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2016 Galois Contest
Solutions
(Grade 10)

Thursday, April 13, 2016
(in North America and South America)

Friday, April 14, 2016
(outside of North American and South America)

©2016 University of Waterloo


    1. The first bucket contains 7 red discs.

      Each bucket after the first contains 3 more red discs than the previous bucket.

      Thus, the second bucket contains 7+3=10 red discs, the third bucket contains 10+3=13 red discs, and the fourth bucket contains 13+3=16 red discs.

    2. Solution 1

      Let b represent the number of buckets after the first.

      Since the first bucket contains 17 green discs and each bucket after the first contains 1 more green disc than the previous bucket, then there are 17+b green discs inside the bucket b after the first.

      Since the first bucket contains 7 red discs and each bucket after the first contains 3 more red discs than the previous bucket, then there are 7+3b red discs inside the bucket b after the first.

      The number of green discs in a bucket is equal to the number of red discs inside the same bucket when 17+b=7+3b or when 2b=10, and so when b=5.

      Thus, there are an equal number of red discs and green discs in the bucket 5 after the first bucket, which is the 6th bucket.

      Solution 2

      Using the fact that the first bucket contains 17 green discs and 7 red discs, and each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket, then we may summarize the number of green and red discs inside each bucket.

      Bucket Number Number of green discs Number of red discs
      1 17 7
      2 18 10
      3 19 13
      4 20 16
      5 21 19
      6 22 22

      Therefore, the 6th bucket contains an equal number of red discs and green discs.

      (Note that since the number of red discs is increasing by 3 each bucket and the number of green discs is increasing by 1 each bucket, then this is the only bucket in which the number of red discs and green discs will be equal.)

      Solution 3

      In the first bucket, there are 17 green discs and 7 red discs. Each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket.

      Since there are 2 more red discs than green being put in, then the difference between the numbers of green and red discs will decrease by 2 for each bucket after the first.

      Since the original difference is 177=10, then it takes 10÷2=5 more buckets to arrive at a bucket where the numbers of green and red discs will be equal.

      Therefore, the 6th bucket contains an equal number of red discs and green discs.

    3. As in part (b), there are 17+b green discs and 7+3b red discs inside the bucket b after the first.

      The number of red discs in a bucket is equal to twice the number of green discs inside the same bucket when 7+3b=2(17+b) or when 7+3b=34+2b, and so when b=27.

      In the 27th bucket after the first (the 28th bucket), there are 17+27=44 green discs and 7+3(27)=88 red discs. (We note that 88 is indeed twice 44.)

      The total number of discs in this bucket is 44+88=132.

    1. A plate with 36 shaded squares has 10 shaded squares along each side of the plate, as shown.

      A large square has sides that are 60 centimeters in length. It is divided into one white square and thiry-six shaded squares. The white square is in the center and the twelve shaded squares form a boarder around the white square.

      We can see this from the diagram, or by considering a plate with s squares along each side.

      In this case, we can count s squares on the top edge and s squares on the bottom edge, plus s2 new squares on each of the left edge and the right edge. (The 2 corner squares on each of these edges are already counted.)

      This means that there are 2s+2(s2)=4s4 squares along the edges.

      Here, we want 4s4=36 or 4s=40, and so s=10.

      There are 10 squares along each side of the plate and the side length of the square plate is 60 cm, thus the side length of each of the shaded squares is 6010=6 cm.

    2. Since the plate is a square, and there are an equal number of identical shaded squares along each edge of the plate, then the unshaded area in the centre of the plate is also a square.

      The area of this unshaded square in the centre of the plate is 1600 cm2, and so each of its sides has length 1600=40 cm, as shown.

      Consider the row of squares along the left edge of the plate.

      Since the side length of the square plate is 60 cm and the side length of the inner square is 40 cm, then the sum of the side lengths of the two shaded corner squares is 6040=20 cm.

      Therefore, each shaded corner square (and thus each shaded square) has side length 202=10 cm.

