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2021 Fermat Contest
Solutions
(Grade 11)

Thursday, February 23, 2021
(in North America and South America)

Wednesday, February 24, 2021
(outside of North American and South America)

©2020 University of Waterloo


  1. A rectangle with width 2 cm and length 3 cm has area 2 cm×3 cm=6 cm2.

    Answer: (E)

  2. Calculating, 2+3×5+2=2+15+2=19.

    Answer: (A)

  3. Expressed as a fraction, 25% is equivalent to 14. Since 14 of 60 is 15, then 25% of 60 is 15.

    Answer: (B)

  4. When x0, we obtain 4xx+2x=4x3x=43. Thus, when x=2021, we have 4xx+2x=43. Alternatively, we could substitute x=2021 to obtain 4xx+2x=80842021+4042=80846063=43.

    Answer: (B)

  5. We note that 6=2×3 and 27=3×9 and 39=3×13 and 77=7×11, which means that each of 6, 27, 39, and 77 can be written as the product of two integers, each greater than 1. Thus, 53 must be the integer that cannot be written in this way. We can check that 53 is indeed a prime number.

    Answer: (C)

  6. We draw an unshaded dot to represent the location of the dot when it is on the other side of the sheet of paper being shown. Therefore, the dot moves as follows:

    First, a shaded dot is in the top right corner; second, the sheet of paper is folded along the diagonal, and an unshaded dot is in the bottom left corner; third, the folded sheet is rotated 90 degrees clockwise and the unshaded dot is in the top left corner; and fourth, the sheet of paper unfolded with a shaded dot is in the bottom right corner.

    It is worth noting that folding and unfolding the paper have no net effect on the figure. Thus, the resulting figure can be determined by rotating the original figure by 90° clockwise.

    Answer: (E)

  7. When x=2, we get x2=4. Here, x<x2.
    When x=12, we get x2=14. Here, x<x2.
    When x=0, we get x2=0. Here, x=x2.
    When x=12, we get x2=14. Here, x>x2.
    When x=2, we get x2=4. Here, x<x2.
    This means that x=12 is the only choice where x>x2.

    Answer: (D)

  8. Suppose that the original integer has tens digit a and ones (units) digit b.
    This integer is equal to 10a+b.
    When the digits are reversed, the tens digit of the new integer is b and the ones digit is a.
    This new integer is equal to 10b+a.
    Since the new two-digit integer minus the original integer is 54, then (10b+a)(10a+b)=54 and so 9b9a=54 which gives ba=6.
    Thus, the positive difference between the two digits of the original integer is 6. An example of a pair of such integers is 71 and 17.

    Answer: (C)

  9. The line with equation y=2x6 has slope 2. When this line is translated, the slope does not change.
    The line with equation y=2x6 has y-intercept 6. When this line is translated upwards by 4 units, its y-intercept is translated upwards by 4 units and so becomes 2.
    This means that the new line has equation y=2x2.
    To find its x-intercept, we set y=0 to obtain 0=2x2 and so 2x=2 or x=1.
    Thus, the x-intercept is 1.

    Answer: (D)

  10. Using exponent laws, 3x+2=3x32=3x9.
    Since 3x=5, then 3x+2=3x9=59=45.

    Answer: (E)

  11. Since the second number being added is greater than 300 and the sum has hundreds digit R, then R cannot be 0.
    From the ones column, we see that the ones digit of R+R is 0. Since R0, then R=5.
    This makes the following sum.

    The addition problem P 7 5 plus 3 9 5 equals 5 Q 0 written vertically, with a 1 above the tens column.

    Since 1+7+9=17, we get Q=7 and then 1+P+3=5 and so P=1, giving the following final sum.

    The sum 175 plus 395 equals 570 written vertically, with a 1 above the tens column and a 1 above the hundreds column.

    Therefore, P+Q+R=1+7+5=13.

    Answer: (A)

  12. A perfect square is divisible by 9 exactly when its square root is divisible by 3.
    In other words, n2 is divisible by 9 exactly when n is divisible by 3.
    In the list 1,2,3,,19,20, there are 6 multiples of 3.
    Therefore, in the list 12,22,32,,192,202, there are 6 multiples of 9.

