A=1 and sinC=1.
    Since C corresponds to an angle in a triangle, 0°<C<180°.
    This and sinC=1 together imply that C=90°. We conclude that ABC has a right angle at C.

    Since A is an angle in a triangle, we have that 0°<A<90°. However, we have already shown that this triangle has a right angle not at A, which means 0°<A<90°.
    Multiplying this inequality through by 3 gives 0°<3A<270°.
    This and sin3A=1 together imply that 3A=90°, which means A=30°.
    It follows that ABC is a 30° - 60° - 90° triangle with hypotenuse AB=10.
    The side BC is opposite the 30° angle, so its length is half that of the hypotenuse, or BC=5.
    The side AC is opposite the 60° angle, so its length is AC=32×AB=53.
    The area of ABC is 12×AC×BC=12×5×53=2532.

    Suppose the length of the altitude from C to AB is h. Then the area of ABC is also equal

    to 12×AB×h=5h.
    Therefore, 5h=2532, and so h=532.

    Answer: 532

  24. Squaring both sides of the equation x+y+z=2 gives x2+y2+z2+2(xy+yz+zx)=4.
    Substituting xy+yz+zx=0 into the above equation gives x2+y2+z2=4.
    From x+y+z=2, we get x+y=2z, and from xy+yz+xz=0, we get xy=z(x+y).
    Substituting x+y=2z into xy=z(x+y), we get xy=z(z2).

    Regardless of the values of x and y, (xy)20. Using this observation as well as x2+y2+z2=4 and xy=z(z2), we get the following equivalent inequalities. 0(xy)20x22xy+y22xyx2+y22xy+z2x2+y2+z22z(z2)+z243z24z40(3z+2)(z2)0

    If z<23, then both 3z+2 and z2 are negative, so their product is positive.
    If 23z2, then 3z+20 and z20, so their product is non-positive.
    If z>2, then 3z+2 and z2 are both positive, so their product is positive.
    Therefore, in order for (3z+2)(z2)0, we need that 23z2.

    We have shown that z is at least 23 and at most 2, but to be sure these are the actual minimum and maximum possible values of z, we must show that there are solutions to the system of equations that have z taking each of these two values.

    If z=2, the equations become x+y+2=2 or x+y=0 and xy+2y+2x=0 which can be factored as xy+2(x+y)=0. Substituting x+y=0 gives xy=0, so x=0 or y=0.
    Since x+y=0 and at least one of x and y is 0, it must be the case that they are both 0. This gives the solution (x,y,z)=(0,0,2).

    If z=23, then the first equation gvies x+y23=2 or x+y=83. The second equation gives xy23(x+y)=0. Substituting x+y=83 into this equation and rearranging gives xy=169.
    Multiplying x+y=83 through by x gives x2+xy=83x, and after substituting xy=169 gives x2+169=83x which is equivalent to 9x224x+16=0.
    Factoring, we get (3x4)2=0 so 3x=4, which means x=43. Substituting this into x+y=43 and solving leads to y=43.
    Therefore, (x,y,z)=(43,43,23) is a soluton.

    We have shown that in any solution, we must have 23z2 and that there are solutions with z=23 and z=2. Thus, b=2 and a=23, so ba=2(23)=83.

    Answer:

    83

  25. With x=1, x13x2=0, and so the given identity implies f(1)+f(0)=1. Since we need to find f(0)+f(1)+f(2), it remains to find the value of f(2).

    With x=2, x13x2=2162=14. Applying the identity gives f(2)+f(14)=2.

    With x=14, x13x2=141342=1438=35. Applying the identity gives f(14)+f(35)=14.

    With x=35, x13x2=351952=35910=2. Applying the identity gives f(35)+f(2)=35.

    Let a=f(2), b=f(14), and c=f(35). We have shown that the following three equations hold: a+b=2b+c=14a+c=35 Adding the first and third equations gives 2a+(b+c)=2+35=135.