    3. Using the same argument as in part (b), the area of the unshaded square in the centre of the plate is 2500 cm2, and so each of its sides has length 2500=50 cm, as shown.

      The side length of the square plate is 60 cm and so the sum of the side lengths of 4 shaded squares (2 stacked vertically in the top two rows and 2 stacked vertically in the bottom two rows) is 6050=10 cm.

      Therefore, each of these shaded squares (and thus each shaded square on the plate) has side length 104=52 cm.

      The side length of the square plate is 60 cm and each shaded square has length 52 cm, and so along an outside edge of the plate there are 60÷52=60×25=12×2=24 shaded squares.

      There are 2 rows that each contain 24 shaded squares along each of the top and bottom of the square, and 2 additional rows that each contain 244=20 shaded squares along the left and right sides of the square.

      That is, the total number of shaded squares on the plate is4×24+4×20=96+80=176.

    1. Triangle ABC is equilateral with side length 6, and so AB=BC=CA=6.

      Since D is the midpoint of BC, then BD=DC=3.

      In ADC, ADC=90 and so by the Pythagorean Theorem AD2=AC2DC2.

      Therefore, h2=6232=369=27 and so h=27=9×3=9×3=33,since h>0.

    2. The shaded region lies inside the circle and outside the hexagon and thus its area is determined by subtracting the area of the hexagon from the area of the circle.

      First we find the area of the hexagon.

      Each vertex of hexagon EFGHIJ lies on the circle.

      Since the circle has centre O and radius 6, then OE=OF=OG=OH=OI=OJ=6.

      Each side length of the hexagon is also 6, and so the hexagon is formed by six congruent equilateral triangles with side length 6. (For example, OGH is one these 6 triangles.)

      Each of these triangles is congruent to ABC from part (a) and thus has height h=33.

      The area of each of the six congruent triangles is 12(6)(33)=93.

      Therefore, the area of hexagon EFGHIJ is 6×93=543.

      The area of the circle with centre O and radius 6 is π(6)2=36π.

      Finally, the area of the shaded region is 36π543.

    3. Let the area of the shaded region that we are required to find be A.

      Let the area of the shaded region in the diagram to the right be S.

      The shaded region is between the line segment MN and the part of the circumference of the circle between M and N.

      We may determine A by subtracting S from the area of the semi-circle with centre P.

      First we determine S.

      Consider the circle with centre O. The shaded region having area S lies inside sector MON of this circle, but outside MON.

      That is, S is determined by subtracting the area of MON from the area of sector MON.

      In MON, MN=ON=OM=r (since ON and OM are radii), and so the triangle is equilateral. Join O to P.

      Since ON=OM and P is the midpoint of MN, then OP is the altitude (height) of MON with base MN.

      In OPN, OPN=90 and so by the Pythagorean Theorem OP2=ON2PN2.

      Since PN=12(MN)=12r, OP2=r2(12r)2=r214r2=34r2, and so OP=34r2=32r.

      Therefore, the area of MON is 12(MN)(OP)=12(r)(32r)=34r2.

      Next, we determine the area of sector MON.

      Since MON is equilateral, then MON=60.

      Thus, the area of sector MON is 60360=16 of the area of the circle with centre O and radius r, or 16πr2.

      Therefore, S=16πr234r2.

      Finally, one-half of the area of the circle with centre P and radius PN=12r is12π(12r)2=18πr2, and so A=18πr2S=18πr2(16πr234r2)=(18π16π+34)r2.

      Simplifying further, the exact area of the shaded region is 63π24r2.

    1. The prime factorization of 126 is 126=213271.

      Given an input of 126=213271, the output from the Barbeau Process is 126(12+23+17)=126(21+28+642)=3(55)=165 .

    2. Given an input of p2q, the output from the Barbeau Process is p2q(2p+1q)=2pq+p2.

      We are told that this output is equal to 135.

      Since 135=335, then 2pq+p2=335 or p(2q+p)=335.

      Since p is a prime number that is a divisor of 335, then p=3 or p=5.