    Answer: (E)

  13. In an isosceles right-angled triangle, the ratio of the length of the hypotenuse to the length of each of the shorter sides is 2:1.
    Consider WZX which is isosceles and right-angled at Z.
    Here, WX:WZ=2:1. Since WX=62, then WZ=622=6.
    Since WZX is isosceles, then XZ=WZ=6.
    Consider XYZ which is isosceles and right-angled at Y.
    Here, YZ=XZ2=62=6222=622=32.
    Since XYZ is isosceles, then XY=YZ=32.
    Therefore, the perimeter of WXYZ is WX+XY+YZ+WZ=62+32+32+6=122+622.97 Of the given choices, this is closest to 23.

    Answer: (C)

  14. Suppose that Natascha runs at r km/h.
    Since she cycles 3 times as fast as she runs, she cycles at 3r km/h.
    In 1 hour of running, Natascha runs (1 h)(r km/h)=r km.
    In 4 hours of cycling, Natascha cycles (4 h)(3r km/h)=12r km.
    Thus, the ratio of the distance that she cycles to the distance that she runs is equivalent to the ratio 12r km:r km which is equivalent to 12:1.

    Answer: (A)

  15. Solution 1:

    Since a is a positive integer, 45a is a positive integer.
    Since b is a positive integer, 45a is less than 2021.
    The largest multiple of 45 less than 2021 is 45×44=1980. (Note that 4545=2025 which is greater than 2021.)
    If a=44, then b=20214544=41.
    Here, a+b=44+41=85.
    If a is decreased by 1, the value of 45a+b is decreased by 45 and so b must be increased by 45 to maintain the same value of 45a+b, which increases the value of a+b by 1+45=44.
    Therefore, if a<44, the value of a+b is always greater than 85.
    If a>44, then 45a>2021 which makes b negative, which is not possible.
    Therefore, the minimum possible value of a+b is 85.

    Solution 2:

    We re-write 45a+b=2021 as 44a+(a+b)=2021.
    Since a and b are positive integers, 44a and a+b are positive integers.
    In particular, this tells us that 44a, which is a multiple of 44, is less than 2021.
    Since the sum of 44a and a+b is constant, to minimize a+b, we can try to maximize 44a.
    Since 4445=1980 and 4446=2024, the largest multiple of 44 less than 2021 is 1980.
    This means that a+b20211980=41.
    However, a+b cannot equal 41 since we would need 44a=1980 and so a=45 (making b=4) to make this possible.
    The next multiple of 44 less than 1980 is 4444=1936.
    If a=44, then a+b=202144a=85.
    If a=44 and a+b=85, then b=41 which is possible.
    Since a+b=41 is not possible and 85 is the next smallest possible value for a+b, then the minimum possible value for a+b is 85.

    Answer: (C)

  16. The first few values of n! are 1!=12!=2(1)=23!=3(2)(1)=64!=4(3)(2)(1)=245!=5(4)(3)(2)(1)=120 We note that 2!1!=14!1!=233!1!=55!1!=119 This means that if a and b are positive integers with b>a, then 1, 3, 5, 9 are all possible ones (units) digits of b!a!.
    This means that the only possible answer is choice (D), or 7.

    To be complete, we explain why 7 cannot be the ones (units) digit of b!a!.
    For b!a! to be odd, one of b! and a! is even and one of them is odd.
    The only odd factorial is 1!, since every other factorial has a factor of 2.
    Since b>a, then if one of a and b is 1, we must have a=1.
    For the ones (units) digit of b!1 to be 7, the ones (units) digit of b! must be 8.
    This is impossible as the first few factorials are shown above and every greater factorial has a ones (units) digit of 0, because it is a multiple of both 2 and 5.

    Answer: (D)

  17. Since the average of the two smallest integers in S is 5, their sum is 25=10.
    Since the average of the two largest integers in S is 22, their sum is 222=44.
    Suppose that the other five integers in the set S are p<q<r<t<u. (Note that the integers in S are all distinct.) The average of the nine integers in S is thus equal to 10+44+p+q+r+t+u9 which equals 6+p+q+r+t+u9.
    We would like this average to be as large as possible.
    To make this average as large as possible, we want p+q+r+t+u9 to be as large as possible, which means that we want p+q+r+t+u to be as large as possible.
    What is the maximum possible value of u?
    Let x and y be the two largest integers in S, with x<y. Since x and y are the two largest integers, then u<x<y.
    Since x+y=44 and x<y and x and y are integers, then x21.
    For u to be as large as possible (which will allow us to make p, q, r, t as large as possible), we set x=21.
    In this case, we can have u=20.
    To make p, q, r, t as large as possible, we can take p=16, q=17, r=18, t=19. Here, p+q+r+t+u=90.
    If x<21, then p+q+r+t+u will be smaller and so not give the maximum possible value.
    This means that the maximum possible average of the integers in S is 6+909=16.

