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2016 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 23, 2016
(in North America and South America)

Thursday, November 24, 2016
(outside of North American and South America)

©2016 University of Waterloo


PART A

  1. Evaluating, 13+23+33(1+2+3)2=1+8+2762=3636=1

    Answer: 1

  2. Solution 1
    Suppose that the full price of the second book was $x.
    Since 50% is equivalent to 12, then the amount that Mike paid, in dollars, for the two books was 33+12x.
    The combined full price, in dollars, of the two books was 33+x.
    Since Mike saved a total of 20% on his purchase, then he paid 80% of the combined full cost.
    Since 80% is equivalent to 45, then, in dollars, Mike paid 45(33+x).
    Therefore, we obtain the equivalent equations 45(33+x)=33+12x45(33)+45x=33+12x310x=15(33)x=10315(33)x=211=22 Therefore, the full cost was $(33+x)=$(33+22)=$55.
    Mike saved 20% of the full cost or 15($55)=$11.

    Solution 2
    Suppose that the full price of the second book was $x.
    The combined full price, in dollars, of the two books was 33+x.
    Since 50% is equivalent to 12, then the amount that Mike saved, in dollars, on the second book was 12x.
    Since 20% is equivalent to 15, then, in dollars, Mike saved a total of 15(33+x) on the purchase of both books.
    Therefore, we obtain the equivalent equations 15(33+x)=12x2(33+x)=5x66+2x=5x3x=66x=22 Therefore, the full cost of the second book was $22.
    Mike saved 50% of the full cost or 12($22)=$11.

    Answer: $11

  3. There are exactly 5 lists with this property, namely 3+5+7+9=1+5+7+11=1+3+5+15=1+3+7+13=1+3+9+11=24 Why are these the only lists?
    If a=3, then since a,b,c,d are odd positive integers with a<b<c<d, then b5 and c7 and d9, and so a+b+c+d3+5+7+9=24.
    Since a+b+c+d=24, then b=5 and c=7 and d=9.
    If a=1, then we cannot have b7, since otherwise c9 and d11, which gives a+b+c+d1+7+9+11=28 Therefore, if a=1, we must have b=3 or b=5 since a<b.
    If b=5, then c+d=24ab=2415=18.
    Since we have 5<c<d, then we must have 18=c+d>2c and so c<9.
    Since c and d are odd positive integers, then c=7 and so d=11.
    If b=3, then c+d=2413=20.
    Therefore, using a similar argument, we get c<10 and so c=9 (which gives d=11) or c=7 (which gives d=13) or c=5 (which gives d=15).
    Therefore, there are exactly 5 lists with the desired properties.

    Answer: 5

  4. Suppose that the rectangular prism has dimensions x×y×z.
    We label the various lengths as shown:

    A description of the labelled net follows.


    (We can think about the four “middle" rectangles in the net as forming the sides of the prism, the “top" rectangle as forming the top, and the “bottom" rectangle as forming the bottom.)
    From the given information, we obtain the equations 2x+2y=38 and y+z=14 and x+z=11.
    Dividing the first equation by 2, we obtain x+y=19.
    Adding the three equations, we obtain (x+y)+(y+z)+(x+z)=19+14+11.
    This gives 2x+2y+2z=44 or x+y+z=22.
    Therefore, x=(x+y+z)(y+z)=2214=8y=(x+y+z)(x+z)=2211=11z=(x+y+z)(x+y)=2219=3 Finally, the volume of the prism is thus xyz=8113=264.

    Answer: 264

  5. Since the first player to win 4 games becomes the champion, then Gary and Deep play at most 7 games. (The maximum number of games comes when the two players have each won 3 games and then one player becomes the champion on the next (7th) game.)
    We are told that Gary wins the first two games.
    For Deep to become the champion, the two players must thus play 6 or 7 games, because Deep wins 4 games and loses at least 2 games. We note that Deep cannot lose 4 games, otherwise Gary would become the champion.
    If Deep wins and the two players play a total of 6 games, then the sequence of wins must be GGDDDD. (Here D stands for a win by Deep and G stands for a win by Gary.)
    If Deep wins and the two players play a total of 7 games, then Deep wins 4 of the last 5 matches and must win the last (7th) match since he is the champion.
    Therefore, the sequence of wins must be GGGDDDD or GGDGDDD or GGDDGDD or GGDDDGD. (In other words, Gary can win the 3rd, 4th, 5th, or 6th game.)
    The probability of the sequence GGDDDD occurring after Gary has won the first 2 games is (12)4=116. This is because the probability of a specific outcome in any specific game is 12, because each player is equally likely to win each game, and there are 4 games with undetermined outcome.
    Similarly, the probability of each of the sequences GGGDDDD, GGDGDDD, GGDDGDD, and GGDDDGD occurring is (12)5=132.
    Therefore, the probability that Gary wins the first two games and then Deep becomes the champion is 116+4132=632=316.