    Since b+c=14, this means 2a=13514=4720, and so f(2)=a=4740.

    We already know that f(0)+f(1)=1, so f(0)+f(1)+f(2)=1+4740=8740.

    Remark: It can be shown that the given condition implies that f(x)=9x3+6x2+x12(3x+1)(3x2) for all x except x=13 and =23 and that f(13) and f(23) are cannot be uniquely determined. However, the values of f(0), f(1), and f(2) are uniquely determined, which is all that was needed for this problem.

    Answer: 8740

Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of t is not initially known, and then t is substituted at the end.)

    1. Since 9=32, the area of the garden is 32 m2 or (3 m)×(3 m).
      Therefore, the side-length of the garden is 3 m.
      The perimeter is 4×3 m=12 m, so N=12.

    2. Each of the nine small squares is divided into two congruent triangular sections, so the square is divided into 9×2=18 triangular sections of equal area.
      Of the 18 sections, 8 are shaded, so 818=49 of ABCD is shaded.
      The area of ABCD is t2, so the shaded area is 4t29.
      Substituting t=12, we get that the area of the shaded region is 4×1229=64.

    3. Expanding, n(n1)(n+1)+n=n(n21)+n=n3n+n=n3.
      This means t=n3 so n=t3.
      Substituting t=64 gives n=643=4.

    Answer: (12,64,4)

    1. Multiplying through by 29(n+1) gives 29>4(n+1) which can be rearranged to get 25>4n or n<254=6.25.
      The positive integers that satisfy the inequality are n=1, n=2, n=3, n=4, n=5, and n=6, so the answer is 6.

    2. Let V be the volume of the tank.
      Using the second given piece of information, we know that when t24 litres are added, the amount of water goes from 2V10 to 5V10, which means t24 is the difference between these two amounts of water.

      Therefore, t24=5V2V10=3V10.
      Solving for V gives V=103×t24=10t212=5t26.
      Since adding t2 litres to the initial x litres leads to the tank having 2V10 litres in it, x=2V10t2.
      Substituting V=5t26 gives x=210×5t26t2=t26t2.

      Substituting t=6, we get x=6263=3.

    3. The line segment OP passes through the origin and P, so its slope is equal to ba.
      Since the slope is 125 and the line passes through the origin, there is some k for which b=12k and a=5k.
      By the Pythagorean theorem, a2+b2=OP2. Substituting b=12k, a=5k, and OP=13t, we get the following equivalent equations. a2+b2=OP2(5k)2+(12k)2=(13t)225k2+144k2=169t2169k2=169t2k2=t2 Since P(a,b) is in the first quadrant, a=5k is positive, so k is positive. As well, since 13t is given to be a length, we can assume t is positive, which means k=t, so a=5t.

      By reasoning similar to earlier, there is some m so that c=4m and d=3m.
      By the Pythagorean theorem, c2+d2=OQ2, so we get the following equivalent equations. c2+d2=OQ2(4m)2+(3m)2=(10t)216m2+9m2=100t225m2=100t2m2=4t2 Since c and t are both positive, m=4t2=2t, so c=8t.
      Therefore, a+c=5t+8t=13t.
      Substituting t=3, we get a+c=13(3)=39.

    Answer: (6,3,39)

    1. The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
      Their sum is 1+2+3+4+6+8+12+24=60

    2. Subtracting the second equation from the first gives (at6b)(at5b)=20(10) which simplifies to (t5t6)b=30 or t30b=30.
      Solving for b gives b=900t. Substituting t=60 gives b=90060=15.

    3. Using that the parabola passes through (0,60) gives 60=a(0)2+b(0)+c or c=60.

      The roots of ax2+bx+c are x=4 and x=t3. [We are implicitly assuming that t12 since otherwise the question would not have a unique answer.]
      The product of the roots of ax2+bx+c is ca=60a.
      Therefore, 60a=4×t3 which can be rearranged to get a=3×604t=45t. Substituting t=15 gives a=3.

    Answer: (60,15,3)