    Answer: (B)

  18. Suppose that PQ=PR=2x and QR=2y.
    The semi-circles with diameters PQ and PR thus have radii x and the radius of the semi-circle with diameter QR is y.
    The area of each semi-circle with radius x is 12πx2 and the area of the semi-circle with radius y is 12πy2.
    Since the sum of the areas of the three semi-circles equals 5 times the area of the semi-circle with diameter QR, then 12πx2+12πx2+12πy2=512πy2 which gives x2+x2+y2=5y2 and so 2x2=4y2 which gives x2=2y2 and so x=2y. Suppose that M is the midpoint of QR and that P is joined to M.

    Since PQR is isosceles with PQ=PR, then PM is perpendicular to QR.
    In other words, PMQ is right-angled at M.
    Therefore, cos(PQR)=cos(PQM)=QMPQ=12QRPQ=y2x=y22y=122=142=18.

    Answer: (B)

  19. Since x+y=7, then x+y+z=7+z.
    Thus, the equation (x+y+z)2=4 becomes (7+z)2=4.
    Since the square of 7+z equals 4, then 7+z=2 or 7+z=2.
    If 7+z=2, then z=5.
    In this case, since xz=180, we get x=1805=36 which gives y=7x=29.
    If 7+z=2, then z=9.
    In this case, since xz=180, we get x=1809=20 which gives y=7x=13.
    We can check by direct substitution that (x,y,z)=(36,29,5) and (x,y,z)=(20,13,9) are both solutions to the original system of equations.
    Since S is the sum of the possible values of y, we get S=(29)+(13)=42 and so S=42.

    Answer: (E)

  20. Let ST=a and let STR=θ.
    Since RST is right-angled at S, then TRS=90°θ.
    Since PRTY and WRSU are squares, then PRT=WRS=90°.
    Thus, QRW+WRT=WRT+TRS and so QRW=TRS=90°θ.
    Since PQXY is a rectangle, then PQX=90°, which means that WQR is right-angled at Q.
    This means that QWR=90°QRW=90°(90°θ)=θ.

    Consider RST.
    Since ST=a and STR=θ, then cosθ=aRT and so RT=acosθ.
    Also, tanθ=RSa and so RS=atanθ.
    Since PRTY is a square, then PY=PR=RT=acosθ.
    Since WRSU is a square, then RW=RS=atanθ.
    Next, consider QRW.
    Since RW=atanθ and QWR=θ, then sinθ=QRatanθ and so QR=atanθsinθ.
    This means that PQ=PRQR=acosθatanθsinθ=acosθasinθcosθsinθ=acosθasin2θcosθ=a(1sin2θ)cosθ=acos2θcosθ(since sin2θ+cos2θ=1)=acosθ(since cosθ0) Since the area of rectangle PQXY is 30, then PQPY=30 and so acosθacosθ=30 which gives a2=30.
    Since a>0, we get a=305.48. Of the given choices, ST=a is closest to 5.5.

    Answer: (C)

  21. Since f(2)=5 and f(mn)=f(m)+f(n), then f(4)=f(22)=f(2)+f(2)=10.
    Since f(3)=7, then f(12)=f(43)=f(4)+f(3)=10+7=17.
    While this answers the question, is there actually a function that satisfies the requirements? The answer is yes.
    One function that satisfies the requirements of the problem is the function f defined by f(1)=0 and f(2p3qr)=5p+7q for all non-negative integers p and q and all positive integers r that are not divisible by 2 or by 3. Can you see why this function satisfies the requirements?

    Answer: (A)

  22. The total surface area of the cone includes the circular base and the lateral surface.
    For the given unpainted cone, the base has area πr2=π(3 cm)2=9π cm2 and the lateral surface has area πrs=π(3 cm)(5 cm)=15π cm2.
    Thus, the total surface area of the unpainted cone is 9π cm2+15π cm2=24π cm2.
    Since the height and base are perpendicular, the lengths s, h and r form a right-angled triangle with hypotenuse s.
    For this cone, by the Pythagorean Theorem, h2=s2r2=(5 cm)2(3 cm)2=16 cm2 and so h=4 cm.
    When the unpainted cone is placed in the container of paint so that the paint rises to a depth of 2 cm, the base of the cone (area 9π cm2) is covered in paint.
    Also, the bottom portion of the lateral surface is covered in paint.

