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2017 Fryer Contest
Solutions
(Grade 9)

Wednesday, April 12, 2017
(in North America and South America)

Thursday, April 13, 2017
(outside of North American and South America)

©2017 University of Waterloo


    1. Red pens are sold in packages of 6 pens.
      Therefore, 5 packages of red pens contain 5×6=30 pens.
      Blue pens are sold in packages of 9 pens.
      Therefore, 3 packages of blue pens contain 3×9=27 pens.
      Altogether, Igor bought 30+27=57 pens.

    2. Robin bought 21 packages of red pens which contain 21×6=126 pens.
      Of the 369 pens that Robin bought, the number of blue pens was 369126=243.
      Blue pens are sold in packages of 9, so the number of packages of blue pens bought by Robin was 243÷9=27.

    3. Solution 1:

      Let the number of packages of red pens bought by Susan be r, for some whole number r.
      Let the number of packages of blue pens bought by Susan be b, for some whole number b.
      Thus, Susan buys 6r red pens and 9b blue pens.
      If Susan bought exactly 31 pens, then 6r+9b=31.
      Factoring the left side of this equation, we get 3(2r+3b)=31.
      Since r and b are whole numbers, then 2r+3b is also a whole number, and so the left side of the equation is a multiple of 3.
      Since the right side, 31, is not a multiple of 3, there is no solution to the equation 6r+9b=31 for whole numbers r and b.
      Therefore, it is not possible for Susan to buy exactly 31 pens.

      Solution 2:

      Let the number of packages of red pens bought by Susan be r, for some whole number r.
      Let the number of packages of blue pens bought by Susan be b, for some whole number b.
      Thus, Susan buys 6r red pens and 9b blue pens.
      If Susan bought exactly 31 pens, then 6r+9b=31.
      The smallest possible value for b is 0, and the largest possible value for b is 3 since if b4, then the number of pens is greater than or equal to 4×9=36.
      Solving the equation 6r+9b=31 for r, we get 6r=319b, and so r=319b6.
      In the table below, we use this equation to determine the value of r given that b can equal 0,1,2, or 3.

      b r=319b6
      0 r=319(0)6=316
      1 r=319(1)6=226
      2 r=319(2)6=136
      3 r=319(3)6=46

      For each of the possible values for b, the resulting value of r is not a whole number.
      Thus, there is no solution to the equation 6r+9b=31 for whole numbers r and b, and so it is not possible for Susan to buy exactly 31 pens.

    1. Expressing 15 and 14 with a common denominator of 40, we get 15=840 and 14=1040.
      We require that n40>840 and n40<1040, thus n>8 and n<10.
      The only integer n that satisfies both of these inequalities is n=9.

    2. Expressing m8 and 13 with a common denominator of 24, we require 3m24>824 and so 3m>8 or m>83.
      Since 83=223 and m is an integer, then m3.
      Expressing m+18 and 23 with a common denominator of 24, we require 3(m+1)24<1624 or 3m+3<16 or 3m<13, and so m<133.
      Since 133=413 and m is an integer, then m4.
      The integer values of m which satisfy m3 and m4 are m=3 and m=4.

    3. At the start of the weekend, Fiona has played 30 games and has w wins, so her win ratio is w30.
      Fiona’s win ratio at the start of the weekend is greater than 0.5=12, and so w30>12.
      Since 12=1530, then we get w30>1530, and so w>15.
      During the weekend Fiona plays five games giving her a total of 30+5=35 games played.
      Since she wins three of these games, she now has w+3 wins, and so her win ratio is w+335.
      Fiona’s win ratio at the end of the weekend is less than 0.7=710, and so w+335<710.
      Rewriting this inequality with a common denominator of 70, we get 2(w+3)70<4970 or 2(w+3)<49 or 2w+6<49 or 2w<43, and so w<432.
      Since 432=2112 and w is an integer, then w21.
      The integer values of w which satisfy w>15 and w21 are w=16,17,18,19,20,21.

    1. Chords DE and FG intersect at X, and so (DX)(EX)=(FX)(GX) or (DX)(8)=(6)(4).
      Solving this equation for DX, we get DX=(6)(4)8=248=3.
      The length of DX is 3.

    2. Chords JK and LM intersect at X, and so (JX)(KX)=(LX)(MX) or(8y)(10)=(16)(y+9).
      Solving this equation, we get 80y=16y+144 or 64y=144, and so y=14464=94.