    Answer: 316

  6. We label the digits of n in order from left to right as n1,n2,n3,n4,n5,n6,n7,n8,n9. Since n is a nine-digit integer, then n10.
    From the given information, the following sums must be multiples of 5: n1+n2+n3+n4+n5 and n2+n3+n4+n5+n6 and n3+n4+n5+n6+n7 and n4+n5+n6+n7+n8 and n5+n6+n7+n8+n9.
    Also, the following sums must be multiples of 4: n1+n2+n3+n4+n5+n6+n7 and n2+n3+n4+n5+n6+n7+n8 and n3+n4+n5+n6+n7+n8+n9.
    We need the following fact, which we label ():

    If a and b are positive integers that are both multiples of the positive integer d, then their difference ab is also a multiple of d.

    This is true because if a=ed and b=fd for some positive integers e and f, then ab=edfd=d(ef) which is a multiple of d.
    By (), (n1+n2+n3+n4+n5) (n2+n3+n4+n5+n6)=n1n6 must be a multiple of 5.
    Since n1 and n6 are taken from the list 0,1,2,3,4,5,6,7,8 and no pair of numbers from this list differs by more than 8, then for n1n6 to be a multiple of 5, we must have that n1 and n6 differ by 5.
    Similarly, n2 and n7 differ by 5, n3 and n8 differ by 5, and n4 and n9 differ by 5.
    The pairs from the list 0,1,2,3,4,5,6,7,8 that differ by 5 are 0,5 and 1,6 and 2,7 and 3,8. These integers must be assigned in some order to the pairs n1,n6 and n2,n7 and n3,n8 and n4,n9.
    Note that n5 is not in the list of positions affected and 4 is not in the list of integers that can be assigned to these slots.
    Therefore, n5=4.
    Using () again, we also see that n1 and n8 differ by a multiple of 4, and n2 and n9 differ by a multiple of 4.
    Suppose that n1=1. Then from above, n6=6 (using divisibility by 5) and n8=5 (using divisibility by 4, since the only integer from our list that differs from 1 by a multiple of 4 is 5).
    Since n8=5, then using divisibility by 5, we have n3=0.
    So if n1=1, the digits of n look like 1A0A46A5A.
    Using similar analysis, we make a table that lists the first digit, and the resulting configurations:

    n1 Configuration
    1 1_0_46_5_
    2 2_1_47_6_
    3 3_2_48_7_
    5 5_6_40_1_
    6 6_7_41_2_
    7 7_8_42_3_
    8 8_5_43_0_

    Note that, in the case where n1=8, we know that n8 differs from 8 by a multiple of 4, but cannot equal 4 since n5=4 already.
    While it is necessary that n1 and n8 differ by a multiple of 4, it is not sufficient to guarantee that the sum of each set of 7 consecutive digits is a multiple of 4.
    Since the sum of the 9 digits of n is 0+1+2+3+4+5+6+7+8=36 (which is a multiple of 4) and n3+n4+n5+n6+n7+n8+n9 is a multiple of 4, then n1+n2 must be a multiple of 4.
    Since each of the digits is between 0 and 8, then the maximum sum of two digits is 15, and so the possible multiples of 4 to which n1+n2 can be equal are 4, 8 and 12.
    If n1=1, then this means that n2=3 or n2=7.
    If n1=2, then n2=6, which is not possible, since n8=6.
    If n1=3, then n2=1 or n2=5.
    If n1=5, then n2=3.
    If n1=6, then n2=2, which is not possible, since n8=6.
    If n1=7, then n2=1 or n2=5.
    If n1=8, then n2=0 or n2=4, which are not possible, as these digits are already used.
    If n2=3, then using a similar analysis to that above, we have n7=8 and n9=7 and n4=2.
    Thus, when n1=1 and n2=3, then n=130246857.
    Using similar analysis on the remaining cases, we obtain the following possible values for n: 130246857,170846253,312048675,352648071, 536240817,576840213,718042635,758642031 We can check that each of these nine-digit integers has the desired properties.
    The sum of these eight nine-digit integers is 3555555552.
    (We note that the sum of the possible values for each digit is 32.)

    Answer: 3555555552

PART B