      If p=3, then 3(2q+3)=335 or 2q+3=45 and so q=21.

      However q is a prime number and 21 is not a prime number, so p3.

      If p=5, then 5(2q+5)=335 or 2q+5=27 and so q=11.

      Therefore, the only pair (p,q) of different prime numbers that satisfies the given conditions is (5,11).

    3. Solution 1

      Given an input of 2a3b5c, the output from the Barbeau Process is 2a3b5c(a2+b3+c5)=2a3b5c(15a+10b+6c30) . We are told that this output is equal to 4×2a3b5c.

      Comparing these gives 15a+10b+6c30=4 or 15a+10b+6c=120.

      Since a,b,c are positive integers, and 15×8=120, 10×12=120, and 6×20=120, then 1a7,1b11, and 1c19.

      Further, 10b,6c and 120 are divisible by 2, and so 15a=12010b6c is divisible by 2.

      Therefore, a is divisible by 2 and since 1a7, then a equals 2,4 or 6.

      Similarly, 15a,6c, and 120 are divisible by 3, and so 10b is divisible by 3.

      Therefore, b is divisible by 3 and since 1b11, then b equals 3,6 or 9.

      Finally, 15a,10b, and 120 are divisible by 5, and so 6c is divisible by 5.

      Therefore, c is divisible by 5 and since 1c19, then c equals 5,10 or 15.

      Since a2,b3, and c5, then 15a15×2=30,10b10×3=30, and 6c6×5=30.

      That is, each of the terms 15a,10b,6c is at least 30, and so each of the terms is at most 1202(30)=60 (for example, 15a=12010b6c1203030=60).

      Therefore, 15a60 and so a is equal to 2 or 4, 10b60 and so b is equal to 3 or 6, and 6c60 and so c is equal to 5 or 10.

      If a=2 and b=3, then 6c=12015(2)10(3)=60, and so c=10.

      If a=2 and b=6, then 6c=12015(2)10(6)=30, and so c=5.

      If a=4 and b=3, then 6c=12015(4)10(3)=30, and so c=5.

      If a=4 and b=6, then 6c=12015(4)10(6)=0, which is not possible.

      Therefore, the triples (a,b,c) of positive integers which satisfy the given conditions are (2,3,10),(2,6,5), and (4,3,5).

      Solution 2

      Given an input of 2a3b5c, the output from the Barbeau Process is 2a3b5c(a2+b3+c5)=2a3b5c(15a+10b+6c30) . We are told that this output is equal to 4×2a3b5c.

      Comparing these gives 15a+10b+6c30=4 or 15a+10b+6c=120.

      Since 10b, 6c and 120 are divisible by 2 and 15a=12010b6c, then 15a is divisible by 2 which means that a is divisible by 2. Thus, we set a=2A for some positive integer A.

      Since 15a, 6c and 120 are divisible by 3 and 10b=12015a6c, then 10b is divisible by 3 which means that b is divisible by 3. Thus, we set b=3B for some positive integer B.

      Since 15a, 10b and 120 are divisible by 5 and 6c=12015a10b, then 6c is divisible by 5 which means that c is divisible by 5. Thus, we set c=5C for some positive integer C.

      Therefore, the equation 15a+10b+6c=120 becomes 30A+30B+30C=120 or A+B+C=4.

      Since A, B and C are positive integers that add to 4, the possible values for (A,B,C) are (2,1,1), (1,2,1), and (1,1,2), because each is at least 1 which only leaves 1 “left over” to make up the total of 4.

      Since (a,b,c)=(2A,3B,5C), then the possible triples (a,b,c) are (4,3,5), (2,6,5), and (2,3,10).

    4. We proceed in a number of steps.

      Step 1: Each exponent in the prime factorization must be a multiple of its prime

      We show that, since the output is an integer multiple of the input, each exponent in the prime factorization must be an integer multiple of its corresponding prime.

      This generalizes what we found in Solution 2 to part (c).

      In other words, suppose that n=p1a1p2a2p3a3pkak where p1,p2,...,pk are different prime numbers and a1,a2,...,ak are integers that are each at least 0.