    3. Chords PQ and ST intersect at U, and so (PU)(QU)=(SU)(TU).
      Since TU=TV+UV, then TU=6+n.
      Substituting values into (PU)(QU)=(SU)(TU), we get (m)(5)=(3)(6+n), and so 5m=18+3n.
      Chords PR and ST intersect at V, and so (PV)(RV)=(TV)(SV).
      Since SV=SU+UV, then SV=3+n.
      Substituting values into (PV)(RV)=(TV)(SV), we get (n)(8)=(6)(3+n), and so 8n=18+6n or 2n=18 or n=9.
      Substituting n=9 into 5m=18+3n, we get 5m=18+3(9) or 5m=45, and so m=9.
      Therefore, m=9 and n=9.

    1. Dave, Yona and Tam have 6, 4 and 8 candies, respectively.
      Since they each have an even number of candies, then no candies are discarded.
      During Step 2, Dave gives half of his 6 candies to Yona and accepts half of Tam’s 8 candies so that he now has 63+4=7 candies.
      Similarly, Yona gives half of her 4 candies to Tam and accepts half of Dave’s 6 candies so that she now has 42+3=5 candies.
      Tam gives half of his 8 candies to Dave and accepts half of Yona’s 4 candies so that he now has 84+2=6 candies.

      Since Dave has 7 candies, and Yona has 5 candies, they each discard one candy while Tam, who has an even number of candies, does nothing.
      These next two steps are summarized in the table to the right.
      Following the given procedure, we continue the table until the procedure ends, as shown.
      When the procedure ends, Dave, Yona and Tam each have 4 candies.

      Dave Yona Tam
      Start 3 7 10
      After Step 1 2 6 10
      After Step 2 6 4 8
      After Step 2 7 5 6
      After Step 1 6 4 6
      After Step 2 6 5 5
      After Step 1 6 4 4
      After Step 2 5 4 4
      After Step 1 4 4 4
    2. Dave, Yona and Tam begin with 16, 0 and 0 candies, respectively.
      The results of each step of the procedure are shown in the table. (We ignore Step 1 when each of the students has an even number of candies.)
      Each student has 4 candies when the procedure ends.

      Dave Yona Tam
      Start 16 0 0
      After Step 2 8 8 0
      After Step 2 4 8 4
      After Step 2 4 6 6
      After Step 2 5 5 6
      After Step 1 4 4 6
      After Step 2 5 4 5
      After Step 1 4 4 4
    3. We begin by investigating the result that Step 2 has on a student’s number of candies.
      Assume Yona has c candies, and Dave (from whom Yona receives candies), has d candies.
      Further, assume that c and d are both even integers.
      During Step 2, Yona will give half of her candies away, leaving her with c2 candies.
      In this same Step 2, Yona will also receive d2 candies from Dave (one half of Dave’s d candies).
      Therefore, Yona completes Step 2 with c2+d2=c+d2 candies, which is the average of the c and d candies that Yona and Dave respectively began the step with.
      On Wednesday, Dave starts with 2n candies, and each of Yona and Tam starts with 2n+3 candies.
      Since 2n+3 is 3 more than a multiple of 2, then 2n+3 is an odd integer for any integer n.
      That is, we begin the procedure by performing Step 1 which leaves Dave with 2n candies (2n is even and so no candies are discarded), and each of Yona and Tam with 2n+2 candies.
      After Step 2, Yona will have the average of her number of candies, 2n+2, and Dave’s number of candies, 2n, or (2n+2)+2n2=4n+22=2n+1.
      Tam will have the average of his number of candies, 2n+2, and Yona’s number of candies, 2n+2, which is 2n+2.

      Dave will have the average of his number of candies, 2n, and Tam’s number of candies,
      2n+2, or (2n+2)+2n2=4n+22=2n+1.
      Since Yona and Dave each now have an odd number of candies, Step 1 is performed.
      The procedure is continued in the table shown.

      Dave Yona Tam
      Start 2n 2n+3 2n+3
      After Step 1 2n 2n+2 2n+2
      After Step 2 2n+1 2n+1 2n+2
      After Step 1 2n 2n 2n+2
      After Step 2 2n+1 2n 2n+1
      After Step 1 2n 2n 2n

      At the end of the procedure, each student has 2n candies.