    A description of the diagram follows.

    The unpainted portion of the cone is itself a cone with height 2.
    When we take a vertical cross-section of the cone through its top vertex and a diameter of the base, the triangle formed above the paint is similar to the original triangle and has half its dimensions. The triangles are similar because both are right-angled and they share an equal angle at the top vertex. The ratio is 2:1 since their heights are 4 cm and 2 cm.
    Therefore, the unpainted cone has radius 1.5 cm (half of the original radius of 3 cm) and slant height 2.5 cm (half of the original slant height of 5 cm).
    Thus, the unpainted lateral surface area is the lateral surface area of a cone with radius 1.5 cm and slant height 2.5 cm, and so has area π(1.5 cm)(2.5 cm)=3.75π cm2.
    This means that the painted lateral surface area is 15π cm23.75π cm2=11.25π cm2. (This is in fact three-quarters of the total surface area. Can you explain why this is true in a different way?)
    Thus, the fraction of the total surface area of the cone that is painted is 9π cm2+11.25π cm224π cm2=20.2524=8196=2732 Since 2732 is in lowest terms, then p+q=27+32=59.

    Answer: (A)


  23. Since each Figure is formed by placing two copies of the previous Figure side-by-side along the base and then adding other pieces above, the number of dots in the base of each Figure is two times as many as in the previous Figure.
    Since each Figure is an equilateral triangle, then the number of dots in the Figure equals the sum of the positive integers from 1 to the number of dots in the base, inclusive. In other words, if the base of a Figure consists of b dots, then the Figure includes 1+2+3++(b1)+b dots. This sum is equal to 12b(b+1). (If this formula for the sum is unfamiliar, can you argue why it is true?)
    Since each Figure is formed by using three copies of the previous Figure and any new dots added are shaded dots, the number of unshaded dots in each Figure is exactly three times the number of unshaded dots in the previous Figure.
    Since each dot is either shaded or unshaded, the number of shaded dots equals the total number of dots minus the number of unshaded dots.
    Using these statements, we construct a table:

    Figure Dots in base Dots in Figure Unshaded dots Shaded dots
    1 2 3 3 0
    2 4 10 9 1
    3 8 36 27 9
    4 16 136 81 55
    5 32 528 243 285
    6 64 2080 729 1351
    7 128 8256 2187 6069
    8 256 32896 6561 26335
    9 512 131328 19683 111645

    Therefore, the smallest value of n for which Figure n includes at least 100 000 dots is n=9.
    We note that since the number of dots in the base of Figure 1 is 2 and the number of dots in the base of each subsequent Figure is double the number of dots in the previous Figure, then the number of dots in the base of Figure n is equal to 2n.
    Since the number of unshaded dots in Figure 1 is 3 and the number of unshaded dots in each subsequent Figure is three times the number of unshaded dots in the previous Figure, then the number of unshaded dots in Figure n is 3n.
    Therefore, a formula for the number of unshaded dots in Figure n is 122n(2n+1)3n which can be re-written as 22n1+2n13n, which agrees with the numbers in the table above.

    Answer: (B)

  24. The curves with equations y=ax2+2bxa and y=x2 intersect exactly when the equation ax2+2bxa=x2 has at least one real solution.
    This equation has at least one real solution when the quadratic equation (a1)x2+2bxa=0 has at least one real solution.
    (Note that when a=1, this equation is actually linear as long as b0 and so will have at least one real solution.)
    This quadratic equation has at least one real solution when its discriminant is non-negative; that is, when (2b)24(a1)(a)0 Manipulating algebraically, we obtain the equivalent inequalities: 4a24a+4b20a2a+b20a2a+14+b214a22a(12)+(12)2+b214(a12)2+b2(12)2 Therefore, we want to determine the probability that a point (a,b) satisfies (a12)2+b2(12)2 given that it satisfies a2+b2(12)2.
    The equation (x12)2+y2=(12)2 represents a circle with centre (12,0) and radius 12.
    The inequality (x12)2+y2(12)2 represents the region outside of this circle.
    The equation x2+y2=(12)2 represents a circle with centre (0,0) and radius 12.
    The inequality x2+y2(12)2 represents the region inside this circle.
    Re-phrasing the problem in an equivalent geometric way, we want to determine the probability that a point (a,b) is outside the circle with centre (12,0) and radius 12 given that it is inside the circle with centre (0,0) and radius 12.
    Note that each of these two circles passes through the centre of the other circle.