      By definition, the output of the Barbeau Process is n(a1p1+a2p2+a3p3++akpk).

      Note that we are allowing each of a1,a2,,ak to be 0, which does not affect the value of the output.

      Here, we are told that the output of the Barbeau process is 3n.

      Therefore, n(a1p1+a2p2+a3p3++akpk)=3n or a1p1+a2p2+a3p3++akpk=3.

      Rearranging, we obtain a1p1=3a2p2a3p3akpk.

      Multiplying both sides by p2p3pk, we obtain a1p2p3pkp1=3p2p3pka2p3pka3p2p4pkakp2p3pk1 Since every term on the right side is an integer, then a1p2p3pkp1 must be an integer.

      Since the prime numbers p1,p2,,pk are all distinct, then a1 must be a multiple of p1 to make this fraction actually equal an integer.

      A similar argument will show that a2 is a multiple of p2, a3 is a multiple of p3, and so on.

      Step 2: n does not have any prime factor larger than 7

      Suppose that n does have a prime factor that is larger than 7.

      In other words, suppose that n has a prime factor pi that is at least 11.

      In this case, the factor piai in the prime factorization of n is at least 1111 since pi11 and ai is a multiple of pi and so must also be at least 11.

      In this case, n1111.

      But n is restricted to be less than 1010, so this is not possible.

      Therefore, n does not have any prime factor larger than 7.

      Step 3: Simplifying the algebra

      Combining Steps 1 and 2, n must be of the form n=2a3b5c7d for some non-negative integers a, b, c, and d that are multiples of 2, 3, 5, and 7, respectively.

      As in Solution 2 to part (c), we set a=2A, b=3B, c=5C, and d=7D for some non-negative integers A,B,C,D.

      Therefore, the equation a1p1+a2p2+a3p3++akpk=3 becomes a2+b3+c5+d7=3, which becomes A+B+C+D=3.

      Therefore, we want to find all of the non-negative integer solutions to A+B+C+D=3 which give n=22A33B55C77D<1010.

      Step 4: Restrictions on A,B,C,D

      Since A+B+C+D=3 and each of A,B,C,D is non-negative, then each of A,B,C,D is at most 3.

      If D=2 or D=3, then n is divisible by 714 or 721, each of which is larger than 1010.

      Therefore, D1.

      If C=3, then n is divisible by 515, which is larger than 1010.

      Therefore, C2.

      Step 5: Determining the values of n

      Since A+B+C+D=3 and each of A,B,C,D is non-negative, then A,B,C,D could be (i) 3,0,0,0 in some order, or (ii) 2,1,0,0 in some order, or (iii) 1,1,1,0, in some order.

      (If one value is 3, the rest must be 0. If one value is 2, then there must be one 1 and two 0s. If no value is 2, then there must be three 1s and one 0.)

      Using the facts that C2 and D1, we enumerate the possibilities:

      Category A B C D n Less than 1010?
      (i) 3 0 0 0 26 Yes
      (i) 0 3 0 0 39 Yes
      (ii) 2 1 0 0 2433 Yes
      (ii) 2 0 1 0 2455 Yes
      (ii) 2 0 0 1 2477 Yes
      (ii) 1 2 0 0 2236 Yes
      (ii) 0 2 1 0 3655 Yes
      (ii) 0 2 0 1 3677 Yes
      (ii) 1 0 2 0 22510 Yes
      (ii) 0 1 2 0 33510 Yes
      (ii) 0 0 2 1 51077 No
      (iii) 1 1 1 0 223355 Yes
      (iii) 1 1 0 1 223377 Yes
      (iii) 1 0 1 1 225577 No
      (iii) 0 1 1 1 335577 No

      In each case, we can check whether the possible value of n is smaller than 1010 using a calculator. (Which of these calculations can you “reason” through without using a calculator?)

      In summary, the possible values of n are 26,39,2433,2455,2477,2236,3655,3677,22510,33510,223355,223377