    4. On Thursday, Dave begins with 22017 candies, gives one half or 12×22017=22016 to Yona, receives 0 from Tam, and thus completes the first Step 2 having 22016 candies.
      In the table shown, we proceed with the first few steps to get a sense of what is happening early in the procedure. (We again ignore Step 1 when each student has an even number of candies.)

      Dave Yona Tam
      Start 22017 0 0
      After Step 2 22016 22016 0
      After Step 2 22015 22016 22015
      After Step 2 22015 22014+22015=22014+2×22014=3×22014 22015+22014=2×22014+22014=3×22014
      After Step 2 22015+22013=22×22013+22013=5×22013 22013+22015=22013+22×22013=5×22013 22015+22014=2×22014+22014=3×22014

      As was demonstrated in part (c), each application of Step 2 gives the average number of candies that two students had prior to the step.
      If each of the three students has a number of candies that is divisible by 2k for some positive integer k, then after performing Step 2, each student will have a number of candies that is divisible by 2k1. Why?
      If Yona has a candies and Dave has b candies, where both a and b are divisible by 2k, then after Step 2, Yona’s number of candies is the average a+b2=a2+b2.
      Since a is divisible by 2k, then a2 is divisible by 2k1 and similarly b2 is divisible by 2k1 and so their sum is at least divisible by 2k1 (and possibly more).
      We proceed by introducing 4 important facts which will lead us to our conclusion.

      Important Fact #1:
      We are starting with 22017, 0 and 0 candies, each of which is divisible by 22017.
      The first application of Step 2 gives three numbers, each of which is divisible by 22016.
      The second application of Step 2 gives three numbers, each of which is divisible by 22015.
      The third application of Step 2 gives three numbers, each of which is divisible by 22014, and so on. (We can verify this in the table above.)
      That is, starting with 22017, 0 and 0 candies, we are able to apply Step 2 2017 times in a row.
      We note that at each of these 2017 steps, the number of candies that each student has is even, and therefore Step 1 is never applied (no candies have been discarded), and so the total number of candies shared by the three students is still 22017.

      Important Fact #2:
      If we begin Step 2 with 2a,2a and 2b candies (exactly two students having an equal number of candies), then the result after applying Step 2 is a+b,2a, and a+b candies.
      That is, there are still exactly two students who have an equal number of candies.

      Important Fact #3:
      If we begin Step 2 with 2a,2a and 2b candies where a<b, then we call this a “2 low, 1 high state” (the two equal numbers are less than the third).
      Applying Step 2 to 2a,2a and 2b (a “2 low, 1 high state”), gives a+b,2a, and a+b which is a “2 high, 1 low state”. (Since a<b, then a+a<b+a or 2a<a+b.)
      Similarly, applying Step 2 again to this “2 high, 1 low state” gives a “2 low, 1 high state”.
      Since we begin with 22017, 0 and 0 candies, which is a “2 low, 1 high state”, then after 2017 applications of Step 2, we will be at a “2 high, 1 low state”.

      Important Fact #4:
      Beginning with 2a,2a and 2b candies, the positive difference between the high number of candies and the low number of candies is 2b2a (or 2a2b if a>b).
      After applying Step 2, we have a+b,2a, and a+b candies and the positive difference between the high and low numbers of candies is ba (or ab if a>b).
      That is, applying Step 2 once decreases the positive difference between the high and low numbers of candies by a factor of 2 (that is, ab=12(2a2b)).
      Therefore, beginning with 22017, 0 and 0 candies, whose positive difference is 22017, and applying Step 2 2017 times gives a “2 high, 1 low state” where the positive difference between the high and low numbers is 1.
      That is, after applying Step 2 2017 times, the number of candies is n+1,n+1 and n for some non-negative integer n.

      Conclusion:
      Since we haven’t applied Step 1, then there are still 22017 candies shared between the three students.
      If n is odd, then the number of candies, 3n+2, is odd.
      Since 3n+2 is equal to 22017, this is not possible and so n is even.
      Since n is even, then n+1 is odd and so we apply Step 1 to n+1,n+1 and n candies so that each student has an equal number of candies, n.
      Two candies were discarded in the application of Step 1 and so there are now 220172 candies remaining.
      Since each student has an equal number of candies, and there are 220172 candies in total, the procedure ends with each student having 2201723 candies.