    A description of the diagram follows.

    Putting this another way, we want to determine the fraction of the area of the circle centred at the origin that is outside the circle centred at (12,0). This region is shaded in the first diagram above.
    Let r=12.
    The area of each circle is πr2.
    To determine the shaded area, we subtract the unshaded area from the area of the circle. By symmetry, the unshaded area below the x-axis will equal the unshaded area above the x-axis.
    In the second diagram above, the origin is labelled O, the point where the circle centred at O intersects the x-axis is labelled B, and the point of intersect in the first quadrant of the two cicles is labelled A. The unshaded region above the x-axis consists of AOB plus two curvilinear regions.
    Since OA and OB are radii of the circle centred at O, then OA=OB=r.
    Since AB is a radius of the circle centred at B, then AB=r.
    This means that OAB is equilateral, and so it has three 60° angles.
    When OAB is combined with the curvilinear region above and to the right of OAB, we thus obtain a sector of the circle with centre O and central angle 60°.
    Since 60° is 16 of 360°, this sector is 16 of the entire circle and so has area 16πr2.
    Similarly, when OAB is combined with the curvilinear region above and to the left of OAB, we obtain a sector of the circle with centre B and central angle 60°; this sector has area 16πr2.
    If K is the area of equilateral OAB, the area of each curvilinear region is 16πr2K and so the unshaded area above the x-axis is equal to 2(16πr2K)+K which simplifies to 13πr2K.
    Thus, the total unshaded area inside the circle of area πr2 is 23πr22K, which means that the shaded area inside the circle is πr2(23πr22K) which is equal to 13πr2+2K.
    Finally, we need to use the area of an equilateral triangle with side length r. This area is equal to 34r2. (To see this, we could use an altitude to divide the equilateral triangle into two 30°-60°-90° triangles. Using the ratio of side lengths in these special triangles, this altitude has length 32r and so the area of the equilateral triangle is 12r(32) or 34r2.)
    Therefore, the fraction of the circle that is shaded is 13πr2+32r2πr2=13π+32π=2π+336π0.609 This means that the probability, p, that a point (a,b) satisfies the given conditions is approximately 0.609 and so 100p60.9.
    Of the given choices, this is closest to 61, or (E).

    Answer: (E)

  25. Suppose that a, b and c are positive integers with abc=22325272112132172192.
    We determine the number of triples (a,b,c) with this property. (We are temporarily ignoring the size ordering condition in the original question.)
    Since the product abc has two factors of 2, then a, b and c have a total of two factors of 2.
    There are 6 ways in which this can happen: both factors in a, both factors in b, both in c, one each in a and b, one each in a and c, and one each in b and c.
    Similarly, there are 6 ways of distributing each of the other squares of prime factors.
    Since abc includes exactly 8 squares of prime factors and each can be distributed in 6 ways, there are 68 ways of building triples (a,b,c) using the prime factors, and so there are 68 triples (a,b,c) with the required product.

    Next, we include the condition no pair of a, b and c should be equal. (We note that a, b and c cannot all be equal, since their product is not a perfect cube.)
    We count the number of triples with one pair equal, and subtract this number from 68.
    We do this by counting the number of these triples with a=b. By symmetry, the number of triples with a=c and with b=c will be equal to this total.
    In order to have a=b and ac and bc, for each of the squared prime factors p2 of abc, either p2 is distributed as p and p in each of a and b, or p2 is distributed to c.
    Thus, for each of the 8 squared prime factors p2, there are 2 ways to distribute, and so 28 triples (a,b,c) with a=b and ac and bc.
    Similarly, there will be 28 triples with a=c and 28 triples with b=c.
    This means that there are 68328 triples (a,b,c) with the required product and with no two of a, b, c equal.

    The original problem asked us to the find the number of triples (x,y,z) with the given product and with x<y<z.
    To convert triples (a,b,c) with no size ordering to triples (x,y,z) with x<y<z, we divide by 6. (Each triple (x,y,z) corresponds to 6 triples (a,b,c) of distinct positive integers with no size ordering.)
    Therefore, the total number of triples (x,y,z) with the required properties is N=16(68328)=6727=279808 When N is divided by 100, the remainder is 8.

    Answer: (C)